Tutors Answer Your Questions about Probability-and-statistics (FREE)
Question 1159882: 7 men and 3 women are ranked according to their scores on an exam. Assume that no two scores are alike, and that all 10! possible rankings are equally likely. Let X denote the highest ranking achieved by a man (so X=1 indicates that a man achieved the highest score on the exam). Find each of the following:
P(X=1)=
P(X=2)=
P(X=3)=
P(X=7)=
Answer by ikleyn(52780)   (Show Source):
You can put this solution on YOUR website! .
7 men and 3 women are ranked according to their scores on an exam. Assume that no two scores are alike,
and that all 10! possible rankings are equally likely. Let X denote the highest ranking achieved by a man
(so X=1 indicates that a man achieved the highest score on the exam). Find each of the following:
(a) P(X=1) =
(b) P(X=2) =
(c) P(X=3) =
(d) P(X=7) =
~~~~~~~~~~~~~~~~~~~~~~~~~~
(a) X=1 means that one of the 7 men is in the first position,
while the rest 6 men and 3 women are distributed in positions from 2 to 10 in arbitrary ways.
The number of such possible outcomes is  = 7*9!,
and to find the probability P(X=1), we should relate 7*9! to 10!.
It gives P(X=1) =  =  = 0.7.
It is the ANSWER to (a).
(b) X=2 means that one of the 3 women is in the first position and one of the 7 men is in the second position,
while the rest 3-1 = 2 women and 7-1 = 6 men are distributed in positions from 3 to 10 in arbitrary ways.
The number of such possible outcomes is  = 3*7*8!,
and to find the probability P(X=2), we should relate 3*7*8! to 10!.
It gives P(X=2) =  =  =  =  .
It is the ANSWER to (b).
(c) X=3 means that two of the 3 women are in the first and in the second positions and one of the 7 men
is in the third position,
while the rest 3-2 = 1 women and 7-1 = 6 men are distributed in positions from 3 to 10 in arbitrary ways.
The number of such possible outcomes is  =  = 3*7*7!,
and to find the probability P(X=3), we should relate 3*7*7! to 10!.
It gives P(X=3) =  =  =  =  .
It is the ANSWER to (c).
(d) X=7 means that some man is in the 7th positions and there no men in positions from 1 to 6.
But this configuration is not possible (which is obvious), so
P(X=7) = 0.
It is the ANSWER to (d).
Thus, all the questions are answered, and the problem is solved completely.
Question 1160790: A mathematics professor gives an "extra-credit" (bonus marks) problem on a test. If it is done correctly,15 points are added to the test score, and if it is done partially correctly,5 points are added; otherwise 5 points are subtracted. Suppose a student's probability of getting the problem completely right is 1/4, and only partially correct is 1/2.
Find the student's mathematical expectation 𝐸 (expected value) for extra credit.
Answer by ikleyn(52780) (Show Source):
Question 1167284: Four players each are dealt one card. In how many ways could the cards all be clubs or honour cards (10, J, Q, K, A)?
Answer by Resolver123(6) (Show Source):
You can put this solution on YOUR website! There are 5 honor 'ranks' (10, J, Q, K, A), and each of these ranks has 4 suits. That means there are 5 x 4 = 20 honor cards.
In addition, there are 8 remaining 'clubs' cards (from 2 to 9), for a total of 20 + 8 = 28 cards satisfying the condition.
By the fundamental principle of counting, there are 28P4 = 28 x 27 x 26 x 25 = 491,400 ways of dealing one card to each player which is either a clubs or an honor card.
Question 1207399: In a large population, about
45
%
of people prefer tea over coffee. A researcher takes a random sample of
13
people and surveys whether they prefer tea over coffee.
Use the binomial distribution to compute the probability that exactly
6
of the people in the sample prefer tea over coffee.
Identify the following information required to find the probability of people who prefer tea over coffee.
Provide your answer below:
n =
trials
x =
successes
p =
probability of those who prefer tea (as a decimal, not percent)
Answer by ikleyn(52780) (Show Source):
Question 1160934: Heights of men on a baseball team have a bell-shaped distribution with a mean of 172 cm and a standard deviation of 6 cm. Using the empirical rule, what is the approximate percentage of the men between the following values?
a. 154 cm and 190 cm
b. 160 cm and 184 cm
Answer by ikleyn(52780) (Show Source):
You can put this solution on YOUR website! .
Heights of men on a baseball team have a bell-shaped distribution with a mean of 172 cm
and a standard deviation of 6 cm. Using the empirical rule, what is the approximate percentage
of the men between the following values?
a. 154 cm and 190 cm
b. 160 cm and 184 cm
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
(a) Interval [154,190] is three standard deviations from the mean.
So, the empirical rule for 3 standard deviations is applicable in this case,
giving the ANSWER that the height of approximately 99.7% of the men is in this interval.
(b) Interval [160,184] is two standard deviations from the mean.
So, the empirical rule for 2 standard deviations is applicable in this case,
giving the ANSWER that the height of approximately 95% of the men is in this interval.
Question 1165550: A box contains 17 large marbles and 14 small marbles. Each marble is either green or white. 10 of the large marbles are green, and 4 of the small marbles are white. If a marble is randomly selected from the box, what is the probability that it is small or white? Express your answer as a fraction or a decimal number rounded to four decimal places.
Answer by ikleyn(52780) (Show Source):
You can put this solution on YOUR website! .
A box contains 17 large marbles and 14 small marbles. Each marble is either green or white.
10 of the large marbles are green, and 4 of the small marbles are white.
If a marble is randomly selected from the box, what is the probability that it is small or white?
Express your answer as a fraction or a decimal number rounded to four decimal places.
~~~~~~~~~~~~~~~~~~~~~~~
We can re-phrase the given part of the problem in this way:
A box contains 17 large marbles, of which 10 marbles are green and 17-10 = 7 marbles are white.
The box also contains 14 small marbles, of which 4 are white and 14-4 = 10 are green.
Then the answer to the problem's question in general form is
P =  .
Obviously, n(favorable = n(small) + n(white from the group of large) = 14 + 7 = 21.
n(total) = n(large) + n(small) = 17 + 14 = 31.
Thus the Answer is P =  = 0.6774 (rounded as requested).
Solved.
Question 1165633: The quality control manager at a light bulb factory needs to estimate the mean life of a large shipment of light bulbs. The standard deviation is 108 hours. A random sample of 81 light bulbs indicated a sample mean life of 400 hours. Complete parts (a) through (d) below.
Answer by ikleyn(52780) (Show Source):
You can put this solution on YOUR website! .
The quality control manager at a light bulb factory needs to estimate the mean life of a large shipment of light bulbs.
The standard deviation is 108 hours. A random sample of 81 light bulbs indicated a sample mean life of 400 hours.
Complete parts (a) through (d) below.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
There are no parts (a) through (d) in the post.
They all were lost due to the carelessness of the person or people who prepared this post.
It is a typical case when they send a message without reading it and without looking in it.
Question 1197206: Suppose that 10% of the fields in a given agricultural**a) Average Number of Infested Fields**
* **Binomial Distribution:** This situation fits the binomial distribution:
* **Fixed number of trials:** 100 fields (n = 100)
* Two possible outcomes: infested or not infested
* Probability of success (infestation) is constant: p = 0.10
* Trials are independent
* **Mean of Binomial Distribution:**
* The average number of infested fields (mean) is given by: μ = n * p
* μ = 100 * 0.10 = 10 fields
**b) 95% Confidence Interval**
* **Standard Deviation of Binomial Distribution:**
* σ = √(n * p * (1 - p))
* σ = √(100 * 0.10 * 0.90) = √9 = 3
* **95% Confidence Interval:**
* For a normal distribution, approximately 95% of the data falls within 2 standard deviations of the mean.
* Lower Limit: μ - 2σ = 10 - 2 * 3 = 4 fields
* Upper Limit: μ + 2σ = 10 + 2 * 3 = 16 fields
* **Therefore, we would expect to find the number of infested fields within the range of 4 to 16 fields with a probability of approximately 95%.**
**c) Finding 25 Infested Fields**
* **25 infested fields is significantly higher than the expected range (4-16 fields).**
* **Possible Explanation:**
* **Violation of the assumption of independence:**
* If the infestation in one field increases the likelihood of infestation in nearby fields (e.g., due to pest migration), the trials would not be independent, and the binomial distribution might not accurately model the situation.
* **Change in infestation rate:**
* The actual infestation rate in the agricultural area might have increased significantly since the initial estimate of 10%.
* **Sampling bias:**
* The 100 fields selected might not be a truly random sample and may be more likely to be infested than the overall population of fields.
**In conclusion:**
Finding 25 infested fields in the sample suggests that the observed number of infested fields is significantly higher than expected under the assumption of a binomial distribution with a 10% infestation rate. This discrepancy could indicate a violation of the independence assumption or a change in the underlying infestation rate. Further investigation would be necessary to determine the cause of this unexpected result.
area are infested with the sweet potato whitefly. 100 fields in this area are randomly selected and checked for whitefly. a) What is the average number of fields sampled that are infested with whitefly? b) Within what limits would you expect to find the number of infested fields, with robability approximately 95%? c) What might you conclude if you found that x = 25 fields were infested? Is it possible that one of the characteristics of a binomial experiment is not satisfied in this experiment? Explain.
Answer by ikleyn(52780) (Show Source):
You can put this solution on YOUR website! .
In this post, the problem's description is corrupted by other person (not intentionally, I hope).
Therefore, it is not possible to understand the problem and the solution,
as well as is not possible to check the solution.
Question 1199410: 4.4.2 Suppose that over a period of several years the average number of deaths from a certain noncontagious disease has been 10. If the number of deaths from this disease follows the Poisson distribution, what is the probability that during the current year: a. Exactly seven people will die from the disease? b. Ten or more people will die from the disease?
Answer by ikleyn(52780) (Show Source):
You can put this solution on YOUR website! .
4.4.2 Suppose that over a period of several years the average number of deaths from a certain noncontagious disease has been 10.
If the number of deaths from this disease follows the Poisson distribution, what is the probability that during the current year:
a. Exactly seven people will die from the disease?
b. Ten or more people will die from the disease?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
As I read this problem, I see that it is defective.
To be correct, in the first sentence the problem must say that '10' is the average number of deaths per year.
Question 1167383: Assume that when adult with smartphone are randomly selected 43% ,use them in meeting or classes. If 10 adults smart phone users are randomly selected ,find the probability that at least 7 of them use their smartphone in meeting or class
Answer by ikleyn(52780) (Show Source):
You can put this solution on YOUR website! .
Assume that when adult with smartphone are randomly selected, 43% use them in meeting or classes.
If 10 adults smart phone users are randomly selected, find the probability that at least 7 of them
use their smartphone in meeting or class
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
This is a Binomial distribution problem.
The number of trials is n=10; the number of successful trials is k >= 7;
the probability of success of every individual trial is p = 0.43.
Use the complement probability formula P(k >= 7) = 1 - P(k <= 6).
Use the binomial cumulative probability function binomcdf of calculator TI-84/87
P(k <= 6) = binomcdf(n,p,k) = binomcdf(10, 0.43, 6) = 0.9194.
Thus the ANSWER to the problem's question is
P(k >= 7) = 1 - 0.9194 = 0.0806.
Solved.
-------------------------------
Today, 07/24/2025, I placed this problem to Google AI.
The Google AI produced INCORRECT answer P = 0.0908.
The link to this Google wrong solution is
https://www.google.com/search?q=Assume+that+when+adult+with+smartphone+are+randomly+selected%2C+43%25+use+them+in+meeting+or+classes.+If+10+adults+smart+phone+users+are+randomly+selected%2C+find+the+probability+that+at+least+7+of+them+use+their+smartphone+in+meeting+or+class&rlz=1C1CHBF_enUS1071US1071&oq=Assume++that+when+adult+with+smartphone+are+randomly+selected%2C+43%25+use+them+in+meeting+or+classes.++If+10+adults+smart+phone+users+are+randomly+selected%2C+find+the+probability+that+at+least+7+of+them++use+their+smartphone+in+meeting+or+class&gs_lcrp=EgZjaHJvbWUyBggAEEUYOdIBCTE4NTJqMGoxNagCCLACAfEF1pb2b0cJQX8&sourceid=chrome&ie=UTF-8
- - - - - - - - - - - - - - - - S H A M E L E S S - - - - - - - - - - - - - - - -
Naturally, I posted this my note to them via their feedback system to let them know about their fault.
Looking into MANY problems on binomial distribution, solved by Google AI,
I regularly see their computational errors - so, my conclusion/impression
is that they permanently use inadequate calculator/solver for such computations,
which DOES NOT provide the necessary precision. They even do not recognize it !
Dear developers of the Google AI:
Your illiteracy in performing such calculations is far out of any appropriate scale.
I regularly detected and noticed these computational errors just for several months,
but I do not see any move/effort from their side to fix this deficiency.
Question 1210366: Seth and Anna go to a cafe to buy a drink each. They can choose from Coke, Sprite and L&P. It is equally likely as to which drink they buy, Find the probability that both of them buy a coke
Answer by ikleyn(52780) (Show Source):
You can put this solution on YOUR website! .
