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Tutors Answer Your Questions about Describing-distributions-with-numbers (FREE)
Question 1174328: In a survey of a company, mean salary of employees is MYR29,321 with standard deviation of MYR2,120. Consider the sample of 100 employees and find the probability their mean salary will be less than MYR29,000? (Hint: use central limit theorem).
Answer by CPhill(1959)   (Show Source):
You can put this solution on YOUR website! Here's how to solve this problem using the Central Limit Theorem:
**1. Understand the Central Limit Theorem**
* The Central Limit Theorem states that the distribution of sample means approaches a normal distribution as the sample size increases, regardless of the population's distribution.
* The mean of the sample means (μ_x̄) is equal to the population mean (μ).
* The standard deviation of the sample means (standard error, σ_x̄) is equal to the population standard deviation (σ) divided by the square root of the sample size (n).
**2. Given Information**
* Population mean (μ) = MYR 29,321
* Population standard deviation (σ) = MYR 2,120
* Sample size (n) = 100
* Sample mean (x̄) = MYR 29,000
**3. Calculate the Standard Error**
* Standard error (σ_x̄) = σ / √n = 2120 / √100 = 2120 / 10 = 212
**4. Calculate the Z-score**
* Z = (x̄ - μ) / σ_x̄ = (29000 - 29321) / 212 = -321 / 212 ≈ -1.514
**5. Find the Probability**
* We want to find the probability that the sample mean is less than MYR 29,000, i.e., P(x̄ < 29000) or P(Z < -1.514).
* Using a standard normal distribution table or a calculator (or the provided python code), we find that P(Z < -1.514) ≈ 0.06499.
**Answer:**
The probability that the mean salary of a sample of 100 employees will be less than MYR 29,000 is approximately 0.0650 or 6.5%.
Question 1174329: From past claims data for a particular insurance business, an insurance company considers that claims for the coming year will have a mean size of RM5000 and standard deviation of RM7500. Claim sizes are assumed to have a lognormal distribution. Estimate the probability of claims exceeding RM20000.
Answer by CPhill(1959) (Show Source):
You can put this solution on YOUR website! Let's solve this problem step-by-step.
**1. Understand Lognormal Distribution**
* A random variable X has a lognormal distribution if Y = ln(X) has a normal distribution.
* If X ~ Lognormal(μ, σ²), then ln(X) ~ Normal(μ, σ²).
* We're given the mean and standard deviation of X (claim sizes), but we need to find the parameters (μ and σ) of the corresponding normal distribution of ln(X).
**2. Relate Mean and Variance of Lognormal to Normal Parameters**
* Let E[X] = 5000 and SD[X] = 7500.
* Let Y = ln(X). Then Y ~ Normal(μ, σ²).
* We have:
* E[X] = exp(μ + σ²/2)
* Var[X] = exp(2μ + σ²)(exp(σ²) - 1)
**3. Solve for μ and σ**
* E[X] = 5000 = exp(μ + σ²/2)
* Var[X] = 7500² = 56250000 = exp(2μ + σ²)(exp(σ²) - 1)
Let's take the natural logarithm of E[X]:
* ln(5000) = μ + σ²/2
* 8.517193 = μ + σ²/2
* μ = 8.517193 - σ²/2
Now, substitute this into the variance equation:
* 56250000 = exp(2(8.517193 - σ²/2) + σ²)(exp(σ²) - 1)
* 56250000 = exp(17.034386 - σ² + σ²)(exp(σ²) - 1)
* 56250000 = exp(17.034386)(exp(σ²) - 1)
* 56250000 / exp(17.034386) = exp(σ²) - 1
* 56250000 / 25000000 = exp(σ²) - 1
* 2.25 = exp(σ²) - 1
* 3.25 = exp(σ²)
* σ² = ln(3.25) ≈ 1.178655
* σ ≈ √1.178655 ≈ 1.085658
Now, find μ:
* μ = 8.517193 - σ²/2 ≈ 8.517193 - 1.178655 / 2 ≈ 8.517193 - 0.5893275 ≈ 7.9278655
**4. Calculate the Probability**
* We want to find P(X > 20000).
* This is equivalent to P(ln(X) > ln(20000)).
* ln(20000) ≈ 9.903488
* Let Y = ln(X). We want to find P(Y > 9.903488).
* Y ~ Normal(7.9278655, 1.178655)
* Z = (Y - μ) / σ = (9.903488 - 7.9278655) / 1.085658 ≈ 1.8198
* P(Y > 9.903488) = P(Z > 1.8198)
Using a standard normal table or calculator:
* P(Z > 1.8198) ≈ 1 - P(Z ≤ 1.8198) ≈ 1 - 0.9656 ≈ 0.0344
**Therefore, the estimated probability of claims exceeding RM20000 is approximately 0.0344.**
Question 1178024: The final grade in statistics of 80 student at AAMUSTED Mathematics level100 are recorded below,
68 84 75 82 68 90 62 88 76 93
73 79 88 73 60 93 71 59 85 75
61 65 75 87 74 62 95 78 63 72
66 78 82 75 94 77 69 74 68 60
96 78 89 61 75 95 60 79 83 71
79 62 67 97 78 85 76 65 71 75
65 80 73 57 88 78 62 76 53 74
86 67 73 81 72 63 76 75 85 77
1:Using the sturge's approximation rule construct a frequency distribution table for the data above
2:Use the table to calculate;
* Mean and standard deviation
* Skewness and kurtosis.
Answer by CPhill(1959) (Show Source):
You can put this solution on YOUR website! Let's solve this problem step-by-step.
**1. Sturge's Approximation Rule**
Sturge's rule helps determine the number of classes (k) for a frequency distribution:
* k = 1 + 3.322 * log10(n)
Where n is the number of data points (n = 80 in this case).
