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Tutors Answer Your Questions about Probability-and-statistics (FREE)
Question 236227: 6. Dewey, Cheetum, and Howe is a law firm that handles divorce cases. A previous study showed that in 34% of the time, their client got everything that was requested in the initial divorce papers. How many divorce proceedings would they have to survey to construct a 99% confidence interval for the true proportion of their clients get everything that they wanted in the divorce? If this interval is high enough, it would be a great advertising piece of information.
I have no idea where to start please help me.
Answer by stanbon(25959)   (Show Source):
You can put this solution on YOUR website!Dewey, Cheetum, and Howe is a law firm that handles divorce cases. A previous study showed that in 34% of the time, their client got everything that was requested in the initial divorce papers. How many divorce proceedings would they have to survey to construct a 99% confidence interval for the true proportion of their clients get everything that they wanted in the divorce? If this interval is high enough, it would be a great advertising piece of information.
------------------------------------------
n = [z*s/E]^2
s = sqrt[pq/n]
---
n = [2.5758*sqrt(0.34*0.66)/0.01]^2
---
n = 122.02^2
n = 14888.7
Rounding up n = 14889
==================================
Cheers,
Stan H.
Question 236189: Find the mean, variance, and standard deviation of n=90, p=0.2
Answer by stanbon(25959) (Show Source):
You can put this solution on YOUR website!Find the mean, variance, and standard deviation of n=90, p=0.2
-------------------
mean = np = 90*0.2 = 18
variance = npq = 18*0.8 = 14.4
std = sqrt(npq) = 3.795
==============================
Cheers,
Stan H.
Question 236120: a manufacturer purchases two machines A and B .the probability that A will last five years is 4/5 and the probability that B will last five years is 3/4 find the probability that atleast one machine will last for five years
Answer by stanbon(25959) (Show Source):
You can put this solution on YOUR website!a manufacturer purchases two machines A and B .the probability that A will last five years is 4/5 and the probability that B will last five years is 3/4 find the probability that atleast one machine will last for five years
------------------
Find the probability that neither will last 5 years:
= (1/5)(1/4) = 1/20
------------------------------
Your Problem:
P(at least will last) = 1 - P(neither will last)
= 1 = 1/20 = 19/20
=============================
Cheers,
Stan H.
Question 236112: In a local college, 60% of the math majors are women. Fifteen math majors are chosen at random. (1 points each)
a. What is the probability that exactly 6 are women?
Answer by stanbon(25959) (Show Source):
You can put this solution on YOUR website!In a local college, 60% of the math majors are women. Fifteen math majors are chosen at random. (1 points each)
a. What is the probability that exactly 6 are women?
P(x=6) = 15C6(0.6)^6(0.4)^9 = binompdf(15,0.6,6) = 0.0612...
Cheers,
Stan H.
Question 236064: A card is drawn from an ordinary deck of 52 cards, and the result is recorded on paper. The card is then returned to the deck and another card is drawn and recorded. Find the probability that the following occurs.
Only the first card is a club
Answer by stanbon(25959) (Show Source):
Question 236105: Three coins are tossed in the air. what is the probability that exactly two of them turn up heads?
{hhh, hht, hth, htt, thh, tht, tth, ttt}
Exactly 2 heads ==> {hht, hth, thh}
P(exactly 2 heads) = 3/8.
Answer by mathsteacher(28) (Show Source):
Question 236066: A card is drawn from an ordinary deck of 52 cards, and the result is recorded on paper. The card is then returned to the deck and another card is drawn and recorded. Find the probability that the following occurs.
Only the first card is a club
Answer by stanbon(25959) (Show Source):
You can put this solution on YOUR website!A card is drawn from an ordinary deck of 52 cards, and the result is recorded on paper. The card is then returned to the deck and another card is drawn and recorded. Find the probability that the following occurs.
Only the first card is a club
-------------------------------------
Since the card is returned to the deck the events are independent.