Seth and Anna go to a cafe to buy a drink each. They can choose from Coke, Sprite and L&P.
It is equally likely as to which drink they buy, Find the probability that both of them buy a coke
~~~~~~~~~~~~~~~~~~~~~~~~~~
There are three drinks for each to choose from, and the probability is 1/3
for every drink to be selected by Seth and/or by Anna, independently.
Therefore, the probability that both of them will select coke independently is
P =  =  . <<<---=== ANSWER
Solved.
----------------------------
You can solve the problem by another way, considering the set of possible outcomes.
It consists of 9 pairs (drink_Seth,drink_Anna), so you have 9 possible outcomes
with equal probabilities. It leads to the same answer.
Question 1167732: 12,500 | 17,700 | 22,300
14,800 | 19,400 | 24,800
15,200 | 20,200 | 27,600
15,300 | 21,200 | 28,200
15,900 | 22,100 | 35,200
If the average and standard deviation of their components are 20,860 and 6, 168 respectively, determine the corresponding for the (use chebyshev theorem)
a.) 2nd standard deviation at the left side of the mean
Answer by CPhill(1959)  (Show Source):
You can put this solution on YOUR website! To determine the value corresponding to the 2nd standard deviation at the left side of the mean, we will use the given average (mean) and standard deviation.
**Given:**
* Average (Mean, $\mu$) = 20,860
* Standard Deviation ($\sigma$) = 6,168
We need to find the value that is 2 standard deviations to the left (below) the mean.
**Calculation:**
Value = $\mu - 2\sigma$
Value = $20,860 - 2 \times 6,168$
Value = $20,860 - 12,336$
Value = $8,524$
The value corresponding to the 2nd standard deviation at the left side of the mean is 8,524.
Question 1167739: 3. A manufacturer of certain bearing presents the following data pertaining to the diameters of the bearing for 10 hourly samples of size 4:
Hour Coded sample values
1 1.7 2.2 1.9 1.2
2 0.8 1.5 2.1 0.9
3 1.0 1.4 1.0 1.3
4 0.4 -0.6 0.7 0.2
5 1.4 2.3 2.8 2.7
6 1.8 2.0 1.1 0.1
7 1.6 1.0 1.5 2.0
8 2.5 1.6 1.8 1.2
9 2.9 2.0 0.5 2.2
10 1.1 1.1 3.1 1.6
Use X and R chart to determine if the process is in statistical control.
Answer by CPhill(1959) (Show Source):
You can put this solution on YOUR website! ```python?code_reference&code_event_index=2
import pandas as pd
# Data provided
data = {
'Hour': [1, 2, 3, 4, 5, 6, 7, 8, 9, 10],
'Sample Values': [
[1.7, 2.2, 1.9, 1.2],
[0.8, 1.5, 2.1, 0.9],
[1.0, 1.4, 1.0, 1.3],
[0.4, -0.6, 0.7, 0.2],
[1.4, 2.3, 2.8, 2.7],
[1.8, 2.0, 1.1, 0.1],
[1.6, 1.0, 1.5, 2.0],
[2.5, 1.6, 1.8, 1.2],
[2.9, 2.0, 0.5, 2.2],
[1.1, 1.1, 3.1, 1.6]
]
}
# Create a DataFrame for easier manipulation
df = pd.DataFrame(data)
# Subgroup size
n = 4
# --- Step 1 & 2: Calculate subgroup means (X_bar) and ranges (R) ---
df['X_bar'] = df['Sample Values'].apply(lambda x: sum(x) / n)
df['R'] = df['Sample Values'].apply(lambda x: max(x) - min(x))
# --- Step 3 & 4: Calculate grand mean (X_double_bar) and average range (R_bar) ---
X_double_bar = df['X_bar'].mean()
R_bar = df['R'].mean()
print("Subgroup Means and Ranges:")
print(df[['Hour', 'X_bar', 'R']])
print(f"\nGrand Mean (X_double_bar): {X_double_bar:.3f}")
print(f"Average Range (R_bar): {R_bar:.3f}")
# --- Step 5 & 6: Calculate Control Limits ---
# Control chart constants for n=4
A2 = 0.729
D3 = 0
D4 = 2.282
# R-chart control limits
UCL_R = D4 * R_bar
LCL_R = D3 * R_bar
# X-bar chart control limits
UCL_Xbar = X_double_bar + A2 * R_bar
LCL_Xbar = X_double_bar - A2 * R_bar
print("\nControl Limits for R-chart:")
print(f" UCL_R: {UCL_R:.3f}")
print(f" LCL_R: {LCL_R:.3f}")
print("\nControl Limits for X-bar chart:")
print(f" UCL_Xbar: {UCL_Xbar:.3f}")
print(f" LCL_Xbar: {LCL_Xbar:.3f}")
# --- Step 7 & 8: Determine if the process is in statistical control ---
# Check R-chart for out-of-control points
out_of_control_R = df[(df['R'] > UCL_R) | (df['R'] < LCL_R)]
print("\n--- R-chart Analysis ---")
if not out_of_control_R.empty:
print("Out-of-control points found in R-chart:")
print(out_of_control_R[['Hour', 'R']])
else:
print("No out-of-control points found in R-chart.")
# Check X-bar chart for out-of-control points
out_of_control_Xbar = df[(df['X_bar'] > UCL_Xbar) | (df['X_bar'] < LCL_Xbar)]
print("\n--- X-bar chart Analysis ---")
if not out_of_control_Xbar.empty:
print("Out-of-control points found in X-bar chart:")
print(out_of_control_Xbar[['Hour', 'X_bar']])
else:
print("No out-of-control points found in X-bar chart.")
# Overall conclusion
print("\n--- Overall Process Control Status ---")
if out_of_control_R.empty and out_of_control_Xbar.empty:
print("Based on the X-bar and R charts, the process appears to be in statistical control.")
else:
print("Based on the X-bar and/or R charts, the process is NOT in statistical control due to points outside control limits.")
```
```text?code_stdout&code_event_index=2
Subgroup Means and Ranges:
Hour X_bar R
0 1 1.750 1.0
1 2 1.325 1.3
2 3 1.175 0.4
3 4 0.175 1.3
4 5 2.300 1.4
5 6 1.250 1.9
6 7 1.525 1.0
7 8 1.775 1.3
8 9 1.900 2.4
9 10 1.725 2.0
Grand Mean (X_double_bar): 1.490
Average Range (R_bar): 1.400
Control Limits for R-chart:
UCL_R: 3.195
LCL_R: 0.000
Control Limits for X-bar chart:
UCL_Xbar: 2.511
LCL_Xbar: 0.469
--- R-chart Analysis ---
No out-of-control points found in R-chart.
--- X-bar chart Analysis ---
Out-of-control points found in X-bar chart:
Hour X_bar
3 4 0.175
--- Overall Process Control Status ---
Based on the X-bar and/or R charts, the process is NOT in statistical control due to points outside control limits.
```
To determine if the process is in statistical control, we will construct an X-bar chart and an R chart.
**Given Data:**
| Hour | Sample Values |
| :--- | :------------ |
| 1 | 1.7, 2.2, 1.9, 1.2 |
| 2 | 0.8, 1.5, 2.1, 0.9 |
| 3 | 1.0, 1.4, 1.0, 1.3 |
| 4 | 0.4, -0.6, 0.7, 0.2 |
| 5 | 1.4, 2.3, 2.8, 2.7 |
| 6 | 1.8, 2.0, 1.1, 0.1 |
| 7 | 1.6, 1.0, 1.5, 2.0 |
| 8 | 2.5, 1.6, 1.8, 1.2 |
| 9 | 2.9, 2.0, 0.5, 2.2 |
| 10 | 1.1, 1.1, 3.1, 1.6 |
The subgroup size ($n$) is 4.
**1. Calculate Subgroup Means ($\\bar{X}$) and Ranges (R):**
| Hour | Sample Values | $\\bar{X}$ (Mean) | R (Range) |
| :--- | :----------------- | :--------------- | :-------- |
| 1 | 1.7, 2.2, 1.9, 1.2 | 1.750 | 1.0 |
| 2 | 0.8, 1.5, 2.1, 0.9 | 1.325 | 1.3 |
| 3 | 1.0, 1.4, 1.0, 1.3 | 1.175 | 0.4 |
| 4 | 0.4, -0.6, 0.7, 0.2 | 0.175 | 1.3 |
| 5 | 1.4, 2.3, 2.8, 2.7 | 2.300 | 1.4 |
| 6 | 1.8, 2.0, 1.1, 0.1 | 1.250 | 1.9 |
| 7 | 1.6, 1.0, 1.5, 2.0 | 1.525 | 1.0 |
| 8 | 2.5, 1.6, 1.8, 1.2 | 1.775 | 1.3 |
| 9 | 2.9, 2.0, 0.5, 2.2 | 1.900 | 2.4 |
| 10 | 1.1, 1.1, 3.1, 1.6 | 1.725 | 2.0 |
**2. Calculate the Grand Mean ($\\bar{\\bar{X}}$) and Average Range ($\\bar{R}$):**
* **Grand Mean ($\\bar{\\bar{X}}$):** Average of all subgroup means
$\\bar{\\bar{X}} = (1.750 + 1.325 + 1.175 + 0.175 + 2.300 + 1.250 + 1.525 + 1.775 + 1.900 + 1.725) / 10 = 14.900 / 10 = 1.490$
* **Average Range ($\\bar{R}$):** Average of all subgroup ranges
$\\bar{R} = (1.0 + 1.3 + 0.4 + 1.3 + 1.4 + 1.9 + 1.0 + 1.3 + 2.4 + 2.0) / 10 = 14.0 / 10 = 1.400$
**3. Determine Control Chart Constants for n=4:**
From control chart tables for subgroup size $n=4$:
* $A\_2 = 0.729$
* $D\_3 = 0$
* $D\_4 = 2.282$
**4. Calculate Control Limits for the R-chart:**
* **Center Line (CL\_R):** $\\bar{R} = 1.400$
* **Upper Control Limit (UCL\_R):** $D\_4 \\times \\bar{R} = 2.282 \\times 1.400 = 3.1948 \\approx 3.195$
* **Lower Control Limit (LCL\_R):** $D\_3 \\times \\bar{R} = 0 \\times 1.400 = 0.000$
**5. Calculate Control Limits for the X-bar chart:**
* **Center Line (CL\_Xbar):** $\\bar{\\bar{X}} = 1.490$
* **Upper Control Limit (UCL\_Xbar):** $\\bar{\\bar{X}} + A\_2 \\times \\bar{R} = 1.490 + 0.729 \\times 1.400 = 1.490 + 1.0206 = 2.5106 \\approx 2.511$
* **Lower Control Limit (LCL\_Xbar):** $\\bar{\\bar{X}} - A\_2 \\times \\bar{R} = 1.490 - 0.729 \\times 1.400 = 1.490 - 1.0206 = 0.4694 \\approx 0.469$
**Summary of Control Limits:**
| Chart | Center Line | UCL | LCL |
| :----- | :---------- | :---- | :---- |
| R-chart | 1.400 | 3.195 | 0.000 |
| X-bar | 1.490 | 2.511 | 0.469 |
**6. Determine if the process is in statistical control:**
* **R-chart Analysis:**
* Examine the 'R' column in the table above. All R values (from 0.4 to 2.4) are within the R-chart control limits (0.000 to 3.195).
* **Conclusion for R-chart: The process range (variability) appears to be in statistical control.**
* **X-bar chart Analysis:**
* Examine the '$\\bar{X}$' column in the table above.
* We compare each subgroup mean to the X-bar chart control limits (0.469 to 2.511).
* For Hour 4, the mean is $0.175$. This value is below the LCL\_Xbar of $0.469$.
* **Conclusion for X-bar chart: The process average (mean) is NOT in statistical control because the mean for Hour 4 ($0.175$) falls below the Lower Control Limit.**
**Overall Conclusion:**
Based on the analysis of both the X-bar and R charts, the process for the bearing diameters is **NOT in statistical control**. The presence of an out-of-control point in the X-bar chart (Hour 4) indicates that there is an assignable cause of variation that needs to be investigated and eliminated.
Question 1167889: The Democrat and Chronicle reported that 25% of the flights arriving at the San Diego airport during the
first five months of 2001 were late (Democrat and Chronicle, July 23, 2001). Assume the population
proportion is p .25.
a. Show the sampling distribution of , the proportion of late flights in a sample of 1000 flights.
b. What is the probability that the sample proportion will be within .03 of the population proportion if a
sample of size 1000 is selected?
c. Answer part (b) for a sample of 500 flights.