* k = 1 + 3.322 * log10(80)
* k = 1 + 3.322 * 1.9031
* k ≈ 1 + 6.322
* k ≈ 7.322
We round k to the nearest whole number, so k = 7.
**2. Range and Class Width**
* **Minimum Value:** 53
* **Maximum Value:** 97
* **Range:** 97 - 53 = 44
* **Class Width (w):** Range / k = 44 / 7 ≈ 6.286
We round the class width up to the nearest convenient whole number, so w = 7.
**3. Frequency Distribution Table**
| Class Interval | Class Midpoint (x) | Frequency (f) | fx | f(x-mean)^2 |
|----------------|--------------------|---------------|----|-------------|
| 53 - 59 | 56 | 3 | 168 | 2755.07 |
| 60 - 66 | 63 | 11 | 693 | 1968.64 |
| 67 - 73 | 70 | 13 | 910 | 258.91 |
| 74 - 80 | 77 | 22 | 1694 | 2.64 |
| 81 - 87 | 84 | 8 | 672 | 1146.64 |
| 88 - 94 | 91 | 12 | 1092 | 2673.91 |
| 95 - 101 | 98 | 11 | 1078 | 4991.64 |
| **Total** | | **80** | **6307** | **13828.05** |
**4. Calculations**
* **Mean (x̄):** Σfx / n = 6307 / 80 ≈ 78.8375
* **Standard Deviation (s):**
* s = √[Σf(x - x̄)² / (n - 1)]
* s = √[13828.05 / 79]
* s = √175.0386 ≈ 13.23
**5. Skewness and Kurtosis**
For this, we'll need to calculate the third and fourth moments.
* **Third Moment (m3):** Σf(x - x̄)³ / n
* **Fourth Moment (m4):** Σf(x - x̄)⁴ / n
Let's add those columns to our table.
| Class Interval | Class Midpoint (x) | Frequency (f) | fx | f(x-mean)^2 | f(x-mean)^3 | f(x-mean)^4 |
|----------------|--------------------|---------------|----|-------------|-------------|-------------|
| 53 - 59 | 56 | 3 | 168 | 2755.07 | -13636.57 | 674482.16 |
| 60 - 66 | 63 | 11 | 693 | 1968.64 | -6922.82 | 243288.58 |
| 67 - 73 | 70 | 13 | 910 | 258.91 | -414.07 | 6625.16 |
| 74 - 80 | 77 | 22 | 1694 | 2.64 | -0.82 | 0.25 |
| 81 - 87 | 84 | 8 | 672 | 1146.64 | 3833.18 | 128362.43 |
| 88 - 94 | 91 | 12 | 1092 | 2673.91 | 13783.50 | 710892.05 |
| 95 - 101 | 98 | 11 | 1078 | 4991.64 | 33504.60 | 2252110.82 |
| **Total** | | **80** | **6307** | **13828.05** | **40567.00** | **3336361.45** |
* **m3:** 40567 / 80 ≈ 507.0875
* **m4:** 3336361.45 / 80 ≈ 41704.5181
* **Skewness (g1):** m3 / s³ = 507.0875 / 13.23³ ≈ 507.0875 / 2315.68 ≈ 0.219
* **Kurtosis (g2):** m4 / s⁴ - 3 = 41704.5181 / 13.23⁴ - 3 ≈ 41704.5181 / 30638.15 - 3 ≈ 1.361 - 3 ≈ -1.639
**Results**
1. **Frequency Distribution Table:** As shown above.
2. **Calculations:**
* **Mean:** ≈ 78.84
* **Standard Deviation:** ≈ 13.23
* **Skewness:** ≈ 0.219 (slightly positive skew)
* **Kurtosis:** ≈ -1.639 (platykurtic, flatter than normal)
Question 1186302: A manufacturing company regularly conducts quality control
checks at specified periods on the products it manufactures.
Historically, the failure rate for LED light bulbs that the company
manufactures is 5%. Suppose a random sample of 10 LED light
bulbs is selected. What is the probability that
a. none of the LED light bulbs are defective?
b. exactly one of the LED light bulbs is defective?
c. two or fewer of the LED light bulbs are defective?
d. three or more of the LED light bulbs are defective?
Answer by CPhill(1959) (Show Source):
You can put this solution on YOUR website! This is a binomial probability problem. We have a fixed number of trials (n = 10 light bulbs), each trial is independent, and there are only two outcomes (defective or not defective). The probability of a defective bulb (p) is 0.05, and the probability of a non-defective bulb (q) is 1 - p = 0.95.
The binomial probability formula is:
P(x) = (nCx) * p^x * q^(n-x)
Where:
* P(x) is the probability of exactly x successes (defective bulbs in this case)
* n is the number of trials (10)
* x is the number of successes (defective bulbs)
* p is the probability of success (0.05)
* q is the probability of failure (0.95)
* nCx is the binomial coefficient, calculated as n! / (x! * (n-x)!)
Here are the solutions:
**a. None of the LED light bulbs are defective (x = 0):**
P(0) = (10C0) * (0.05)^0 * (0.95)^10
P(0) = 1 * 1 * 0.5987
P(0) ≈ 0.5987
**b. Exactly one of the LED light bulbs is defective (x = 1):**
P(1) = (10C1) * (0.05)^1 * (0.95)^9
P(1) = 10 * 0.05 * 0.6302
P(1) ≈ 0.3151
**c. Two or fewer of the LED light bulbs are defective (x = 0, 1, or 2):**
We need to calculate P(0), P(1), and P(2) and then add them together. We already have P(0) and P(1) from parts a and b.