-----------------------------------
P(1st card is a club) = 1/4
--------------
P(2nd card is not a club) = 3/4
---------------
P(club and not club) = (1/4)(3/4) = 3/16
==============================================
Cheers,
Stan H.
Question 235071: Find the area under the standard normal curve between z = -1.53 and z = 0. Round the answer to 4 decimal places.
Answer by stanbon(25959) (Show Source):
Question 236051: Contruct a binominal distribution using n=6 and p=0.36
X P(x)
o
1
2
3
4
5
6
Answer by stanbon(25959) (Show Source):
You can put this solution on YOUR website!Contruct a binominal distribution using n=6 and p=0.36
X P(x)
o
1
2
3
4
5
6
-------
For each different value of "x"
P(x=k) = 6Ck(0.36)^k*(0.64)^(n-k)
---
For example, if k = 4 you get:
p(x=4) = 6C4(0.36)^4*(0.64)^2 = 15*0.0168*0.4096
= 15*.0069 = 0.1032
===========================
Cheers,
Stan H.
Question 236031:
Can you PLEASE help me with this homework problem?? I appreciate any help you can give!
Table I
White Black Hispanic Asian Foreign Total
Dept. 1. 4 0 1 1 2 8
Dept. 2. 2 4 1 1 0 8
Dept. 3. 1 2 3 1 1 8
Dept. 4. 4 0 1 1 2 8
Dept. 5. 3 2 1 0 2 8
Total. 14 8 7 4 7 40
Table II
Female Male Total
Black 6 2 8
White 6 8 14
Hispanic 3 4 7
Asian 2 2 4
Foreign 5 2 7
Total 22 18 40
From Table 1: What is the probabilty of choosing a White employee from Dept. 2 and then a Hispanic employee from Dept. 3?
What is the probability of choosing a Foreigner or an Hispanic both from Department 5?
From Table 2: What is the probability of choosing a Black male or a Foreign male?
Answer by stanbon(25959) (Show Source):
You can put this solution on YOUR website!Table I
White Black Hispanic Asian Foreign Total
Dept. 1. 4 0 1 1 2 8
Dept. 2. 2 4 1 1 0 8
Dept. 3. 1 2 3 1 1 8
Dept. 4. 4 0 1 1 2 8
Dept. 5. 3 2 1 0 2 8
Total. 14 8 7 4 7 40
Table II
Female Male Total
Black 6 2 8
White 6 8 14
Hispanic 3 4 7
Asian 2 2 4
Foreign 5 2 7
Total 22 18 40
From Table 1: What is the probabilty of choosing a White employee from Dept. 2 and then a Hispanic employee from Dept. 3?
(2/40) * (3/38) = (1/20)*(3/38) = 3/760
-----------------------------------------------------
What is the probability of choosing a Foreigner or an Hispanic both from Department 5?
(1+2)/40 = 3/40
-----------------------
From Table 2: What is the probability of choosing a Black male or a Foreign male?
(2+2)/40 = 1/10
----------------------------
Cheers,
Stan H.
Question 235963: I have answered some of these but am not for sure if they are correct, the others, I am not for sure. Thanks in advance
1. Decide whether the random variable is discrete or continuous.
a.The number of bottles of water drunk in Florida in August 2008
Discrete
b.The number of wasps in the lab test nest
Discrete
c.The height of sophomores in a required US History class at Ridgemont High in 2008
Continuous
d.The number of players on the Washington Redskins
Discrete
e.The speed of cars starting in the Darlington 500 in 2008
Continuous
2. Decide whether the distribution is a probability distribution. If it is not a probability distribution, identify the property that is not satisfied.
x P(x)
1 0.200
2 0.238
3 0.245
4 0.307
No it is not a probability distribution. It does not equal 1.