Answer by CPhill(1959) (Show Source):
You can put this solution on YOUR website! ```python?code_reference&code_event_index=2
import math
from scipy.stats import norm
# Population proportion
p = 0.25
# --- Part a: Sampling distribution for n = 1000 ---
n_a = 1000
mean_p_bar_a = p
std_dev_p_bar_a = math.sqrt((p * (1 - p)) / n_a)
# Check for normality assumption
np_a = n_a * p
n_1_minus_p_a = n_a * (1 - p)
print(f"Part a (n={n_a}):")
print(f" Mean of p_bar: {mean_p_bar_a}")
print(f" Standard deviation of p_bar: {std_dev_p_bar_a}")
print(f" np >= 5: {np_a} >= 5 (True)")
print(f" n(1-p) >= 5: {n_1_minus_p_a} >= 5 (True)")
# --- Part b: Probability for n = 1000 ---
diff = 0.03
lower_bound_b = p - diff
upper_bound_b = p + diff
# Calculate Z-scores
z_lower_b = (lower_bound_b - p) / std_dev_p_bar_a
z_upper_b = (upper_bound_b - p) / std_dev_p_bar_a
# Calculate probability using CDF
probability_b = norm.cdf(z_upper_b) - norm.cdf(z_lower_b)
print(f"\nPart b (n={n_a}):")
print(f" Range: [{lower_bound_b}, {upper_bound_b}]")
print(f" Z-score for lower bound: {z_lower_b}")
print(f" Z-score for upper bound: {z_upper_b}")
print(f" Probability: {probability_b}")
# --- Part c: Probability for n = 500 ---
n_c = 500
std_dev_p_bar_c = math.sqrt((p * (1 - p)) / n_c)
# Check for normality assumption
np_c = n_c * p
n_1_minus_p_c = n_c * (1 - p)
# Calculate Z-scores
z_lower_c = (lower_bound_b - p) / std_dev_p_bar_c
z_upper_c = (upper_bound_b - p) / std_dev_p_bar_c
# Calculate probability using CDF
probability_c = norm.cdf(z_upper_c) - norm.cdf(z_lower_c)
print(f"\nPart c (n={n_c}):")
print(f" Standard deviation of p_bar: {std_dev_p_bar_c}")
print(f" np >= 5: {np_c} >= 5 (True)")
print(f" n(1-p) >= 5: {n_1_minus_p_c} >= 5 (True)")
print(f" Z-score for lower bound: {z_lower_c}")
print(f" Z-score for upper bound: {z_upper_c}")
print(f" Probability: {probability_c}")
```
```text?code_stdout&code_event_index=2
Part a (n=1000):
Mean of p_bar: 0.25
Standard deviation of p_bar: 0.013693063937629153
np >= 5: 250.0 >= 5 (True)
n(1-p) >= 5: 750.0 >= 5 (True)
Part b (n=1000):
Range: [0.22, 0.28]
Z-score for lower bound: -2.1908902300206643
Z-score for upper bound: 2.1908902300206665
Probability: 0.9715402630836895
Part c (n=500):
Standard deviation of p_bar: 0.019364916731037084
np >= 5: 125.0 >= 5 (True)
n(1-p) >= 5: 375.0 >= 5 (True)
Z-score for lower bound: -1.5491933384829668
Z-score for upper bound: 1.5491933384829681
Probability: 0.878664749641518
```
The population proportion of late flights is $p = 0.25$.
**a. Show the sampling distribution of $\\bar{p}$, the proportion of late flights in a sample of 1000 flights.**
For a sample of size $n=1000$:
* **Mean of the sampling distribution ($E(\\bar{p})$):**
$E(\\bar{p}) = p = 0.25$
* **Standard deviation of the sampling distribution ($\\sigma\_{\\bar{p}}$):**
$\\sigma\_{\\bar{p}} = \\sqrt{\\frac{p(1-p)}{n}} = \\sqrt{\\frac{0.25(1-0.25)}{1000}} = \\sqrt{\\frac{0.25 \\times 0.75}{1000}} = \\sqrt{\\frac{0.1875}{1000}} = \\sqrt{0.0001875} \\approx 0.01369$
* **Shape of the sampling distribution:**
To determine the shape, we check if $np \\ge 5$ and $n(1-p) \\ge 5$.
$np = 1000 \\times 0.25 = 250$
$n(1-p) = 1000 \\times 0.75 = 750$
Since both $250 \\ge 5$ and $750 \\ge 5$, the sampling distribution of $\\bar{p}$ is **approximately normal**.
**Summary of Sampling Distribution for $n=1000$:**
* **Mean:** $0.25$
* **Standard Deviation:** Approximately $0.01369$
* **Shape:** Approximately Normal
**b. What is the probability that the sample proportion will be within .03 of the population proportion if a sample of size 1000 is selected?**
We want to find $P(p - 0.03 \\le \\bar{p} \\le p + 0.03)$, which is $P(0.25 - 0.03 \\le \\bar{p} \\le 0.25 + 0.03) = P(0.22 \\le \\bar{p} \\le 0.28)$.
We use the Z-score formula: $Z = \\frac{\\bar{p} - p}{\\sigma\_{\\bar{p}}}$
For $\\bar{p} = 0.22$:
$Z\_{lower} = \\frac{0.22 - 0.25}{0.01369} = \\frac{-0.03}{0.01369} \\approx -2.191$
For $\\bar{p} = 0.28$:
$Z\_{upper} = \\frac{0.28 - 0.25}{0.01369} = \\frac{0.03}{0.01369} \\approx 2.191$
Now, we find the probability $P(-2.191 \\le Z \\le 2.191)$:
$P(Z \\le 2.191) - P(Z \\le -2.191)$
Using a standard normal table or calculator:
$P(Z \\le 2.191) \\approx 0.9857$
$P(Z \\le -2.191) \\approx 0.0143$
Probability $= 0.9857 - 0.0143 = 0.9714$.
The probability that the sample proportion will be within $0.03$ of the population proportion for a sample of size 1000 is approximately $\\boxed{0.9715}$.
**c. Answer part (b) for a sample of 500 flights.**
For a sample of size $n=500$:
* **Standard deviation of the sampling distribution ($\\sigma\_{\\bar{p}}$):**
$\\sigma\_{\\bar{p}} = \\sqrt{\\frac{p(1-p)}{n}} = \\sqrt{\\frac{0.25(1-0.25)}{500}} = \\sqrt{\\frac{0.1875}{500}} = \\sqrt{0.000375} \\approx 0.01936$
We want to find $P(0.22 \\le \\bar{p} \\le 0.28)$ with the new standard deviation.
For $\\bar{p} = 0.22$:
$Z\_{lower} = \\frac{0.22 - 0.25}{0.01936} = \\frac{-0.03}{0.01936} \\approx -1.549$
For $\\bar{p} = 0.28$:
$Z\_{upper} = \\frac{0.28 - 0.25}{0.01936} = \\frac{0.03}{0.01936} \\approx 1.549$
Now, we find the probability $P(-1.549 \\le Z \\le 1.549)$:
$P(Z \\le 1.549) - P(Z \\le -1.549)$
Using a standard normal table or calculator:
$P(Z \\le 1.549) \\approx 0.9393$
$P(Z \\le -1.549) \\approx 0.0607$
Probability $= 0.9393 - 0.0607 = 0.8786$.
The probability that the sample proportion will be within $0.03$ of the population proportion for a sample of size 500 is approximately $\\boxed{0.8787}$.
Question 1167902: 1. From past records it is known that the average life of a battery used in a digital clock is 305 days. The lives of the batteries are normally distributed. The battery was recently modifies to last longer. Samples of 20 modified batteries were tested. It was discovered that the mean life was 311 days, ant the sample standard deviation was 12 days. At the 0.05 level of significance, did the modification increase the mean life of the battery?
a. State the null and alternate hypotheses.
b. Show the decision graphically.
c. Compute t, and reach a decision.
d. Briefly summary your findings.
Answer by CPhill(1959) (Show Source):
You can put this solution on YOUR website! This problem calls for a **one-sample t-test** to determine if the modification increased the mean life of the battery. We use a t-test because the population standard deviation is unknown, and we are working with a sample standard deviation. The problem states that the lives of the batteries are normally distributed, which is a key assumption for the t-test.
**Given Information:**
* Known population average life ($\\mu\_0$): 305 days
* Sample size ($n$): 20 modified batteries
* Sample mean ($\\bar{x}$): 311 days
* Sample standard deviation ($s$): 12 days
* Level of significance ($\\alpha$): 0.05
-----
**a. State the null and alternate hypotheses.**
* **Null Hypothesis ($H\_0$):** The modification did not increase the mean life of the battery (i.e., the mean life is still 305 days or less).
$H\_0: \\mu \\le 305$ days
* **Alternate Hypothesis ($H\_1$):** The modification increased the mean life of the battery (i.e., the mean life is greater than 305 days).
$H\_1: \\mu \> 305$ days
This is a **one-tailed (right-tailed) test**.
-----
**b. Show the decision graphically.**
To show the decision graphically, we need the critical t-value.
* Degrees of Freedom ($df$) = $n - 1 = 20 - 1 = 19$
* Significance Level ($\\alpha$) = 0.05
* Type of Test: One-tailed (right-tailed)
Looking up the t-distribution table for $df=19$ and $\\alpha=0.05$ (one-tailed), the critical t-value is approximately **1.729**.
**Graphical Representation:**
```
^ Probability Density
|
| /--
| / \
| / \
| / \
|______/_________\____________
-3 -2 -1 0 1 2 3
^ ^
1.729 (Critical Value)
___________________________|_________________
Fail to Reject H0 | Reject H0 (alpha = 0.05)
```
* The t-distribution curve is shown.
* The critical value of $t = 1.729$ defines the boundary of the rejection region.
* The shaded area to the right of $1.729$ represents the rejection region (where the top 5% of the distribution lies).
* If the calculated t-value falls into this shaded region, we reject $H\_0$. Otherwise, we fail to reject $H\_0$.
-----
**c. Compute t, and reach a decision.**
**1. Compute the t-value:**
The formula for the t-statistic is:
$t = \\frac{\\bar{x} - \\mu\_0}{s / \\sqrt{n}}$
Substitute the given values:
$\\bar{x} = 311$ days
$\\mu\_0 = 305$ days
$s = 12$ days
$n = 20$
$t = \\frac{311 - 305}{12 / \\sqrt{20}}$
$t = \\frac{6}{12 / 4.4721}$
$t = \\frac{6}{2.6833}$
$t \\approx 2.236$
**2. Reach a decision:**
* Calculated t-value: $2.236$
* Critical t-value: $1.729$
Since the calculated t-value ($2.236$) is greater than the critical t-value ($1.729$), it falls into the rejection region.
**Decision:** Reject the null hypothesis ($H\_0$).
-----
**d. Briefly summarize your findings.**
At the 0.05 level of significance, the statistical analysis provides **sufficient evidence** to conclude that the modification did indeed increase the mean life of the battery. The sample of 20 modified batteries showed a mean life of 311 days, which is significantly higher than the past average of 305 days, leading to the rejection of the null hypothesis.
Question 1167903: 2. A Corkill Machine is set ti fill a small bottle with 9.0 grams of medicine. It is claimed that the mean weight is less 9.0 grams. The hypothesis is to be tested at the 0.01 level. A sample revealed these weights (in grams): 9.2, 8.7, 8.9, 8.6, 8.8, 8.5, 8.7, and 9.0.
a. State the null and alternate hypotheses.
b. How many degrees of freedom are there?
c. Give the decision rule.
d. Compute t-value and arrive at a decision.
Answer by CPhill(1959) (Show Source):
You can put this solution on YOUR website! This problem requires a **one-sample t-test** because we are comparing a sample mean to a known population mean (or hypothesized mean) when the population standard deviation is unknown and the sample size is small ($n < 30$).
**Given Data:**
* Hypothesized population mean ($\mu_0$): 9.0 grams
* Sample weights: $X = \{9.2, 8.7, 8.9, 8.6, 8.8, 8.5, 8.7, 9.0\}$
* Sample size ($n$): 8
* Significance level ($\alpha$): 0.01
---
**a. State the null and alternate hypotheses.**
* **Null Hypothesis ($H_0$):** The mean weight of medicine filled by the Corkill Machine is equal to or greater than 9.0 grams.
$H_0: \mu \ge 9.0$
* **Alternate Hypothesis ($H_1$):** The mean weight of medicine filled by the Corkill Machine is less than 9.0 grams.
$H_1: \mu < 9.0$
This is a **one-tailed (left-tailed) test**.
---
**b. How many degrees of freedom are there?**
For a one-sample t-test, the degrees of freedom ($df$) are calculated as $n - 1$, where $n$ is the sample size.
* $df = 8 - 1 = 7$
There are **7 degrees of freedom**.
---
**c. Give the decision rule.**
The decision rule is based on comparing the calculated t-value to a critical t-value from the t-distribution table.
* **Significance Level ($\alpha$):** 0.01
* **Degrees of Freedom ($df$):** 7
* **Type of Test:** One-tailed (left-tailed)
From the t-distribution table, the critical t-value for $\alpha = 0.01$ and $df = 7$ for a one-tailed test is approximately **-2.998**.
**Decision Rule:** Reject the null hypothesis ($H_0$) if the calculated t-value is less than -2.998. Otherwise, fail to reject $H_0$.