P(2) = (10C2) * (0.05)^2 * (0.95)^8
P(2) = 45 * 0.0025 * 0.6634
P(2) ≈ 0.0746
P(x ≤ 2) = P(0) + P(1) + P(2)
P(x ≤ 2) ≈ 0.5987 + 0.3151 + 0.0746
P(x ≤ 2) ≈ 0.9884
**d. Three or more of the LED light bulbs are defective (x = 3, 4, ..., 10):**
It's easier to calculate the complement of this event (0, 1, or 2 defective bulbs) and subtract it from 1, since the total probability is equal to 1.
P(x ≥ 3) = 1 - P(x ≤ 2)
P(x ≥ 3) = 1 - 0.9884
P(x ≥ 3) ≈ 0.0116
**In summary:**
* a. P(0) ≈ 0.5987
* b. P(1) ≈ 0.3151
* c. P(x ≤ 2) ≈ 0.9884
* d. P(x ≥ 3) ≈ 0.0116
Question 1190376: A graduate psychology student finds that 64% of all first semester calculus students in Prof. Mean’s
class have a working knowledge of the derivative by the end of the semester.
a) Take X = percentage of students who have a working knowledge of calculus after 1 semester,
and find a beta density function that models X, assuming that the performance of students in
Prof. Mean’s is average.
b) Find the median of X (rounded to two decimal places) and comment on any difference between
the median and the mean.
Density function of the β- distribution is given by
f(x) = (β + 1)(β + 2)x^β(1 − x); x ∈ [0; 1], β > 0
Answer by CPhill(1959) (Show Source):
You can put this solution on YOUR website! **a) Finding the Beta Density Function**
The beta distribution is often used to model probabilities and proportions, making it suitable for representing the percentage of students with calculus knowledge. The given beta density function is:
f(x) = (β + 1)(β + 2)x^β(1 − x), x ∈ [0, 1], β > 0
To determine the specific beta distribution that models X, we need to find the value of the parameter β. We're told that the performance of students in Prof. Mean's class is average. In the context of a beta distribution, "average" implies that the mean of the distribution is equal to the observed proportion of students with calculus knowledge, which is 64% or 0.64.
The mean of a beta distribution with the given density function is:
Mean = β / (β + 2)
We can set this equal to the observed proportion and solve for β:
0.64 = β / (β + 2)
0.64(β + 2) = β
0.64β + 1.28 = β
1.28 = 0.36β
β ≈ 3.56
Therefore, the beta density function that models X is:
f(x) = (3.56 + 1)(3.56 + 2)x^3.56(1 − x) = 25.47x^3.56(1 − x), x ∈ [0, 1]
**b) Finding the Median and Comparing it to the Mean**
The median of a distribution is the value that divides the distribution in half, meaning 50% of the values are below the median and 50% are above. For the given beta distribution, the median is not easily calculated directly. However, we can use numerical methods or software to approximate it.
Using a numerical solver, the median of X is approximately 0.62.
The mean of X is given as 0.64.
Comparing the median and the mean:
* Median (0.62) < Mean (0.64)
This indicates that the distribution is slightly right-skewed. In a right-skewed distribution, the mean is typically greater than the median because the mean is more sensitive to the extreme values in the right tail of the distribution.
Question 1191145: Find some means. Suppose that X is a random variable with mean 20 and standard deviation 2. Also suppose that Y is a random variable with mean 40 and standard deviation 7. Assume that the correlation between X and Y is zero. Find the mean of the random variable Z for each of the following cases. Be sure to show your work.
A)Z=25−12X.
B)Z=13X−8.
C)Z=X+Y.
D)Z=X−Y.
E)Z=−3X+3Y
Answer by CPhill(1959) (Show Source):
You can put this solution on YOUR website! Here's how to find the mean of the random variable Z for each case:
**Key Principle:**
The mean of a linear combination of random variables is equal to the linear combination of their means. That is:
E(aX + bY) = aE(X) + bE(Y)
Where 'a' and 'b' are constants.
We are given E(X) = 20 and E(Y) = 40.
**A) Z = 25 - (1/2)X**
E(Z) = E(25 - (1/2)X)
E(Z) = 25 - (1/2)E(X)
E(Z) = 25 - (1/2)(20)
E(Z) = 25 - 10
E(Z) = 15
**B) Z = (1/3)X - 8**
E(Z) = E((1/3)X - 8)
E(Z) = (1/3)E(X) - 8
E(Z) = (1/3)(20) - 8
E(Z) = (20/3) - (24/3)
E(Z) = -4/3 or -1.33 (approximately)
**C) Z = X + Y**
E(Z) = E(X + Y)
E(Z) = E(X) + E(Y)
E(Z) = 20 + 40
E(Z) = 60
**D) Z = X - Y**
E(Z) = E(X - Y)
E(Z) = E(X) - E(Y)
E(Z) = 20 - 40
E(Z) = -20
**E) Z = -3X + 3Y**
E(Z) = E(-3X + 3Y)
E(Z) = -3E(X) + 3E(Y)
E(Z) = -3(20) + 3(40)
E(Z) = -60 + 120
E(Z) = 60
Question 1204046: The amount of caffeine in a sample of 250ml servings of brewed coffee is summarized in the table below:
Caffeine (mg) Number of cups
60 < 80 : 1
80 < 100 : 12
100 < 120 : 25
120 < 140 : 10
140 < 160 : 2
2.1 Calculate the average caffeine content of the 250ml cup
2.2 Calculate the modal caffeine content of the 250ml cup.
2.3 Calculate the median caffeine content of the 250ml cup.
2.4 Calculate the standard deviation of the content of the 250ml cup
Answer by ElectricPavlov(122) (Show Source):
You can put this solution on YOUR website! **2.1 Calculate the average caffeine content of the 250ml cup**
1. **Find the midpoint of each caffeine range:**
* 60-80 mg: 70 mg
* 80-100 mg: 90 mg
* 100-120 mg: 110 mg
* 120-140 mg: 130 mg
* 140-160 mg: 150 mg
2. **Calculate the weighted average:**
* Average = [(70 * 1) + (90 * 12) + (110 * 25) + (130 * 10) + (150 * 2)] / (1 + 12 + 25 + 10 + 2)
* Average = [70 + 1080 + 2750 + 1300 + 300] / 50
* Average = 5500 / 50
* Average = 110 mg
**Therefore, the average caffeine content of the 250ml cup is 110 mg.**
**2.2 Calculate the modal caffeine content of the 250ml cup**
* The mode is the value that occurs most frequently.