0.200 + 0.238 + 0.245 + 0.307 = 0.99
Section 4.2: Binomial Probability
4. In a local college, 60% of the math majors are women. Fifteen math majors are chosen at random. (1 points each)
a. What is the probability that exactly 6 are women?
b. What is the probability that 5 or less women are selected?
c. What is the probability that 10 women are selected?
d. Find the mean
e. Find the variance 2
5. A multiple choice test has 20 questions with each having 4 possible answers with one correct. Assume a student randomly guesses the answer to every question. (2.5 points each)
a. What is the probability of getting exactly 11 correct answers?
b. What is the probability of getting less than 9 correct answers?
Answer by stanbon(25959) (Show Source):
You can put this solution on YOUR website!1. Decide whether the random variable is discrete or continuous.
a.The number of bottles of water drunk in Florida in August 2008
Discrete
OK
-----------------------
b.The number of wasps in the lab test nest
Discrete
OK
-----------------------
c.The height of sophomores in a required US History class at Ridgemont High in 2008
Continuous
OK
-----------------------------
d.The number of players on the Washington Redskins
Discrete
OK
--------------
e.The speed of cars starting in the Darlington 500 in 2008
Continuous
OK
--------------------------
2. Decide whether the distribution is a probability distribution. If it is not a probability distribution, identify the property that is not satisfied.
x P(x)
1 0.200
2 0.238
3 0.245
4 0.307
No it is not a probability distribution. It does not equal 1.
0.200 + 0.238 + 0.245 + 0.307 = 0.99
OK
---------------------
Section 4.2: Binomial Probability
4. In a local college, 60% of the math majors are women. Fifteen math majors are chosen at random. (1 points each)
a. What is the probability that exactly 6 are women?
P(x=6) = 15C6(0.6)^6*(0.4)^9 = binompdf(15,0.6,6)= 0.0612
==========================================
b. What is the probability that 5 or less women are selected?
P(0<= x <= 5) = binomcdf(16,0.6,5) = 0.0338
==========================================
c. What is the probability that 10 women are selected?
P(x=10) = binompdf(15,0.6,10) = 0.1859
==========================================
d. Find the mean
mean = np = 15*0.6 = 9
=========================================
e. Find the variance
variance = npq = 9*0.4 = 3.6
======================================
5. A multiple choice test has 20 questions with each having 4 possible answers with one correct. Assume a student randomly guesses the answer to every question. (2.5 points each)
a. What is the probability of getting exactly 11 correct answers?
binompdf(20,0.25,11) = 0.0030
b. What is the probability of getting less than 9 correct answers?
binomcdf(20,0.25,8) = 0.9591
===================================
Cheers,
Stan H.
Question 235378: A factory worker places 86 newly created circuits on a shelf to be checked for quality. of these 6 will not work correctly. Suppose that she is asked to randomly select 2 circuits, without replacement. what is the chance that both circuits she selects will be defective? approximate to the nearest ten thousandth.
Answer by stanbon(25959) (Show Source):
You can put this solution on YOUR website!A factory worker places 86 newly created circuits on a shelf to be checked for quality. of these 6 will not work correctly. Suppose that she is asked to randomly select 2 circuits, without replacement. what is the chance that both circuits she selects will be defective? approximate to the nearest ten thousandth.
---------------------------
# of ways to pick 2 defective: 6C2
# of ways to pick 2 or the 86: 86C2
---------------
Ans: 6C2/86C2 = 15/3655 = 0.0041
===============================
Cheers,
Stan H.
Question 235922: In how many ways can a doubles game of tennis be arranged from five boys and eleven girls if each side must have one boy and one girl?
Answer by solver91311(4818)  (Show Source):
Question 235920: A decade-old study found that the proportion of high school seniors who felt that "getting rich" was an important personal goal was 66%. Suppose that we have reason to believe that this proportion has changed, and we wish to carry out a hypothesis test to see if our belief can be supported. State the null hypothesis Ho and the alternative hypothesis H1 that we would use for this test.