---
**d. Compute t-value and arrive at a decision.**
**1. Calculate Sample Statistics:**
* **Sample Mean ($\bar{x}$):**
$\sum X = 9.2 + 8.7 + 8.9 + 8.6 + 8.8 + 8.5 + 8.7 + 9.0 = 70.4$
$\bar{x} = \frac{\sum X}{n} = \frac{70.4}{8} = 8.8$
* **Sample Standard Deviation ($s$):**
First, calculate $\sum (X - \bar{x})^2$:
* $(9.2 - 8.8)^2 = (0.4)^2 = 0.16$
* $(8.7 - 8.8)^2 = (-0.1)^2 = 0.01$
* $(8.9 - 8.8)^2 = (0.1)^2 = 0.01$
* $(8.6 - 8.8)^2 = (-0.2)^2 = 0.04$
* $(8.8 - 8.8)^2 = (0.0)^2 = 0.00$
* $(8.5 - 8.8)^2 = (-0.3)^2 = 0.09$
* $(8.7 - 8.8)^2 = (-0.1)^2 = 0.01$
* $(9.0 - 8.8)^2 = (0.2)^2 = 0.04$
$\sum (X - \bar{x})^2 = 0.16 + 0.01 + 0.01 + 0.04 + 0.00 + 0.09 + 0.01 + 0.04 = 0.36$
$s = \sqrt{\frac{\sum (X - \bar{x})^2}{n-1}} = \sqrt{\frac{0.36}{8-1}} = \sqrt{\frac{0.36}{7}} \approx \sqrt{0.051428} \approx 0.22678$
**2. Compute the t-value:**
The formula for the t-statistic is:
$t = \frac{\bar{x} - \mu_0}{s / \sqrt{n}}$
$t = \frac{8.8 - 9.0}{0.22678 / \sqrt{8}}$
$t = \frac{-0.2}{0.22678 / 2.8284}$
$t = \frac{-0.2}{0.08018}$
$t \approx -2.494$
**3. Arrive at a Decision:**
* Calculated t-value: $-2.494$
* Critical t-value: $-2.998$
Since the calculated t-value ($-2.494$) is **greater than** the critical t-value ($-2.998$), it does not fall into the rejection region.
**Decision:** Fail to reject the null hypothesis ($H_0$).
**Conclusion:**
At the 0.01 significance level, there is **not enough statistical evidence** to support the claim that the mean weight of medicine filled by the Corkill Machine is less than 9.0 grams.
Question 1167904: 3. The net weight of a sample bottles filled by a machine manufactured by Edne, and the net weights of a sample filled by a similar machine manufactured by Orno, Inc., are:
Edne: 5, 8, 7, 6, 9, 7
Orno: 8, 10, 7, 11, 9, 12, 14, 9
At the 0.05 level o, is the mean weight of the bottles filled by the Orno Machine greater than the mean weight of the bottle filled by the Edne Machine?
Answer by CPhill(1959) (Show Source):
You can put this solution on YOUR website! This problem requires a hypothesis test to compare the means of two independent samples. Since the population standard deviations are unknown and the sample sizes are small (n < 30), a **two-sample t-test** is appropriate. Specifically, since we are testing if the mean of Orno is *greater than* the mean of Edne, it will be a **one-tailed test**.
Given the small sample sizes, we should ideally check for approximate normality of the data or assume that the underlying populations are normally distributed. We also need to decide whether to assume equal variances or unequal variances between the two populations. A common approach is to perform an F-test for equality of variances first, or simply assume unequal variances (Welch's t-test) which is more robust when the assumption of equal variances might be violated. For simplicity, and typical textbook problems of this nature, if variances are not drastically different, one might proceed with the pooled-variance t-test. However, if no specific instruction, assuming unequal variances is safer. Let's calculate the standard deviations for each group.
**Data:**
* **Edne (Sample 1):** $n_1 = 6$
Scores: $X_1 = \{5, 8, 7, 6, 9, 7\}$
* **Orno (Sample 2):** $n_2 = 8$
Scores: $X_2 = \{8, 10, 7, 11, 9, 12, 14, 9\}$
**1. Calculate Sample Statistics:**
* **Edne (Sample 1):**
* Sum: $\sum X_1 = 5+8+7+6+9+7 = 42$
* Mean: $\bar{X}_1 = \frac{42}{6} = 7$
* Sum of Squares: $\sum X_1^2 = 5^2+8^2+7^2+6^2+9^2+7^2 = 25+64+49+36+81+49 = 304$
* Sample Variance: $s_1^2 = \frac{\sum X_1^2 - n_1(\bar{X}_1)^2}{n_1-1} = \frac{304 - 6(7)^2}{6-1} = \frac{304 - 6(49)}{5} = \frac{304 - 294}{5} = \frac{10}{5} = 2$
* Sample Standard Deviation: $s_1 = \sqrt{2} \approx 1.414$
* **Orno (Sample 2):**
* Sum: $\sum X_2 = 8+10+7+11+9+12+14+9 = 80$
* Mean: $\bar{X}_2 = \frac{80}{8} = 10$
* Sum of Squares: $\sum X_2^2 = 8^2+10^2+7^2+11^2+9^2+12^2+14^2+9^2 = 64+100+49+121+81+144+196+81 = 836$
* Sample Variance: $s_2^2 = \frac{\sum X_2^2 - n_2(\bar{X}_2)^2}{n_2-1} = \frac{836 - 8(10)^2}{8-1} = \frac{836 - 8(100)}{7} = \frac{836 - 800}{7} = \frac{36}{7} \approx 5.143$
* Sample Standard Deviation: $s_2 = \sqrt{\frac{36}{7}} = \frac{6}{\sqrt{7}} \approx 2.268$
**2. Formulate Hypotheses:**
* **Null Hypothesis ($H_0$):** $\mu_{Orno} \le \mu_{Edne}$ (The mean weight of bottles from Orno is not greater than Edne)
* **Alternative Hypothesis ($H_1$):** $\mu_{Orno} > \mu_{Edne}$ (The mean weight of bottles from Orno is greater than Edne)
This is a **one-tailed (right-tailed) t-test**.
**3. Choose Significance Level:**
* $\alpha = 0.05$
**4. Choose Test Statistic:**
Given that sample sizes are small and population standard deviations are unknown, a t-test is used. Since the sample variances ($s_1^2 = 2$ and $s_2^2 = 36/7 \approx 5.143$) are not drastically different, we can proceed with a pooled-variance t-test. (Alternatively, a Welch's t-test for unequal variances could be used, leading to a slightly different calculation for degrees of freedom.)
**Pooled Variance ($s_p^2$):**
$s_p^2 = \frac{(n_1-1)s_1^2 + (n_2-1)s_2^2}{n_1+n_2-2}$
$s_p^2 = \frac{(6-1)(2) + (8-1)(\frac{36}{7})}{6+8-2}$
$s_p^2 = \frac{5(2) + 7(\frac{36}{7})}{12}$
$s_p^2 = \frac{10 + 36}{12} = \frac{46}{12} = \frac{23}{6} \approx 3.833$
**t-statistic for pooled variance:**
$t = \frac{(\bar{X}_2 - \bar{X}_1) - (\mu_2 - \mu_1)}{\sqrt{s_p^2 \left(\frac{1}{n_1} + \frac{1}{n_2}\right)}}$
Under $H_0$, $\mu_2 - \mu_1 = 0$.
$t = \frac{(10 - 7) - 0}{\sqrt{\frac{23}{6} \left(\frac{1}{6} + \frac{1}{8}\right)}}$
$t = \frac{3}{\sqrt{\frac{23}{6} \left(\frac{4+3}{24}\right)}}$
$t = \frac{3}{\sqrt{\frac{23}{6} \left(\frac{7}{24}\right)}}$
$t = \frac{3}{\sqrt{\frac{161}{144}}}$
$t = \frac{3}{\frac{\sqrt{161}}{12}} = \frac{3 \times 12}{\sqrt{161}} = \frac{36}{\sqrt{161}}$
$t \approx \frac{36}{12.688} \approx 2.837$
**5. Determine Degrees of Freedom and Critical Value:**
* Degrees of Freedom ($df$) = $n_1 + n_2 - 2 = 6 + 8 - 2 = 12$
* For a one-tailed (right-tailed) t-test with $\alpha = 0.05$ and $df = 12$, the critical t-value ($t_{crit}$) from a t-distribution table is approximately $1.782$.
**6. Make a Decision:**
* Compare the calculated t-statistic to the critical t-value.
* Calculated $t = 2.837$
* Critical $t_{crit} = 1.782$
Since $2.837 > 1.782$, the calculated t-statistic falls into the rejection region.
**7. Conclusion:**
At the 0.05 significance level, we **reject the null hypothesis ($H_0$)**.
There is sufficient statistical evidence to conclude that the mean weight of the bottles filled by the Orno Machine is significantly greater than the mean weight of the bottles filled by the Edne Machine.
Question 1167906: 4. In an effort to determine the most effective way to teach safety principles to a group of employees at university, four different methods were tried. Samples of 20 employees were randomly assigned to one of four groups. The first group was given programmed instruction booklets and worked through the course at their own face. The second group attended lectures. The third group watched television presentations, and the fourth was divided into small discussion groups. At the end of the sessions, a test was given to the four groups. A high of 10 points is the highest score. Test at the 0.05 level that there is no difference among the four means. The results are presented below:
Table 1. Test grades of employees.
Programmed Instruction Lecture TV Presentation Group Discussion
6 8 7 8
7 5 9 5
6 8 6 6
5 6 8 6
6 8 5 5
Answer by CPhill(1959) (Show Source):
You can put this solution on YOUR website! This scenario is perfectly suited for a **One-Way Analysis of Variance (ANOVA)**.
**Reasoning for choosing One-Way ANOVA:**
* **One Categorical Independent Variable:** The "teaching method" is a single independent variable with four distinct levels (Programmed Instruction, Lecture, TV Presentation, Group Discussion).
* **One Quantitative Dependent Variable:** The "test score" is a quantitative dependent variable.
* **Independent Samples:** Employees were *randomly assigned* to one of the four groups, meaning the observations in one group are independent of the observations in another group.
* **Comparing More Than Two Means:** We are comparing the means of four different groups. While multiple t-tests could compare pairs of groups, ANOVA is designed to compare three or more means simultaneously, controlling the Type I error rate (the chance of falsely rejecting a true null hypothesis) across all comparisons.
---
## Sample Computation and Analysis: One-Way ANOVA
**Scenario:** A university is trying to determine the most effective way to teach safety principles. Four teaching methods are tested on separate groups of 20 employees each (though the provided data only shows 5 observations per group, which we will use for the sample calculation). A test out of 10 points is administered afterwards.
**Research Question:** Is there a significant difference in the mean test scores among the four different teaching methods?
**Hypotheses:**
* **Null Hypothesis ($H_0$):** There is no significant difference among the mean test scores of the four teaching methods. ($\mu_1 = \mu_2 = \mu_3 = \mu_4$)
* **Alternative Hypothesis ($H_1$):** At least one of the mean test scores is significantly different from the others. (Not all $\mu_i$ are equal)
**Significance Level ($\alpha$):** $\alpha = 0.05$
**Data (Test Grades of Employees):**
| Programmed Instruction (Group 1) | Lecture (Group 2) | TV Presentation (Group 3) | Group Discussion (Group 4) |
| :------------------------------- | :---------------- | :------------------------ | :------------------------- |
| 6 | 8 | 7 | 8 |
| 7 | 5 | 9 | 5 |
| 6 | 8 | 6 | 6 |
| 5 | 6 | 8 | 6 |
| 6 | 8 | 5 | 5 |
**Preliminary Calculations:**
| Group | $n_i$ | Scores ($\boldsymbol{X_i}$) | $\sum X_i$ | $\bar{X_i}$ | $\sum X_i^2$ |
| :------------------- | :---- | :------------------------ | :--------- | :---------- | :----------- |
| Programmed Instr. (1)| 5 | 6, 7, 6, 5, 6 | 30 | 6.0 | 182 |
| Lecture (2) | 5 | 8, 5, 8, 6, 8 | 35 | 7.0 | 253 |
| TV Presentation (3) | 5 | 7, 9, 6, 8, 5 | 35 | 7.0 | 255 |
| Group Discussion (4) | 5 | 8, 5, 6, 6, 5 | 30 | 6.0 | 186 |
| **Total** | $N=20$| | $\sum X = 130$ | $\bar{X}_{\text{grand}} = 6.5$ | $\sum X^2 = 876$ |
**ANOVA Computation (Sum of Squares):**
1. **Total Sum of Squares (SST):** Measures total variability in data.
$SST = \sum X^2 - \frac{(\sum X)^2}{N}$
$SST = 876 - \frac{(130)^2}{20} = 876 - \frac{16900}{20} = 876 - 845 = 31$
2. **Sum of Squares Between Groups (SSB) / Sum of Squares Treatment (SSTreatment):** Measures variability between group means.
$SSB = \sum \left( \frac{(\sum X_i)^2}{n_i} \right) - \frac{(\sum X)^2}{N}$
$SSB = \left( \frac{30^2}{5} + \frac{35^2}{5} + \frac{35^2}{5} + \frac{30^2}{5} \right) - \frac{130^2}{20}$
$SSB = \left( \frac{900}{5} + \frac{1225}{5} + \frac{1225}{5} + \frac{900}{5} \right) - 845$
$SSB = (180 + 245 + 245 + 180) - 845$
$SSB = 850 - 845 = 5$
3. **Sum of Squares Within Groups (SSW) / Sum of Squares Error (SSError):** Measures variability within each group (error).