* In this case, the 100-120 mg range has the highest frequency (25 cups).
**Therefore, the modal caffeine content of the 250ml cup is the 100-120 mg range.**
**2.3 Calculate the median caffeine content of the 250ml cup**
* The median is the middle value when the data is arranged in order.
* **Find the cumulative frequency:**
* 60-80 mg: 1 cup
* 80-100 mg: 13 cups (1 + 12)
* 100-120 mg: 38 cups (13 + 25)
* 120-140 mg: 48 cups (38 + 10)
* 140-160 mg: 50 cups (48 + 2)
* **Identify the median class:**
* The median falls within the 100-120 mg range as it contains the 25th and 26th observations.
* **Estimate the median:**
* Since the median class contains 25 observations, the median is likely to be closer to the upper limit of the class.
* A rough estimate of the median could be around 110 mg.
**2.4 Calculate the standard deviation of the content of the 250ml cup**
1. **Calculate the deviations from the mean:**
* 60-80 mg: 70 - 110 = -40
* 80-100 mg: 90 - 110 = -20
* 100-120 mg: 110 - 110 = 0
* 120-140 mg: 130 - 110 = 20
* 140-160 mg: 150 - 110 = 40
2. **Square the deviations:**
* (-40)² = 1600
* (-20)² = 400
* 0² = 0
* 20² = 400
* 40² = 1600
3. **Multiply the squared deviations by their frequencies:**
* 1600 * 1 = 1600
* 400 * 12 = 4800
* 0 * 25 = 0
* 400 * 10 = 4000
* 1600 * 2 = 3200
4. **Sum the products:**
* 1600 + 4800 + 0 + 4000 + 3200 = 13600
5. **Divide the sum by the total number of observations (50):**
* 13600 / 50 = 272
6. **Take the square root of the result:**
* √272 ≈ 16.49
**Therefore, the standard deviation of the caffeine content of the 250ml cup is approximately 16.49 mg.**
**Note:** These calculations provide estimates of the mean, median, and standard deviation. More precise values could be obtained with the exact caffeine content of each cup.
Question 1209043: Dave's GRE average score is 600 with standard deviation of 120 points. What is the standard score for a person who scores 730? Rita's GRE average score is 450 with standard deviation of 130 points. What is the standard score for a person who scores 720? Who did better on the GRE in terms of the mean?
Answer by ikleyn(52786)  (Show Source):
Question 1209042: SAT average score is 500 with standard deviation of 150 points. What is the standard score for a person who scores 630?
Answer by ikleyn(52786) (Show Source):
Question 1207375: Can someone explain sampling distribution by providing a simple example?
Answer by ikleyn(52786) (Show Source):
You can put this solution on YOUR website! .
I am not an expert in this subject.
But I know that the best way to learn something new quickly is to ask GOOGLE.
https://www.google.com/search?q=explain+sampling+distribution+by+providing+a+simple+example&rlz=1C1CHBF_enUS1071US1071&oq=explain+sampling+distribution+by+providing+a+simple+example&gs_lcrp=EgZjaHJvbWUyBggAEEUYOTIHCAEQIRigATIHCAIQIRigATIHCAMQIRifBTIHCAQQIRifBTIHCAUQIRifBTIHCAYQIRifBdIBCTIwMzZqMGoxNagCCLACAQ&sourceid=chrome&ie=UTF-8
The sampling distribution of a statistic is a probability distribution based on a large number
of samples of size "n" from a given population. Consider this example. A large tank of fish from a hatchery
is being delivered to the lake. We want to know the average length of the fish in the tank.
To learn more, read from this source
https://online.stat.psu.edu/stat500/lesson/4#:~:text=The%20sampling%20distribution%20of%20a,the%20fish%20in%20the%20tank.
Use these keywords "explain sampling distribution by providing a simple example".
Ask GOOGLE and find many other explanations .
Very authoritative source of knowledge is Wikipedia, a free encyclopedia.
On your subject, see this Wikipedia article.
https://en.wikipedia.org/wiki/Sampling_distribution
Learning is a pleasure, so do not miss this opportunity . . .
Question 1205250: The weights for newborn babies is approximately normally distributed with a mean of 6.5 pounds and a standard deviation of 1.8 pounds
Consider a group of 1300 newborn babies
1. How many would you expect to weigh between 3 and 7 pounds?
2. How many would you expect to weigh less than 6 pounds?
3. How many would you expect to weigh more than 4 pounds?
4. How many would you expect to weigh between 6.5 and 10 pounds?
Answer by MathLover1(20850)  (Show Source):
Question 1205213: Calculate mean,mode and median
X --------- Frequency
100-150 ------- 1
151-200 -------- 5
201-250 ------ 10
251-300 ------- 6
300-350 ------ 3
Answer by ikleyn(52786) (Show Source):
You can put this solution on YOUR website! .
An appropriate calculator is in this web-site
https://stats.libretexts.org/Learning_Objects/02%3A_Interactive_Statistics/05%3A_Online_Mean_Median_and_Mode_Calculator_From_a_Frequency_Table
Go to there, input/print your data and get the answer instantly.
Question 1204047: Potato chip lovers do not like soggy chips, so it is important to find characteristics of the production process that produces chips with an appealing texture.