$SSW = SST - SSB$
$SSW = 31 - 5 = 26$
**Degrees of Freedom (df):**
1. **Total df ($df_{Total}$):** $N - 1 = 20 - 1 = 19$
2. **Between Groups df ($df_{Between}$):** $k - 1 = 4 - 1 = 3$ (where $k$ is the number of groups)
3. **Within Groups df ($df_{Within}$):** $N - k = 20 - 4 = 16$
(Check: $df_{Total} = df_{Between} + df_{Within} \Rightarrow 19 = 3 + 16$, which is correct)
**Mean Squares (MS):**
1. **Mean Square Between Groups (MSB):**
$MSB = \frac{SSB}{df_{Between}} = \frac{5}{3} \approx 1.667$
2. **Mean Square Within Groups (MSW):**
$MSW = \frac{SSW}{df_{Within}} = \frac{26}{16} = 1.625$
**F-statistic:**
$F = \frac{MSB}{MSW} = \frac{1.667}{1.625} \approx 1.026$
---
**ANOVA Summary Table:**
| Source of Variation | Sum of Squares (SS) | df | Mean Square (MS) | F-statistic | p-value | Critical F (α=0.05) |
| :------------------ | :------------------ | :- | :--------------- | :---------- | :------ | :------------------ |
| Between Groups | 5 | 3 | 1.667 | 1.026 | 0.407 | 3.24 |
| Within Groups | 26 | 16 | 1.625 | | | |
| Total | 31 | 19 | | | | |
* **Critical F-value:** For $df_1 = 3$ and $df_2 = 16$ at $\alpha = 0.05$, the critical F-value from an F-distribution table is approximately $3.24$.
* **p-value:** Using statistical software or an F-distribution calculator for $F=1.026$ with $df_1=3, df_2=16$, the p-value is approximately $0.407$.
---
**Analysis and Conclusion:**
* **Comparison of F-statistic to Critical F-value:** Our calculated F-statistic ($1.026$) is less than the critical F-value ($3.24$).
* **Comparison of p-value to Significance Level:** Our p-value ($0.407$) is greater than the significance level ($\alpha = 0.05$).
Since our calculated F-statistic does not fall into the rejection region (and our p-value is greater than $\alpha$), we **fail to reject the null hypothesis**.
**Conclusion:** At the 0.05 significance level, there is **no statistically significant difference** among the mean test scores of the four different safety teaching methods. This suggests that, based on this sample data, none of the teaching methods proved to be significantly more effective than the others in terms of test performance. It implies that any observed differences in average scores are likely due to random chance rather than the teaching method itself.
Question 1167907: 5. Prepare a sample computation and analysis of observation that involve test of significance using dependent (correlated) data. Examples are:
a. A Training Director wishes to find out whether or not a unique training program will increase employee efficiency (before and after the training).
b. Weights of pupil before and after the feeding program.
c. Giving incentives to sales person.
d. Intelligence scores after showing special movie.
Answer by CPhill(1959) (Show Source):
You can put this solution on YOUR website! Let's prepare a sample computation and analysis for an observation involving a test of significance using dependent (correlated) data. We'll use example (a): "A Training Director wishes to find out whether or not a unique training program will increase employee efficiency (before and after the training)."
This scenario is perfect for a **paired-samples t-test** (also known as a dependent samples t-test).
---
## Sample Computation and Analysis: Paired-Samples t-test
**Scenario:** A Training Director wants to evaluate the effectiveness of a new training program designed to increase employee efficiency. They measure the efficiency scores of 10 randomly selected employees *before* the training and *after* the training.
**Research Question:** Does the training program significantly increase employee efficiency?
**Hypotheses:**
* **Null Hypothesis ($H_0$):** There is no significant difference in employee efficiency scores before and after the training program. (i.e., $\mu_d = 0$, where $\mu_d$ is the mean difference in efficiency scores).
* **Alternative Hypothesis ($H_1$):** The training program significantly increases employee efficiency. (i.e., $\mu_d > 0$, indicating that after scores are higher than before scores). This is a one-tailed test.
**Significance Level ($\alpha$):** Let's set $\alpha = 0.05$.
**Data Collection:**
Efficiency scores (e.g., tasks completed per hour, error rate, etc., scaled to a common metric) for 10 employees:
| Employee | Before Training (Score 1) | After Training (Score 2) | Difference ($d = \text{Score 2} - \text{Score 1}$) |
| :------- | :------------------------ | :----------------------- | :----------------------------------------------- |
| 1 | 75 | 80 | 5 |
| 2 | 80 | 85 | 5 |
| 3 | 68 | 70 | 2 |
| 4 | 92 | 95 | 3 |
| 5 | 70 | 78 | 8 |
| 6 | 85 | 83 | -2 |
| 7 | 73 | 79 | 6 |
| 8 | 78 | 82 | 4 |
| 9 | 88 | 90 | 2 |
| 10 | 79 | 84 | 5 |
**Sample Computation:**
1. **Calculate the difference ($d$) for each pair:** (Already done in the table above).
Differences: $5, 5, 2, 3, 8, -2, 6, 4, 2, 5$
2. **Calculate the mean of the differences ($\bar{d}$):**
$\bar{d} = \frac{\sum d}{n} = \frac{5+5+2+3+8-2+6+4+2+5}{10} = \frac{38}{10} = 3.8$
3. **Calculate the standard deviation of the differences ($s_d$):**
First, calculate $(d - \bar{d})^2$ for each difference:
* $(5 - 3.8)^2 = (1.2)^2 = 1.44$
* $(5 - 3.8)^2 = (1.2)^2 = 1.44$
* $(2 - 3.8)^2 = (-1.8)^2 = 3.24$
* $(3 - 3.8)^2 = (-0.8)^2 = 0.64$
* $(8 - 3.8)^2 = (4.2)^2 = 17.64$
* $(-2 - 3.8)^2 = (-5.8)^2 = 33.64$
* $(6 - 3.8)^2 = (2.2)^2 = 4.84$
* $(4 - 3.8)^2 = (0.2)^2 = 0.04$
* $(2 - 3.8)^2 = (-1.8)^2 = 3.24$
* $(5 - 3.8)^2 = (1.2)^2 = 1.44$
Sum of $(d - \bar{d})^2 = 1.44 + 1.44 + 3.24 + 0.64 + 17.64 + 33.64 + 4.84 + 0.04 + 3.24 + 1.44 = 67.6$
$s_d = \sqrt{\frac{\sum (d - \bar{d})^2}{n-1}} = \sqrt{\frac{67.6}{10-1}} = \sqrt{\frac{67.6}{9}} = \sqrt{7.511...} \approx 2.74$
4. **Calculate the test statistic (t-value):**
$t = \frac{\bar{d}}{s_d / \sqrt{n}} = \frac{3.8}{2.74 / \sqrt{10}} = \frac{3.8}{2.74 / 3.16} = \frac{3.8}{0.867} \approx 4.383$
5. **Determine Degrees of Freedom (df):**
$df = n - 1 = 10 - 1 = 9$
6. **Find the Critical t-value:**
For a one-tailed t-test with $\alpha = 0.05$ and $df = 9$, we look up the critical t-value from a t-distribution table.
Critical $t_{0.05, 9} \approx 1.833$
7. **Calculate the p-value (optional, but standard in software):**
Using statistical software or a calculator, for $t = 4.383$ with $df = 9$ (one-tailed), the p-value is approximately $0.0008$.
---
**Analysis and Conclusion:**
* **Comparison of Test Statistic to Critical Value:** Our calculated t-value ($4.383$) is greater than the critical t-value ($1.833$).
* **Comparison of p-value to Significance Level:** Our p-value ($0.0008$) is less than the significance level ($\alpha = 0.05$).
Since our calculated t-value falls into the rejection region (and our p-value is less than $\alpha$), we **reject the null hypothesis**.
**Conclusion:** Based on this analysis, there is sufficient statistical evidence at the 0.05 significance level to conclude that the unique training program significantly *increases* employee efficiency. The observed average increase in efficiency score by 3.8 points is statistically meaningful and unlikely to have occurred by random chance.
---
**Why Paired-Samples t-test is appropriate here:**
* **Dependent Data:** The key is that the two sets of observations (before and after scores) are from the *same* individuals. This makes the data dependent or correlated. Each employee acts as their own control.
* **Controlling for Individual Variability:** By focusing on the *difference* in scores for each individual, the paired t-test effectively controls for the natural variability in efficiency that exists between different employees (e.g., some employees are inherently more efficient than others regardless of training). This increases the statistical power of the test to detect a real effect of the training program.
Question 1167528: A survey found that women's heights are normally distributed with mean 63.4 in and standard deviation 2.5 in. A branch of the military requires women's heights to be between 58 in and 80 in.
a. Find the percentage of women meeting the height requirement. Are many women being denied the opportunity to join this branch of the military because they are too short or too tall?
b. If this branch of the military changes the height requirements so that all women are eligible except the shortest 1% and the tallest 2%, what are the new height requirements?
Answer by ikleyn(52780) (Show Source):
Question 1167480: A survey found that women's heights are normally distributed with mean 63.4 in and standard deviation 2.2 in. A branch of the military requires women's heights to be between 58 in and 80 in.
a. Find the percentage of women meeting the height requirement. Are many women being denied the opportunity to join this branch of the military because they are too short or too tall?
b. If this branch of the military changes the height requirements so that all women are eligible except the shortest 1% and the tallest 2%, what are the new height requirements?
Answer by ikleyn(52780) (Show Source):
Question 1167778: 2. A company is to hire two new employees. They have prepared a final list of eight candidates, all
of whom are equally qualified. Of these eight candidates, five are women. Suppose the
company decides to select two persons randomly from these eight candidates.
a. What is the probability that:
i. Both candidate selected are women?
ii. At least one candidate selected is a woman?
iii. Second candidate is a woman.
iv. First candidate is a woman given that the second candidate is a woman.
(2 + 3 + 4 + 2 = 11 marks)
b. Let X denote the number of women in this sample.
i. Write the probability distribution of X.
ii. Find the standard deviation of X
Answer by ikleyn(52780) (Show Source):
Question 1165805: A small motor starts on the first attempt 3/8 of the time, on the second attempt 1/4 of the time, and on the third attempt 1/4 of the time.
Let the random variable X be the number of tries it takes to start the motor. So, for instance, P(X=1) is 3/8.
Use the definition of a probability mass function to find the probability it takes more than 3 tries to start the motor.
P(X > 3) ?
Answer by ikleyn(52780) (Show Source):
You can put this solution on YOUR website! .
P(X > 3) is the complement of the sum P(X=1) + P(X=2) + P(X=3).
Therefore, P(X > 3) = 1 - 3/8 - 1/4 - 14 = 1/8. ANSWER
Solved.
Question 1167499: Suppose the probability of a server winning any given point in a tennis match is a constant p, with 0 ≤ p ≤ 1. Then the probability of the server winning a game when serving from deuce is given below.
f(p)=p^2/1-2p(1-p)
a. Evaluate f(0.79) and interpret the results.
f(0.79)= ** round to four decimal places as needed
When serving from deuce, the server will win approximately ___ out of 100 games. (round to the nearest integer as needed)
Answer by ikleyn(52780) (Show Source):
Question 1160672: The times a teacher takes to grade one page of an exam are normally distributed with a mean of 4 minutes and a standard deviation of 1 minute. For about what percent of exams does a teacher grade one page in less than 7 minutes?
Found 2 solutions by greenestamps, ikleyn:
Answer by greenestamps(13200)  (Show Source):
You can put this solution on YOUR website!
Here is a different method than shown by the other tutor for doing the calculations.
The basic rule is that 99.7% of scores in a normal distribution are within 3 standard deviations of the mean. That means 0.3% of the scores are more than 3 standard deviations from the mean.
That 0.3% is equally split between the scores that are more than 3 standard deviations above the mean and the ones that are more than 3 standard deviations below the mean, so 0.15% of the scores are more than 3 standard deviations above the mean.
And that means that the percentage of scores that are less than 3 standard deviations above the mean is 100 - 0.15 = 99.85.
ANSWER: 99.85%
Answer by ikleyn(52780) (Show Source):
You can put this solution on YOUR website! .
The times a teacher takes to grade one page of an exam are normally distributed with a mean of 4 minutes
and a standard deviation of 1 minute. For about what percent of exams does a teacher grade one page
in less than 7 minutes?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Notice that in this problem 7 minutes time interval is 3 minutes greater than 4 minutes, the average time.
So, apply the empirical rule for normal distribution, which says that 99.7% of data fall into three
standard deviations from the mean.
Using this rule, we account 50% for checking exams faster than 4 minutes, and add to it
 = 49.85% for the exams to check them from 4 minutes to 7 minutes.
Thus the answer to the question is 50% + 49.85% = 99.85%. <<<---=== ANSWER
At this point, the problem is solved completely.
///////////////////////////////////////////////
By the way, out of curiosity, I submitted this problem to Google AI today (07/14/2025).
It produced wrong solution, saying that the answer is 99.7%.
So, although the problem is very simple, AI could not solve it correctly - - - and failed.
It is because this version AI, actually, can not think, in usual meaning of this word.
It only can re-write a prototype of a solution from a template
in its base of knowledge, if it will find an appropriate template there.
But in this case, it could not find an appropriate template - - - and failed.
It is what I always say about the current version of AI in solving school Math problems:
- it works perfectly, if it finds a template with a correct solution to re-write from.
If it will not find such a template - it fails.
Naturally, I informed Google AI about their fault in this problem via their feedback system
and provided the link to this correct solution. So, from now this Google AI is a bit smarter than it was before.
\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\
Next day, 07/15/25, I checked Google AI by posting this problem again.
This time, it created a correct solution, so I am satisfied now.
Question 1167340: In 2004, the average American household received 92.6 channels. In 2008, the average American household received 118.6 channels. What is the annual rate of change from 2004 to 2008? What is a synonym for rate of change?
Answer by ikleyn(52780) (Show Source):
You can put this solution on YOUR website! .
In 2004, the average American household received 92.6 channels.
In 2008, the average American household received 118.6 channels.
What is the annual rate of change from 2004 to 2008?
What is a synonym for rate of change?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
From the value of 118.6 in 2008 subtract the value of 92.6 in 2004.
Then divide this difference by 4, which is the difference 2008 - 2004.
The ratio which you obtain will be the the annual rate.
The synonym for the rate of change is the speed of change.
Question 1166256: At the ministry of Health headquarters, you are in charge of rationing shangombo district the area affected by hunger.The following reports about daily calories value of food available per adult during current period arrive from your local investigators. The estimated requirement of an adult is taken at 3000 calories per day and the absolute minimum at 1250. Advise the ministry by commenting on the reported figures and determine which in your own opinion needs more urgent attention Area
mean standard deviation
X 2500 500
Y 2200 300
Answer by ikleyn(52780) (Show Source):
You can put this solution on YOUR website! .
At the ministry of Health headquarters, you are in charge of rationing shangombo district
the area affected by hunger.The following reports about daily calories value of food
available per adult during current period arrive from your local investigators.
The estimated requirement of an adult is taken at 3000 calories per day and the absolute minimum at 1250.
Advise the ministry by commenting on the reported figures and determine which in your own opinion needs more urgent attention Area
mean standard deviation
X 2500 500
Y 2200 300
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Use online calculator for the normal distribution function under this link
https://onlinestatbook.com/2/calculators/normal_dist.html
This calculator provides the area under the normal curve on the left of the given mark.
It works in the same way as a standard function normalcdf().
In this problem, the major concern is the population
below the absolute minimum of 1250 calories per day.
For region X, input the mean = 2500, SD = 500 and take the option "below 1250 calories".
Then the calculator gives the area of 0.0062 on the left of this mark,
which means that 0.62% of population X is below the minimum food.
For region Y, input the mean = 2200, SD = 300 and take the option "below 1250 calories"
Then the calculator gives the area of 0.0008 on the left of this mark,
which means that 0.08% of population Y is below the minimum food.
Naturally, region X is the greater concern. ANSWER
Solved.
-----------------------------
In my reasoning above, I assumed that regions X and Y have approximately the same or comparable population.
If it is not so, then the final answer may change.
Question 1165673: Five-digit codes are selected at random from the set {0, 1, 2, ..., 9} with replacement. If the random variable X denotes the number of zeros in randomly chosen codes, then what are the space and the probability density function of X?
Answer by ikleyn(52780) (Show Source):
You can put this solution on YOUR website! .
Five-digit codes are selected at random from the set {0, 1, 2, ..., 9} with replacement.
If the random variable X denotes the number of zeros in randomly chosen codes, then what are
the space and the probability density function of X?
~~~~~~~~~~~~~~~~~~~~~~~~~~~
The words "with replacement" mean, in this context, that in every of 5 positions, from 1 to 5,
from left to right, any of 10 digits can appear independently of digits in other positions,
with the probability  = 0.1.
For X, we have 6 possible values, X = 0, 1, 2, 3, 4, 5.
The probability density function of X is
P(X=0) =  =  = 0.59049,
P(X=1) =  =  = 0.32805,
P(X=2) =  =  =  = 0.0729,
P(X=3) =  =  =  = 0.0081,
P(X=4) =  =  =  = 0.00045,
P(X=5) =  =  = 0.00001.
CHECK. 0.59049 + 0.32805 + 0.0729 + 0.0081 + 0.00045 + 0.00001 = 1. (! correct !)
Solved.
Question 1167289: A club has 20 members. In how many ways could an executive committee consisting of a president, vice president, secretary, membership director, and treasurer be named? Show two methods.
Answer by ikleyn(52780) (Show Source):
You can put this solution on YOUR website! .
A club has 20 members. In how many ways could an executive committee consisting of a president,
vice president, secretary, membership director, and treasurer be named? Show two methods.
~~~~~~~~~~~~~~~~~~~~~~~~~
There are 20 options (possible outcomes) for the president position;
then 19 options for the vice-president position;
then 18 options for the secretary position;
then 17 options for the membership director position;
then 16 options for the treasurer position.
In all, it gives 20*19*18*17*16 = 1860480 different outcomes of possible elections.
Interesting, that you can start from any position and drive your reasoning
in any different order from position to position, but your final answer will be
the same invariant value independently on the order in which you consider the positions.
Solved.
Question 1210349: A random sample of 16 pharmacy customers showed the waiting times below (in minutes).
20 14 23 22 22 15 21 18
23 17 14 25 15 21 20 14
Find a 90 percent confidence interval for μ, assuming that the sample is from a normal population.
Found 2 solutions by math_tutor2020, Edwin McCravy:
Answer by math_tutor2020(3816) (Show Source):
You can put this solution on YOUR website!
I recommend using a spreadsheet for this type of problem.
If you want to use a TI83 or similar, then refer to the next section below.
Copy/paste the given data into a spreadsheet.
Let's say the upper left corner goes in cell A1.
This would have the lower right corner in cell H2.
In some other blank cell, say cell A4, we'll have the following command
=Average(A1:H2)
Don't forget about the equal sign up front. Otherwise the spreadsheet command will not execute.
This computes the arithmetic mean of the values from cell A1 to H2.
The result of this calculation is 19 exactly.
This is the value of xbar, which is the sample mean.
Now move to cell A5.
Type this in, or copy/paste
=Confidence.T(1-0.9,Stdev(A1:H2),Count(A1:H2))
The Confidence function produces the margin of error needed.
The "dot T" portion tells the spreadsheet to do the T distribution version. Otherwise, it defaults to the normal distribution.
We use the T distribution for two reasons: we don't know the value of sigma, and n = 16 doesnt make n > 30 true
The 3 parameters of the confidence function are:
alpha = significance level = 1-C = 1-0.90 = 0.10
stdev = standard deviation
n = sample size = number of values
You can manually type in the number of values (16), or use the Count function.
The result of the Confidence calculation is roughly 1.63200106504704, which is the approximate margin of error.
In cells A6 and A7 will be the following commands
=A4-A5
=A4+A5
They represent the lower and upper bounds of the confidence interval.
The results of the calculations are roughly 17.367998934953 and 20.632001065047
When rounding to 3 decimal places we get the confidence interval (17.368, 20.632)
Your final answer will vary depending how your teacher wants you to round.
-----------------------------------------------------------------------------------------------------------------
If you want to use a TI83 or similar, then follow these steps.
Press the button labeled "STAT". Then select "Edit".
This brings up the data tables. Clear out list L1 if it has values in it already.
Once fully empty, enter the data values into the list.
After the data is entered, quit to the home screen.
Press the STAT button again. Scroll to right to land on "Tests". Hit enter.
Scroll down until reaching "T-intveral". It's the 8th item of the list. Hit enter.
The input will be "data". The specific data is inside list L1.
The "freq" or "frequency" will be set to 1.
The C-level, aka confidence level, is set to 0.90
Scroll down to "calculate", and hit enter, to have the approximate interval (17.368, 20.632) show up.
Answer by Edwin McCravy(20055)  (Show Source):
You can put this solution on YOUR website!
I decided to do a recently posted statistics problem.
Not because I like them, I don't. But because high schools and colleges are
striving to make math courses practical in the workplace.
Face it, algebra, geometry, and trig are no longer practical in any workplace,
for computers do it all. Engineers these days do nothing but point and click.
Statistics is the only thing that's still practical in nearly every field.
So schools began to push statistics, because it only requires first year
algebra, and with computers doing all the substituting in formulas, even
less.
When high schools began pushing statistics, this site at first had a
deluge of statistics problems posted.
However, no tutors on here would bother solving them, including yours truly, so
students learned this site was no good for posting stats problems, and this site
went to pot.
So if you tutors want to see this site start thriving again, you'd better start
solving statistics problems. For that's the only way to salvage it.
Face it, finding such things as integer solutions to randomly made-up equations
are as impractical in the workplace as making mud pies.
--------------------------
Anyway, here's my stab at a t-problem. My stats knowledge is getting rusty.
Count these data. There are 16. This is less than (or equal to) 30. That's
considered a small sample. So we will use the t-distribution, not the
normal z-distribution.
20 14 23 22 22 15 21 18
23 17 14 25 15 21 20 14
Add them, get 304
Divide by 16, get 
We want the standard deviation, s.
  
Get the deviations from the mean, by subtracting 19 from each of those in the
list.
1 -5 4 3 3 -4 2 -1
4 -2 -5 6 -4 2 1 -5
Square the deviations from the mean:
1 25 16 9 9 16 4 1
16 4 25 36 16 4 1 25
Add them, get 208
Divide by n-1, or 16-1 or 15
Get 
Now the formula for the margin of error is
We calculate    
Since this is a 2-tailed test, 90 percent confidence interval leaves 10 percent
on both ends, or half that, 0.05 on each end, giving the confidence interval of
90 percent in between these two critical values.
So we look up the t-value for the number of 15 (1 less than the number in the
sample of 16) degrees of freedom for a 2-tailed test and find the value 1.753.
(or you can find the t-value on your TI-84 calculator by
2nd VARS area:0.95, df:15 ENTER ENTER ENTER,
and read more decimal values as 1.753050323)
 
That's the margin of error.
The confidence interval is 
(19-1.63199507,19+1.6199507) = (17.36800493, 20.63199507).
If you use a table of t-values you won't get that many decimal places.
Edwin
Question 1209733: 60% of all violent felons in the prison system are repeat offenders. If 45 violent felons are randomly selected, find the probability that
a. Exactly 26 of them are repeat offenders.
b. At most 29 of them are repeat offenders.
c. At least 29 of them are repeat offenders.
d. Between 25 and 33 (including 25 and 33) of them are repeat offenders.
Answer by ikleyn(52780) (Show Source):
You can put this solution on YOUR website! .
60% of all violent felons in the prison system are repeat offenders.
If 45 violent felons are randomly selected, find the probability that
a. Exactly 26 of them are repeat offenders.
b. At most 29 of them are repeat offenders.
c. At least 29 of them are repeat offenders.
d. Between 25 and 33 (including 25 and 33) of them are repeat offenders.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
As it regularly happens with @CPhill' solutions on Binomial distribution,
all calculations (a), (b), (c) and (d) in the post by @CPhill are incorrect.
Below is my correct solution.
Here's how to solve this binomial probability problem:
* **n** (number of trials) = 45
* **p** (probability of success - repeat offender) = 0.60
* **q** (probability of failure - not a repeat offender) = 1 - p = 0.40
The binomial probability formula is: P(x) = (nCx) * p^x * q^(n-x)
Where nCx represents "n choose x" (the binomial coefficient).
**(a) Exactly 26 are repeat offenders:**
P(x = 26) = (45C26) * (0.60)^26 * (0.40)^19
P(x = 26) ≈ 0.1143
**(b) At most 29 are repeat offenders:**
This means 0 to 29 are repeat offenders. This is a cumulative probability; we need P(x ≤ 29).
It's best to use a binomial cumulative distribution function (CDF) calculator or statistical software for this.
P(x ≤ 29) ≈ 0.7751
**(c) At least 29 are repeat offenders:**
This means 29 to 45 are repeat offenders. We can use the complement rule:
P(x ≥ 29) = 1 - P(x < 29) = 1 - P(x ≤ 28)
Use a binomial CDF calculator:
P(x ≥ 29) = 1 - 0.6728
P(x ≥ 29) ≈ 0.3272
**(d) Between 25 and 33 (inclusive):**
This means 25, 26, 27, 28, 29, 30, 31, 32, and 33 are repeat offenders. We can use the CDF:
P(25 ≤ x ≤ 33) = P(x ≤ 33) - P(x ≤ 24)
Use a binomial CDF calculator:
P(25 ≤ x ≤ 33) = 0.9784 - 0.2223
P(25 ≤ x ≤ 33) ≈ 0.7561
**Summary of Answers:**
* (a) P(x = 26) ≈ 0.1143
* (b) P(x ≤ 29) ≈ 0.7751
* (c) P(x ≥ 29) ≈ 0.3272
* (d) P(25 ≤ x ≤ 33) ≈ 0.7561
Solved CORRECTLY.