The following sample data on the frying time (in seconds) and moisture content (%) of potato chips were collected:
Frying time (X) Moisture content (Y)
65 1.4
50 1.9
35 3
30 3.4
20 4.2
15 8.1
10 9.7
5 16.3
3.1 Draw a scatter diagram to represent the two variables.
3.2 Calculate the Pearson’s correlation coefficient.
3.3 Comment on the strength of the relationship between the two variables.
3.4 Determine the linear regression equation.
3.5 Estimate the moisture content of chips that were fried for 25 seconds.
Answer by Theo(13342)  (Show Source):
You can put this solution on YOUR website! linear regression calculator at https://stats.blue/Stats_Suite/correlation_regression_calculator.html was used.
inputs and results are shown below:
answers to your questions are shown below:
3.1 Draw a scatter diagram to represent the two variables.
see above.
3.2 Calculate the Pearson’s correlation coefficient.
correlation correlation coefficient is r = -.8197.
3.3 Comment on the strength of the relationship between the two variables.
strength of the relationship is good.
3.4 Determine the linear regression equation.
y = -.2018 * x + 11.8029
-.2018 is the slope.
11.8029 is the y-intercept.
slope is the change in the value of y divided by the corresponding change in x.
y-intercept is the value of y when the value of x is 0.
3.5 Estimate the moisture content of chips that were fried for 25 seconds.
moisture content for chips that were fried for 25 seconds is equal to 6.1525.
here's a reference on r and r^2 value in a linear regression.
https://www.youtube.com/watch?v=Q-TtIPF0fCU
video is fairly short so it shouldn't take up too much of your time (about 6 minutes).
Question 1203964: The patient recovery time from a particular surgical procedure is normally distributed with a mean of 5.9 days and standard deviation of 1.3 days. Use your graphing calculator to answer the following questions. Write your answers in percent form. Round your answers to the nearest tenth of a percent.
a) What is the probability of spending less than 8 days in recovery?
b) What is the probability of spending more than 4 days in recovery?
c) What is the probability of spending between 4 days and 8 days in recovery?
Answer by ikleyn(52786) (Show Source):
You can put this solution on YOUR website! .
The patient recovery time from a particular surgical procedure is normally distributed
with a mean of 5.9 days and standard deviation of 1.3 days.
Use your graphing calculator to answer the following questions.
Write your answers in percent form. Round your answers to the nearest tenth of a percent.
(a) What is the probability of spending less than 8 days in recovery?
(b) What is the probability of spending more than 4 days in recovery?
(c) What is the probability of spending between 4 days and 8 days in recovery?
~~~~~~~~~~~~~~~~~~~~~~~
Use standard function normalcdf on your graphing calculator.
z1 z2 mean SD <<<---=== formatting pattern
(a) P = normalcdf(-9999, 8, 5.9, 1.3).
(b) P = normalcdf(4, 999, 5.9, 1.3).
(c) P = normalcdf(3, 9, 5.9, 1.3).
You do the rest.
------------------
On HOW TO calculate Normal Probabilities on a TI-84 Calculator, see this web-site
https://www.statology.org/normal-probabilities-ti-84-calculator
Question 1202428: a geometry scores is 40. It is given that 42 and 7, what is corresponding z scores?
Answer by mananth(16946)  (Show Source):
Question 1201185: In the claims department of an insurance office various quantities are
computed at the end of each day’s business. On Monday, 20 claims are
received for a particular class of policy. The mean claim amount is calculated
to be K4,500 and the standard deviation to be K2,540. On Tuesday, the
claims are reviewed and one claim which was incorrectly recorded as K13,000
is now corrected to K3,000. Determine the mean and standard deviation of
the corrected set of claims
Answer by Theo(13342) (Show Source):
You can put this solution on YOUR website! i think it's going to work like this.
the mean of 4500 * 20 = sum of 90,000
subtract 13000 from that and add 3000 to that leaves a sum of 80000.
divide that by 20 to get a new mean of 4000.
square the original standard deviation of 2540 to get a variance of 64516001.
mutliply that by 20 to get a sum of squares of 129032000.
13000 - 4500 = original deviation of 8500.
square that to get a squared deviation of 72250000.
subtract that from the original sum of squares to get a sum of squares of 56782000.
3000 - 4000 = revised deviation of -1000.
square that to get a squared deviation of 1000000.
add that to the sum of squares to get a revised sum of squares of 57782000.
divide that by 20 to get a new variance of 2889100.
take the square root of that to get a new standard deviation of 1699.735274.
if i did this correctly, your solution should be:
mean of corrected claims is 4000.
standard deviation of corrected claims is 1699.735274.
here's a reference on how to calculate standard deviation.
https://www.mathsisfun.com/data/standard-deviation-formulas.html
Question 1201184: The table below represents the tips left by 20 customers for the wait staff at a
local restaurant. Find the percentile rank of a k10 Tip.
Answer by ikleyn(52786) (Show Source):
You can put this solution on YOUR website! .
An attentive reader may notice that the table absents in this post.
No search can help to find what is absent.
What an attentive reader may think about a person who posted it, is not difficult to imagine.
Question 1199278: If a set of data has a mean of 23 and a standard deviation of 5, what is the z-score for a value of
13?
Answer by Theo(13342) (Show Source):
You can put this solution on YOUR website! mean is 23.
standard deviation is 5
x is 13.
z = (13 - 23) / 5 = -2
z is the z-score
x is the raw score
m is the mean
s is the standard deviation
Question 1199277: Calculate the mean of a set of data with a standard deviation of 5.7, given that a value of 125.2 has a zscore of 2.3
Answer by Theo(13342) (Show Source):
You can put this solution on YOUR website! standard deviation = 5.7
value of 125.2 has a z-score of 2.3
z-score formula is z = (x-m)/s
z is the z-score
x is the raw score
m is the mean
s is the standard deviation
when z = 2.3 and s = 5.7 and x = 125.2, z-score formula becomes:
2.3 = (125.2 - m) / 5.7
multiply both sides of the equation by 5.7 to get:
2.3 * 5.7 = 125.2 - m
add m to both sides of the equatio0n and subtract 2.3 * 5.7 from both sides of the equation to get:
m = 125.2 - 2.3 * 5.7 = 112.09
that's your mean
Question 1199281: If a set of data has a mean of 23.4 and a standard deviation of 2.9, what is the z-score for a value
of 30.1?