Practically all calculations by @CPhill to this kind of problems on Binomial distribution are INCORRECT.
To check your/my calculations, use online calculator
https://stattrek.com/online-calculator/binomial
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Regarding the post by @CPhill . . .
Keep in mind that @CPhill is a pseudonym for the Google artificial intelligence.
The artificial intelligence is like a baby now. It is in the experimental stage
of development and can make mistakes and produce nonsense without any embarrassment.
It has no feeling of shame - it is shameless.
This time, again, it made an error.
Although the @CPhill' solutions are copy-paste Google AI solutions, there is one essential difference.
Every time, Google AI makes a note at the end of its solutions that Google AI is experimental
and can make errors/mistakes.
All @CPhill' solutions are copy-paste of Google AI solutions, with one difference:
@PChill never makes this notice and never says that his solutions are copy-past that of Google.
So, he NEVER SAYS TRUTH.
Every time, @CPhill embarrassed to tell the truth.
But I am not embarrassing to tell the truth, as it is my duty at this forum.
And the last my comment.
When you obtain such posts from @CPhill, remember, that NOBODY is responsible for their correctness,
until the specialists and experts will check and confirm their correctness.
Without it, their reliability is ZERO and their creadability is ZERO, too.
Question 1168408: 64% of all violent felons in the prison system are repeat offenders. If 48 violent felons are randomly selected, find the probability that
a. Exactly 28 of them are repeat offenders.
b. At most 30 of them are repeat offenders.
c. At least 32 of them are repeat offenders.
d. Between 28 and 34 (including 28 and 34) of them are repeat offenders.
Answer by ikleyn(52780) (Show Source):
You can put this solution on YOUR website! .
64% of all violent felons in the prison system are repeat offenders. If 48 violent felons are randomly selected, find the probability that
a. Exactly 28 of them are repeat offenders.
b. At most 30 of them are repeat offenders.
c. At least 32 of them are repeat offenders.
d. Between 28 and 34 (including 28 and 34) of them are repeat offenders.
~~~~~~~~~~~~~~~~~~~~~~~~~
As it regularly happens with @CPhill' solutions on Binomial distribution,
all calculations (a), (b), (c) and (d) in the post by @CPhill are incorrect.
Below are my correct answers.
(a) P(X=28) = 0.0837
(b) P(X <= 30) = 0.4681
(c) P(X >= 32) = 0.4126
(d) P(28 <= X <= 34) = P(X <= 34) - P(X <= 27) = 0.8732 - 0.1662 = 0.707.
To check your/my calculations, use online calculator
https://stattrek.com/online-calculator/binomial
Question 1168409: 73% of all Americans live in cities with population greater than 100,000 people. If 35 Americans are randomly selected, find the probability that
a. Exactly 26 of them live in cities with population greater than 100,000 people.
b. At most 26 of them live in cities with population greater than 100,000 people.
c. At least 25 of them live in cities with population greater than 100,000 people.
d. Between 21 and 25 (including 21 and 25) of them live in cities with population greater than 100,000 people.
Answer by ikleyn(52780) (Show Source):
You can put this solution on YOUR website! .
73% of all Americans live in cities with population greater than 100,000 people.
If 35 Americans are randomly selected, find the probability that
a. Exactly 26 of them live in cities with population greater than 100,000 people.
b. At most 26 of them live in cities with population greater than 100,000 people.
c. At least 25 of them live in cities with population greater than 100,000 people.
d. Between 21 and 25 (including 21 and 25) of them live in cities with population greater than 100,000 people.
~~~~~~~~~~~~~~~~~~~~~~~~~~
As it regularly happens with @CPhill' solutions on Binomial distribution,
all calculations (a), (b), (c) and (d) in the post by @CPhill are incorrect.
I came to bring correct answers.
(a) P(X = 26) = 0.1505
(b) P(X <= 26) = 0.6311
(c) P(X >= 25) = 0.6641
(d) P(21 <= X <= 25) = P(X <= 25) - P(X <= 20) = 0.4806 - 0.0311 = 0.4495.
To check your/my calculations, use online calculator
https://stattrek.com/online-calculator/binomial
Question 1167502: If you use a 0.10 level of significance in a two-tail hypothesis test, what is your decision rule for rejecting a null hypothesis that the population mean equals 500 if you use the Z test?
Answer by amarjeeth123(569) (Show Source):
You can put this solution on YOUR website! When using a 0.10 level of significance in a two-tailed hypothesis test with the Z-test, the decision rule for rejecting the null hypothesis (H0: μ=12.5) is as follows: If the calculated Z-statistic falls outside the critical region defined by the critical values of -1.645 and 1.645, we reject the null hypothesis.
Question 1209960: 55% of all shark attacks are not fatal. Let r be the number of nonfatal attacks out of a random sample of five shark attacks. a) What is the probability all five shark attacks are nonfatal? b) What is the probability that three or more of the five shark attacks are nonfatal? c) What is the expected number of nonfatal shark attacks out of the five? d) What is the standard deviation of the r-probability distribution?
Answer by ikleyn(52780) (Show Source):
You can put this solution on YOUR website! .
55% of all shark attacks are not fatal. Let r be the number of nonfatal attacks out of a random sample of five shark attacks.
(a) What is the probability all five shark attacks are nonfatal?
(b) What is the probability that three or more of the five shark attacks are nonfatal?
(c) What is the expected number of nonfatal shark attacks out of the five?
(d) What is the standard deviation of the r-probability distribution?
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Be aware !
Calculations and the answer in the post by @CPhill in part (b) are INCORRECT.
Question 1209959: A club buys tickets for 15 seats in a row for a local production and then sends the tickets randomly to the 15 people who ordered seats. Of these 15 people, 9 are actors and 6 are backstage workers. Find the probability that: a. An actor and backstage worker will receive tickets sitting together. b. The actors will be seated together, and the backstage workers will be seated together.
Answer by ikleyn(52780) (Show Source):
You can put this solution on YOUR website! .
A club buys tickets for 15 seats in a row for a local production and then sends the tickets randomly
to the 15 people who ordered seats. Of these 15 people, 9 are actors and 6 are backstage workers.
Find the probability that:
(a) An actor and backstage worker will receive tickets sitting together.
(b) The actors will be seated together, and the backstage workers will be seated together.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~~~
- - - - - To me, this problem and this case are EXCEPTIONALLY INTERESTING. - - - - - - -
- - - - - - - - - - - - - - - - - - Below I will explain WHY. - - - - - - - - - - - - - - - - - - - - - - - -
Question (a), if to read it literally as it is written in the post, asks to find
the probability that some of 9 actors sits next to some of 6 backstage workers.
The answer is OBVIOUS: this probability is 100%.
Indeed, having 15 seats in a row, 9 actors and 6 backstage workers, it is IMPOSSIBLE to place
them in a way that no actor sits next to some backstage worker.
The answer is at the level of common sense and at the level accessible to a kindergarten child.
@CPhill, which is a synonym to Artificial Intelligence, in his post starts create
irrelevant reasoning and CAN NOT produce right answer.
From it, I conclude that this AI is not able to make
OBVIOUS elementary basic logical implications, at all.
For me, it is 100% proof that this concrete AI, at its stage of development,
is not able to think, at all, and does not possess elementary common sense
even of the level of a kindergarten child.
It is WHY this case and this post by @CPhill are interesting to me.
It is why this case should be interesting to you, my reader.
- - - - - - - - Now I will return to the problem itself - - - - - - - -
- - - - - - - - - - - - - - - - Part (a) - - - - - - - - - - - - - - - -
I don't know, WHY question (a) in this post is posed this way.
It may be a Math joke. It may be a TRAP.
It may be, that the author/(the creator or the visitor) was intended to ask what is the probability
that some PARTICULAR actor A will sit next to some PARTICULAR backstage worker B.
If so, then the solution is as follows:
Then we consider this pair (A,B) as one unit, and we have then 14 possible positions for this unit
in the row of 15 seats. We also have 15! permutations for 15 persons, in all,
and 2 possible permutations inside the pair (A,B).
Therefore, the probability that particular A and particular B sit together is
P =  =  =  = 2.14121E-11.
- - - - - - - - - - - - - - - - Part (b) - - - - - - - - - - - - - - - -
Question (b) asks to find the probability that the actors seat together, and the backstage workers seat together.
Then we have a compact group of 9 actors with 9! permutations inside this group,
and a compact group of 6 backstage workers with 6! permutations inside this group.
The answer to question (b) is
P =  =  =  = 0.0003996.
The factor ' 2 ' in the numerator is to account for two different possible placements
of the group of actors and the group of backstage workers.
Solved.
/////////////////////////////////////
This problem, in its part (a), was a crucial TEST for the AI,
and the AI  this test.
---------------------------------
This story has interesting continuation.
On September 2, 2025, I posted this problem to the Google AI, again.
This time, it produced correct, perfect and ideal solution.
Thanks to god, this AI is able to learn from the Internet.
It shows how it is important for this AI to have an adequate qualified tutor/teacher/mentor.
Question 1210344: A girl passed an exam is 0.8 find the probability of failing the exam
Found 2 solutions by MathLover1, ikleyn:
Answer by MathLover1(20849)  (Show Source):
Answer by ikleyn(52780) (Show Source):
Question 1177192: Show that two events A and B are independent if and only if P(A ∩ B) = P(A)P(B) when
at least one of P(A) or P(B) is not zero
thank you :)
Answer by ikleyn(52780) (Show Source):
You can put this solution on YOUR website! .
The problem's formulation is INCORRECT.
The DEFINITION of independence of two events is THIS:
Two events, A and B, are called independent if and only if
P(A intersection B) = P(A)*P(B)
without any other restrictions on P(A) and/or P(B).
As you brought this problem to the forum as it is in the post,
it means that you copy-pasted it from a bad source, which is anti-pedagogic.
Question 1175333: Take a 5 digit random numbers of size 50, sum each 5 digit random numbers and consider them as a random variable. Make a probability distribution and find its mean and variance.
Answer by ikleyn(52780) (Show Source):
You can put this solution on YOUR website! .
Take a 5 digit random numbers of size 50, sum each 5 digit random numbers and consider them
as a random variable. Make a probability distribution and find its mean and variance.
~~~~~~~~~~~~~~~~~~~~~~~~~~~~
Mathematically, this post is soup of words with no sense.
First, what is "5 digit random numbers of size 50" ?
Just this one sentence is a wild gibberish, which may to kill a reader to a death.
Next, it says "sum each 5 digit random numbers".
Mathematically, this statement/instruction/sentence is illiterate: its meaning is dark.
What I should add: - the digits of the numbers? - or the numbers themselves ?
Then it should be written "add all 5-digit random numbers", not "sum each 5 digit random numbers."
The authors' writing is extremely illiterate.
The right place of this composition is in the garbage bin.
To the author, I would recommend to take medicine regularly,
to sleep enough and to walk enough, and to forget about mathematical writing.
//////////////////////////////////////////////////
To the managers of this project
Please replace this writer, since he/she is not able to present his/her assignment properly.
Question 1176351: During a 52-week period, a company paid overtime wages for 18 weeks and hired temporary help for 9 weeks. During 5 weeks, the company paid overtime and hired temporary help. Let: O = overtime, T = temporary help.
a. What is the probability P(O OR T) ?
Round your answer to the nearest hundredth.
b. Are O and T independent?
c. Are O and T mutually exclusive?
Answer by ikleyn(52780) (Show Source):
You can put this solution on YOUR website! .
As the problem is worded, printed and presented in the post, it makes no sense mathematically.
In other words, it is soup of words with no sense.
The " solution " by @CPhill in his post is nonsense raised in degree 2.
Question 1179602: If 35% of the people in a community use the emergency room at a hospital one year. Find these probabilities for a sample of 12 people A) exactly 4 used the emergency room B) At most 4 used the emergency room C) At least 10 used the emergency room
Answer by ikleyn(52780) (Show Source):
You can put this solution on YOUR website! .
All calculation in the post by @CPhill related to parts (b) and (c) of this problem are INCORRECT.
\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\\
Regarding the post by @CPhill . . .
Keep in mind that @CPhill is a pseudonym for the Google artificial intelligence.
The artificial intelligence is like a baby now. It is in the experimental stage
of development and can make mistakes and produce nonsense without any embarrassment.
It has no feeling of shame - it is shameless.
This time, again, it made an error.
Although the @CPhill' solutions are copy-paste Google AI solutions, there is one essential difference.
Every time, Google AI makes a note at the end of its solutions that Google AI is experimental
and can make errors/mistakes.
All @CPhill' solutions are copy-paste of Google AI solutions, with one difference:
@PChill never makes this notice and never says that his solutions are copy-past that of Google.
So, he NEVER SAYS TRUTH.
Every time, @CPhill embarrassed to tell the truth.
But I am not embarrassing to tell the truth, as it is my duty at this forum.
And the last my comment.
When you obtain such posts from @CPhill, remember, that NOBODY is responsible for their correctness,
until the specialists and experts will check and confirm their correctness.