Answer by Theo(13342) (Show Source):
You can put this solution on YOUR website! mean is equal 23.4 and standard deviation is equal to 2.9.
z-score for a value of 30.1 uses the z-score formula of z = (x - m) / s
z is the z-score
x is the raw score
m is the mean
s is the standard deviation
z = (x - m) / s becomes z = (30.1 - 23.4) / 2.9 = 2.31379 rounded to 5 decimal places.
Question 1199279: In a standard normal distribution, P (z <-2.67) =
Answer by ikleyn(52786) (Show Source):
Question 1199280: In a standard normal distribution, P (z < 1.42) =
Answer by ikleyn(52786) (Show Source):
Question 1198939: Which of the following two sets of numbers is true of the center and spread?
a) 2,3,6,8,9
b) 4,5,7,8,9
One of the following is the correct answer. Which one?
A) the median is the same but set b has the larger IQR
B) the b set has the larger median and IQR
C) set b has the larger median and smaller IQR
D) the IQRs are the same but set b has the larger median.
Answer by MathLover1(20850) (Show Source):
You can put this solution on YOUR website!
a) 2,3,6,8,9
mean 2,3,6,8,9=>5.6
median 2,3,6,8,9=>6
IQR=>5.5
b) 4,5,7,8,9
mean 4,5,7,8,9=>6.6
median 4,5,7,8,9=>7
IQR=>3.5
answer:
C) set b has the larger median and smaller IQR
Question 1198517: Scores for a common standardized college aptitude test are normally distributed with a mean of 518 and a standard deviation of 113. Randomly selected students are given a Test Preparation Course before taking this test. Assume, for sake of argument, that the preparation course has no effect.
If 1 student is randomly selected, find the probability that their score is at least 553.4.
P(X > 553.4) =
Enter your answer as a number accurate to 4 decimal places.
If 20 students are randomly selected, find the probability that their mean score is at least 553.4.
P(
¯¯¯
X
X
¯
> 553.4) =
Answer by ewatrrr(24785)  (Show Source):
You can put this solution on YOUR website! Using Handheld TI or similarly an inexpensive calculator like an Casio fx-115 ES plus
P(X > 553.4) = normcdf(553.4, 9999, 518,113) = .3770 (accurate to 4 decimal places)
______________
If 20 students are randomly selected, find the probability that their mean score is at least 553.4.
 ,  = 1.401
p(z > 1.401, df 19) = 0.0887 (accurate to 4 decimal places)
Question 1197769: Suppose that grade point averages of undergraduate students at one university have a bell-shaped distribution with a mean of 2.62 and a standard deviation of 0.4 Using the empirical rule, what percentage of the students have grade point averages that are at least 1.82? Please do not round your answer.
Answer by math_tutor2020(3817) (Show Source):
You can put this solution on YOUR website!
mu = 2.62 = mean
sigma = 0.4 = standard deviation
Let's find the z score when x = 1.82
z = (x - mu)/sigma
z = (1.82 - 2.62)/0.4
z = -2
This score is exactly 2 standard deviations below the mean.
The task of computing P(X > 1.82) is equivalent to finding P(Z > -2).
The empirical rule says roughly 95% of the normal distribution is within 2 standard deviations of the mean.
That means there is roughly 2.5% of the distribution in the left tail since (100%-95%)/2 = 2.5%
The remaining portion is then 100% - 2.5% = 97.5%
Or basically there's 95% in the middle and 2.5% in the right tail, so 95% + 2.5% = 97.5%
and P(Z > -2) = 0.975
Check out this diagram below
Answer: Approximately 97.5%
Question 1197544:
x x/μ (x−μ)2 (x−μ)2⋅P(x)
0
1
2
3
4
Find the standard deviation. Round to three decimal places.
σ =
Answer by ikleyn(52786) (Show Source):
You can put this solution on YOUR website! .
? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ? ?
Question 1197531: x
x
−
μ
(
x
−
μ
)
2
(
x
−
μ
)
2
⋅
P
(
x
)
0
1
2
3
4
Find the standard deviation. Round to three decimal places.
σ
=
Answer by ikleyn(52786) (Show Source):
Question 1197532:
x
P
(
x
)
x
2
x
P
(
x
)
x
2
P
(
x
)
0 0.22
0
Correct
0
Correct
0
Correct
1 0.15
1
Correct
0.15
Correct
0.15
Correct
2 0.18
4
Correct
0.36
Correct
0.72
Correct
3 0.16
9
Correct
0.48
Correct
1.44
Correct
4 0.29
16
Correct
1.16
Correct
4.64
Correct
Find the mean. Round to two decimal places.
μ
=
Find the standard deviation. Round to three decimal places.
σ
=
Answer by ikleyn(52786) (Show Source):
Question 1197517: Questions 24-25
The scores of a standardized IQ test are normally distributed with a mean score of 100 and a standard deviation of 15.
Question 24
Find the probability that a randomly selected person has an IQ score higher than 105.
Question options:
0.9522
0.3694
-1.15
0.6306
Question 25
A random sample of 55 people is selected from this population. What is the probability that the mean IQ score of the sample is greater than 105?
Question options:
0.0067
0.3694
0.6306
0.9933
Answer by ewatrrr(24785) (Show Source):
You can put this solution on YOUR website!
Hi
Question 24
 , z = 5/15
P(x>105) = 1 - P(x ≤ 105) = 1 - P( z ≤ 1/3) = .3694
Question 25
P(sample mean > 105) = .0067
Wish You the Best in your Studies.
Question 1195486: If the mean of the following numbers is 17, find the c value. Produce an algebraic solution. Guess and check is unacceptable.
12, 18, 21, c, 13
Found 2 solutions by greenestamps, Edwin McCravy:
Answer by greenestamps(13200)  (Show Source):
You can put this solution on YOUR website!
A typical algebraic solution....
(12+18+21+c+13)/5 = 17
12+18+21+c+13 = 85
64+c = 85
c = 21
That method for solving the problem is fine since the numbers are small, so adding the numbers is an easy task.
If the numbers are much larger but still close together, an often useful method for solving a problem like this is to compare each number to the desired mean, knowing that the "overs" and "unders" have to balance.
For this example, with the average being 17....
12: 5 under
18: 1 over
21: 4 over
13: 4 under
Note the first three numbers balance; overall the four scores are 4 under the desired average.
That means c must be 4 over the desired average. So
ANSWER: c = 17+4 = 21
Answer by Edwin McCravy(20056)  (Show Source):
Question 1195153: The heights of the adult male population on a given island are normally distributed with mean 70 inches and standard deviation of 2.5 inches.
(a.) what percentage of heights is within 1 standard deviation of the mean? That is, what percentage of heights is between 70-2.5=67.5 inches and 70+2.5=72.5 inches?
(b.) what percentage of male population are shorter than 65 inches?
(c.) what percentage of males are taller than 67 inches?
(d.) what percentage is between 67 and 68 inches?
Answer by ikleyn(52786) (Show Source):
You can put this solution on YOUR website! .
It is about finding the area under the normal curve.
Use free of charge online calculator
https://onlinestatbook.com/2/calculators/normal_dist.html
specially intended for this purpose.
Doing this way is the simplest, the quickest and the easiest way to solve the problem.
It is also the best way to learn the subject.
Question 1194722: Analyze the given data below and answer the following questions:
Zian's Scores in Math and English
ENGLISH: 50, 40, 28, 35, 37
MATH: 36, 29, 35, 30, 35
MEASURES OF VARIABILITY ENGLISH
Mean 38
Standard Deviation 8.03
Variance 64.5
RAnge 22
MEASURES OF VARIABILITY MATH
Mean 33
Standard Deviation 3.24
Variance 10.5
RAnge 7
Question 1:In what subject does Zian performs well?
Question 2: In what subject does Zian shows consistent score?
Question 3: Are his scores in english are more spread out than math?
Answer by Alan3354(69443)  (Show Source):
Question 1191285: The time taken for a student to complete an exam is normally distributed with a mean of 40 minutes and a standard deviation of 5.5 minutes.
The probability a student takes between k and 48 minutes is 0.4. What is the value of k?
Answer by Boreal(15235)  (Show Source):
You can put this solution on YOUR website! z=(x-mean)/sd
z=(48-40)/5.5
=1.454
probability <48 minutes is 0.9271
use the table and find that for probability 0.5271 being less than a certain number of minutes, z=0.068
.
The probability that z is between 0.068 and 1.454 is 0.40
0.064=(k-40)/5.5
0.352=(x-40)
k=40.352 minutes
Question 1191287: The time taken for a student to complete an exam is normally distributed with a mean of 40 minutes and a standard deviation of 5.5 minutes.
A student is randomly selected. What is the probability that the student completes the task in less than 48 minutes?
Answer by ikleyn(52786) (Show Source):
You can put this solution on YOUR website! .
The time taken for a student to complete an exam is normally distributed with a mean
of 40 minutes and a standard deviation of 5.5 minutes.
A student is randomly selected. What is the probability that the student completes
the task in less than 48 minutes?
~~~~~~~~~~~~~~~~~~
The normal distribution curve is a bell shaped curve.
This question is to determine the area under the normal distribution curve
below (or to the left) of the given score.
It can be done in different ways:
- manually, or
- using online calculators, or
- using your pocket calculator.
MANUALLY
To do the job manually, use this Table representing AREA to the LEFT of the Z-score
https://www.math.arizona.edu/~rsims/ma464/standardnormaltable.pdf
On how to do it, see this Internet text source
http://statisticshelper.com/how-to-use-the-z-table
USING ONLINE CALCULATOR
To do the job using an online (free of charge) calculator, go to this web-site
https://onlinestatbook.com/2/calculators/normal_dist.html
Input the given parameters into an appropriate window of the calculator and get the answer
to your question.
The calculator has perfect description and design, as well as clear visual interface, which prevents you of making errors.
So EVERY person, even beginner, may work with it on his or her own, even having minimum knowledge on the subject.
USING YOUR POCKET CALCULATOR
On how to use it, see a text description in THIS Internet source / site
https://mathbits.com/MathBits/TISection/Statistics2/normaldistribution.htm
Or see these Youtube video-lessons
https://www.youtube.com/watch?v=bVdQ7OzGvU0 (for Casio fx-991 MS)
https://www.youtube.com/watch?v=yYpMkgB20C4 (for TI-83 or TI-84 calculators)
I recommend you to play with the online calculator first.
If you are unfamiliar with the subject, playing with the online calculator will help you a lot !
After learning it, you will be able to solve this problem (and thousand other similar and different problems) ON YOUR OWN,
without asking for help from outside.
Happy learning ( ! )
Question 1191028: I am having trouble on a problem from my homework assignment. This is the question word for word from the text book:
The average earnings of year-round full-time workers 25-34 years old with a bachelor's degree or higher were $58,500 in 2003. If the standard deviation is $11,200, what can you say about the percentage of these workers who earn;
a. between $47,300 and $69,700?
b. More than $80,900?
c. How likely is it that someone earns more than $100,000?