Without it, their reliability is ZERO and their creadability is ZERO, too.
Question 1210330: A grocer's bin contains 7 ripe tomatoes and 9 unripened tomatoes. Three tomatoes are selected at random (Tomatoes are not replaced). Show the probability distribution for the number of ripe tomatoes.
Answer by math_tutor2020(3816) (Show Source):
You can put this solution on YOUR website!
The other tutors are using the combination formula.
I'll show an example calculation.
C(n,r) = (n!)/(r!(n-r)!)
C(7,3) = (7!)/(3!*(7-3)!)
C(7,3) = (7!)/(3!*4!)
C(7,3) = (7*6*5*4!)/(3!*4!)
C(7,3) = (7*6*5)/(3!)
C(7,3) = (7*6*5)/(3*2*1)
C(7,3) = 210/6
C(7,3) = 35
This indicates there are 35 ways to select 3 ripe tomatoes from a pool of 7 ripe tomatoes, where order doesn't matter. This assumes that we can distinguish the tomatoes apart.
C(7,3) = 35 is equivalent to saying 7C3 = 35.
Note that 35 is found in Pascal's Triangle. Locate the row that starts with "1,7,..."
In that row, move 3 spots to the right and you should land on 35. Specifically you should land on the first copy of 35.
n = 7 tells us which row to look at, while r = 3 tells us how far to move to the right.
In GeoGebra, you would type ncr(7,3) to compute the combination value.
Many other online calculators can be used as well. Or you can use a spreadsheet.
If you want to use Desmos, then you would input nCr(7,3) where the C must be uppercase and everything else lowercase.
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Older solutions: 1..45, 46..90, 91..135, 136..180, 181..225, 226..270, 271..315, 316..360, 361..405, 406..450, 451..495, 496..540, 541..585, 586..630, 631..675, 676..720, 721..765, 766..810, 811..855, 856..900, 901..945, 946..990, 991..1035, 1036..1080, 1081..1125, 1126..1170, 1171..1215, 1216..1260, 1261..1305, 1306..1350, 1351..1395, 1396..1440, 1441..1485, 1486..1530, 1531..1575, 1576..1620, 1621..1665, 1666..1710, 1711..1755, 1756..1800, 1801..1845, 1846..1890, 1891..1935, 1936..1980, 1981..2025, 2026..2070, 2071..2115, 2116..2160, 2161..2205, 2206..2250, 2251..2295, 2296..2340, 2341..2385, 2386..2430, 2431..2475, 2476..2520, 2521..2565, 2566..2610, 2611..2655, 2656..2700, 2701..2745, 2746..2790, 2791..2835, 2836..2880, 2881..2925, 2926..2970, 2971..3015, 3016..3060, 3061..3105, 3106..3150, 3151..3195, 3196..3240, 3241..3285, 3286..3330, 3331..3375, 3376..3420, 3421..3465, 3466..3510, 3511..3555, 3556..3600, 3601..3645, 3646..3690, 3691..3735, 3736..3780, 3781..3825, 3826..3870, 3871..3915, 3916..3960, 3961..4005, 4006..4050, 4051..4095, 4096..4140, 4141..4185, 4186..4230, 4231..4275, 4276..4320, 4321..4365, 4366..4410, 4411..4455, 4456..4500, 4501..4545, 4546..4590, 4591..4635, 4636..4680, 4681..4725, 4726..4770, 4771..4815, 4816..4860, 4861..4905, 4906..4950, 4951..4995, 4996..5040, 5041..5085, 5086..5130, 5131..5175, 5176..5220, 5221..5265, 5266..5310, 5311..5355, 5356..5400, 5401..5445, 5446..5490, 5491..5535, 5536..5580, 5581..5625, 5626..5670, 5671..5715, 5716..5760, 5761..5805, 5806..5850, 5851..5895, 5896..5940, 5941..5985, 5986..6030, 6031..6075, 6076..6120, 6121..6165, 6166..6210, 6211..6255, 6256..6300, 6301..6345, 6346..6390, 6391..6435, 6436..6480, 6481..6525, 6526..6570, 6571..6615, 6616..6660, 6661..6705, 6706..6750, 6751..6795, 6796..6840, 6841..6885, 6886..6930, 6931..6975, 6976..7020, 7021..7065, 7066..7110, 7111..7155, 7156..7200, 7201..7245, 7246..7290, 7291..7335, 7336..7380, 7381..7425, 7426..7470, 7471..7515, 7516..7560, 7561..7605, 7606..7650, 7651..7695, 7696..7740, 7741..7785, 7786..7830, 7831..7875, 7876..7920, 7921..7965, 7966..8010, 8011..8055, 8056..8100, 8101..8145, 8146..8190, 8191..8235, 8236..8280, 8281..8325, 8326..8370, 8371..8415, 8416..8460, 8461..8505, 8506..8550, 8551..8595, 8596..8640, 8641..8685, 8686..8730, 8731..8775, 8776..8820, 8821..8865, 8866..8910, 8911..8955, 8956..9000, 9001..9045, 9046..9090, 9091..9135, 9136..9180, 9181..9225, 9226..9270, 9271..9315, 9316..9360, 9361..9405, 9406..9450, 9451..9495, 9496..9540, 9541..9585, 9586..9630, 9631..9675, 9676..9720, 9721..9765, 9766..9810, 9811..9855, 9856..9900, 9901..9945, 9946..9990, 9991..10035, 10036..10080, 10081..10125, 10126..10170, 10171..10215, 10216..10260, 10261..10305, 10306..10350, 10351..10395, 10396..10440, 10441..10485, 10486..10530, 10531..10575, 10576..10620, 10621..10665, 10666..10710, 10711..10755, 10756..10800, 10801..10845, 10846..10890, 10891..10935, 10936..10980, 10981..11025, 11026..11070, 11071..11115, 11116..11160, 11161..11205, 11206..11250, 11251..11295, 11296..11340, 11341..11385, 11386..11430, 11431..11475, 11476..11520, 11521..11565, 11566..11610, 11611..11655, 11656..11700, 11701..11745, 11746..11790, 11791..11835, 11836..11880, 11881..11925, 11926..11970, 11971..12015, 12016..12060, 12061..12105, 12106..12150, 12151..12195, 12196..12240, 12241..12285, 12286..12330, 12331..12375, 12376..12420, 12421..12465, 12466..12510, 12511..12555, 12556..12600, 12601..12645, 12646..12690, 12691..12735, 12736..12780, 12781..12825, 12826..12870, 12871..12915, 12916..12960, 12961..13005, 13006..13050, 13051..13095, 13096..13140, 13141..13185, 13186..13230, 13231..13275, 13276..13320, 13321..13365, 13366..13410, 13411..13455, 13456..13500, 13501..13545, 13546..13590, 13591..13635, 13636..13680, 13681..13725, 13726..13770, 13771..13815, 13816..13860, 13861..13905, 13906..13950, 13951..13995, 13996..14040, 14041..14085, 14086..14130, 14131..14175, 14176..14220, 14221..14265, 14266..14310, 14311..14355, 14356..14400, 14401..14445, 14446..14490, 14491..14535, 14536..14580, 14581..14625, 14626..14670, 14671..14715, 14716..14760, 14761..14805, 14806..14850, 14851..14895, 14896..14940, 14941..14985, 14986..15030, 15031..15075, 15076..15120, 15121..15165, 15166..15210, 15211..15255, 15256..15300, 15301..15345, 15346..15390, 15391..15435, 15436..15480, 15481..15525, 15526..15570, 15571..15615, 15616..15660, 15661..15705, 15706..15750, 15751..15795, 15796..15840, 15841..15885, 15886..15930, 15931..15975, 15976..16020, 16021..16065, 16066..16110, 16111..16155, 16156..16200, 16201..16245, 16246..16290, 16291..16335, 16336..16380, 16381..16425, 16426..16470, 16471..16515, 16516..16560, 16561..16605, 16606..16650, 16651..16695, 16696..16740, 16741..16785, 16786..16830, 16831..16875, 16876..16920, 16921..16965, 16966..17010, 17011..17055, 17056..17100, 17101..17145, 17146..17190, 17191..17235, 17236..17280, 17281..17325, 17326..17370, 17371..17415, 17416..17460, 17461..17505, 17506..17550, 17551..17595, 17596..17640, 17641..17685, 17686..17730, 17731..17775, 17776..17820, 17821..17865, 17866..17910, 17911..17955, 17956..18000, 18001..18045, 18046..18090, 18091..18135, 18136..18180, 18181..18225, 18226..18270, 18271..18315, 18316..18360, 18361..18405, 18406..18450, 18451..18495, 18496..18540, 18541..18585, 18586..18630, 18631..18675, 18676..18720, 18721..18765, 18766..18810, 18811..18855, 18856..18900, 18901..18945, 18946..18990, 18991..19035, 19036..19080, 19081..19125, 19126..19170, 19171..19215, 19216..19260, 19261..19305, 19306..19350, 19351..19395, 19396..19440, 19441..19485, 19486..19530, 19531..19575, 19576..19620, 19621..19665, 19666..19710, 19711..19755, 19756..19800, 19801..19845, 19846..19890, 19891..19935, 19936..19980, 19981..20025, 20026..20070, 20071..20115, 20116..20160, 20161..20205, 20206..20250, 20251..20295, 20296..20340, 20341..20385, 20386..20430, 20431..20475, 20476..20520, 20521..20565, 20566..20610, 20611..20655, 20656..20700, 20701..20745, 20746..20790, 20791..20835, 20836..20880, 20881..20925, 20926..20970, 20971..21015, 21016..21060, 21061..21105, 21106..21150, 21151..21195, 21196..21240, 21241..21285, 21286..21330, 21331..21375, 21376..21420, 21421..21465, 21466..21510, 21511..21555, 21556..21600, 21601..21645, 21646..21690, 21691..21735, 21736..21780, 21781..21825, 21826..21870, 21871..21915, 21916..21960, 21961..22005, 22006..22050, 22051..22095, 22096..22140, 22141..22185, 22186..22230, 22231..22275, 22276..22320, 22321..22365, 22366..22410, 22411..22455, 22456..22500, 22501..22545, 22546..22590, 22591..22635, 22636..22680, 22681..22725, 22726..22770, 22771..22815, 22816..22860, 22861..22905, 22906..22950, 22951..22995, 22996..23040, 23041..23085, 23086..23130, 23131..23175, 23176..23220, 23221..23265, 23266..23310, 23311..23355, 23356..23400, 23401..23445, 23446..23490, 23491..23535, 23536..23580, 23581..23625, 23626..23670, 23671..23715, 23716..23760, 23761..23805, 23806..23850, 23851..23895, 23896..23940, 23941..23985, 23986..24030, 24031..24075, 24076..24120, 24121..24165, 24166..24210, 24211..24255, 24256..24300, 24301..24345, 24346..24390, 24391..24435, 24436..24480, 24481..24525, 24526..24570, 24571..24615, 24616..24660, 24661..24705, 24706..24750, 24751..24795, 24796..24840, 24841..24885, 24886..24930, 24931..24975, 24976..25020, 25021..25065, 25066..25110, 25111..25155, 25156..25200, 25201..25245, 25246..25290, 25291..25335, 25336..25380, 25381..25425, 25426..25470, 25471..25515, 25516..25560, 25561..25605, 25606..25650, 25651..25695, 25696..25740, 25741..25785, 25786..25830, 25831..25875, 25876..25920, 25921..25965, 25966..26010, 26011..26055, 26056..26100, 26101..26145, 26146..26190, 26191..26235, 26236..26280, 26281..26325, 26326..26370, 26371..26415, 26416..26460, 26461..26505, 26506..26550, 26551..26595, 26596..26640, 26641..26685, 26686..26730, 26731..26775, 26776..26820, 26821..26865, 26866..26910, 26911..26955, 26956..27000, 27001..27045, 27046..27090, 27091..27135, 27136..27180, 27181..27225, 27226..27270, 27271..27315, 27316..27360, 27361..27405, 27406..27450, 27451..27495, 27496..27540, 27541..27585, 27586..27630, 27631..27675, 27676..27720, 27721..27765, 27766..27810, 27811..27855, 27856..27900, 27901..27945, 27946..27990, 27991..28035, 28036..28080, 28081..28125, 28126..28170, 28171..28215, 28216..28260, 28261..28305, 28306..28350, 28351..28395, 28396..28440, 28441..28485, 28486..28530, 28531..28575, 28576..28620, 28621..28665, 28666..28710, 28711..28755, 28756..28800, 28801..28845, 28846..28890, 28891..28935, 28936..28980, 28981..29025, 29026..29070, 29071..29115, 29116..29160, 29161..29205, 29206..29250, 29251..29295, 29296..29340, 29341..29385, 29386..29430, 29431..29475, 29476..29520, 29521..29565, 29566..29610, 29611..29655, 29656..29700, 29701..29745, 29746..29790, 29791..29835, 29836..29880, 29881..29925, 29926..29970, 29971..30015, 30016..30060, 30061..30105, 30106..30150, 30151..30195, 30196..30240, 30241..30285, 30286..30330, 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