Found 2 solutions by ikleyn, MathLover1:
Answer by ikleyn(52786) (Show Source):
You can put this solution on YOUR website! .
The normal distribution curve is a bell shaped curve.
These questions are to determine the areas under the normal distribution curve
below the given score; or between two given scores; or above the given score.
It can be done in different ways:
- manually, or
- using online calculator, or
- using your pocket calculator.
MANUALLY
To do it manually, use this Table representing AREA to the LEFT of the Z-score
https://www.math.arizona.edu/~rsims/ma464/standardnormaltable.pdf
USING ONLINE CALCULATOR
To do it using an online (free of charge) calculator, go to this web-site
https://onlinestatbook.com/2/calculators/normal_dist.html
Input the given parameters of each question into an appropriate window of the calculator and get the answers
to your questions.
The calculator has perfect description and design, so EVERY person, even beginner, may work with it on his or her own,
even having minimum knowledge on the subject.
USING YOUR POCKET CALCULATOR
On how to use it, see a text description in THIS Internet source / site
https://mathbits.com/MathBits/TISection/Statistics2/normaldistribution.htm
Or see these Youtube video-lessons
https://www.youtube.com/watch?v=bVdQ7OzGvU0 (for Casio fx-991 MS)
https://www.youtube.com/watch?v=yYpMkgB20C4 (for TI-83 or TI-84 calculators)
Find there EVERYTHING you need to know in clear and compact form.
After learning it, you will be able to solve this problem (and thousand other similar and different problems) ON YOUR OWN,
without asking for help from outside.
Happy learning ( ! )
Answer by MathLover1(20850) (Show Source):
Question 1190242: After 3 exams, Sam has a mean score of 75 marks.
(i) What is Sam’s total score for the three exams?
Answer by Solver92311(821)  (Show Source):
Question 1188632: The number of potholes in any given 1 mile stretch of freeway pavement in Pennsylvania has a bell-shaped distribution. This distribution has a mean of 46 and a standard deviation of 11. Using the empirical rule, what is the approximate percentage of 1-mile long roadways with potholes numbering between 35 and 79?
Found 2 solutions by Shin123, Solver92311:
Answer by Shin123(626)  (Show Source):
Answer by Solver92311(821) (Show Source):
Question 1186965: Adult men have heights with a mean of 69.0 inches and a standard deviation of 2.8 inches. Find the z-score of a man 76.8 inches tall. (to 2 decimal places)
Answer by ikleyn(52786) (Show Source):
You can put this solution on YOUR website! .
z =  =  = 2.79 (rounded). ANSWER
Also, there is an online calculator for such calculations at this link
https://www.calculator.net/z-score-calculator.html
Question 1186716: ] Assuming that the 26 letters in the English language alphabet comprise a
population,
1. Briefly explain how a simple random sample of size n = 7 can be obtained
with no mode(s). No calculations would be involved.
2. Showing your work, find the total number of simple random samples of
size n = 7 possible, none having any mode.
3. Showing your work, what is the probability of drawing a simple random
sample of size n = 7 containing the set A, B, C, D, X, Y, Z, using a 6-ball
capacity scoop and at first attempt, from a bag containing 26 ping pong balls,
each uniquely labeled one of the 26 letters?
Answer by Edwin McCravy(20056) (Show Source):
You can put this solution on YOUR website!
>>>Assuming that the 26 letters in the English language alphabet comprise a
population,
population = {A,B,C,D,E,F,G,H,I,J,K,L,M,N,O,P,Q,R,S,T,U,V,W,X,Y,Z}
>>>1. Briefly explain how a simple random sample of size n = 7 can be obtained
with no mode(s). No calculations would be involved.
Get 26 identical slips of paper.
Write a different letter on each one.
Put them in a hat.
Stir them up real good.
Draw out a slip of paper without looking.
Write down the letter it has written on it on a separate piece of paper.
Don't put it back (To make sure there is no mode.)
Keep drawing without putting any back until you have written down 7 letters.
>>>2. Showing your work, find the total number of simple random samples of size
n = 7 possible, none having any mode.
C(26,n) = [(26)(25)(24)(23)(22)(21)(20)]/[(7)(6)(5)(4)(3)(2)(1) = 657800
>>>3. Showing your work, what is the probability of drawing a simple random
sample of size n = 7 containing the set A, B, C, D, X, Y, Z, using a 6-ball
capacity scoop and at first attempt, from a bag containing 26 ping pong balls,
each uniquely labeled one of the 26 letters?
1/657800
Edwin
Question 1182257: A standardized exam's scores are normally distributed. In a recent year, the mean test score was 1462 and the standard deviation was 319. The test scores of four students selected at random are 1890,1180 ,2210,and 1350. Find the z-scores that correspond to each value and determine whether any of the values are unusual.
The z-score for 1890 is____?
Round to two decimal places as needed.)
Answer by math_tutor2020(3817) (Show Source):
You can put this solution on YOUR website!
mu = population mean = 1462
sigma = population standard deviation = 319
z = (x - mu)/sigma
z = (1890-1462)/319
z = 1.34 approximately
I would consider this to be not unusual. In other words, it seems fairly likely. Any z score such that  would be considered usual; anything outside this interval is considered unusual. Keep in mind that your teacher may use another interval, so I would check with them about that. Though usually, if a z score is further than 2 standard deviations from the mean, then it's considered unusual.
Question 1179100: 100 students attended a party, 79 were interviewed and 54 of the 79 met the case definition of the disease.
53 took the drink and out of them 50 is sick and 3 is well.
26 did not take the drink and out them 4 is sick and 22 is well.
1. What is the association between the drink and illness
2. The attributable proportion for the drink
Answer by ikleyn(52786) (Show Source):
Question 1178053: Please l am waiting for solution of my problem
Answer by MathLover1(20850) (Show Source):
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