Answer by stanbon(25959) (Show Source):
You can put this solution on YOUR website!A decade-old study found that the proportion of high school seniors who felt that "getting rich" was an important personal goal was 66%.
--------------------------------
Suppose that we have reason to believe that this proportion has changed, and we wish to carry out a hypothesis test to see if our belief can be supported.
--------------------------------
State the null hypothesis Ho and the alternative hypothesis H1 that we would use for this test.
----------------
Ho: p = 0.66
H1: p is not equal to 0.66
===============================
Cheers,
Stan H.
Question 235921: Suppose that a researcher is interested in estimating the mean systolic blood pressure,u, of executives of major corporations. He plans to use the blood pressures of a random sample of executives of major corporations to estimate u. Assuming that the standard deviation of the population of systolic blood pressures of executives of major corporations is 25 mm Hg, what is the minimum sample size needed for the researcher to be 90% confident that his estimate is within 5 mm Hg of u?
Answer by stanbon(25959) (Show Source):
You can put this solution on YOUR website!Suppose that a researcher is interested in estimating the mean systolic blood pressure,u, of executives of major corporations. He plans to use the blood pressures of a random sample of executives of major corporations to estimate u. Assuming that the standard deviation of the population of systolic blood pressures of executives of major corporations is 25 mm Hg, what is the minimum sample size needed for the researcher to be 90% confident that his estimate is within 5 mm Hg of u?
---------------------------
n = [z*s/E]^2
----
n = [1.645*25/5]^2
---
n = [1.645*5]^2
---
n = 67.65
---
Rounding up you get n = 68
================================
Cheers,
Stan H.
Question 235907: Let two cards be dealt successively, without replacement, from a standard 52-card deck. Find the probability of each event:
a)no face cards
b)club second, given diamond first
Answer by josmiceli(2979) (Show Source):
You can put this solution on YOUR website!The face cards are Jack, Queen, and King
There are 4 of each, d, h, s, & c
That makes 12 face cards
 non-face cards
The 1st card has  probability
of not being a face card
The 2nd card has  probability
of not being a face card
Getting both has probability 
-------------------------------
Getting a certain diamond from a deck
of 52 cards is 
Then getting a club is 
Getting both has probability 
Question 235408: The US Mint selects ten pennies from the production line to test the hypothesis that the mean weight of each penny is at least 6 grams. The normally-distributed weights (in grams) of these pennies are as follows: 6, 6, 8, 5, 9, 5, 9, 2, 3, 8. Assume = 0.01.
· State the null and alternate hypotheses
· Calculate the sample mean and standard deviation
· Determine which test statistic is appropriate (z or t), and calculate its value.
· Determine the critical value(s).
· State your decision: Should the null hypothesis be rejected?
I submitted this question on yesterday and did not get a response. I would like to thank you tutor for all of your hard work.
Answer by stanbon(25959) (Show Source):
You can put this solution on YOUR website!The US Mint selects ten pennies from the production line to test the hypothesis that the mean weight of each penny is at least 6 grams. The normally-distributed weights (in grams) of these pennies are as follows: 6, 6, 8, 5, 9, 5, 9, 2, 3, 8. Assume alpha = 0.01.
· State the null and alternate hypotheses
Ho: u >=6
Ha: u < 6
---------------------
· Calculate the sample mean and standard deviation
s = 6.1 ; std = 2.4244
--------------------------------
· Determine which test statistic is appropriate (z or t), and calculate its value.
t(6.1) = (6.1-6)/[2.4244/sqrt(10)] = 0.1304
--------------------
· Determine the critical value(s).
For a left-tail test with alpha = 1% and df = 9 the critical value
is invT(0.01,9) = -2.8214
----------------------------
· State your decision: Should the null hypothesis be rejected?
Since the test statistic is not in the reject interval,
do not reject Ho based on these these results.
======================================
Cheers,
Stan H.
Question 235409: I need your help in working this problem out. Thank you very much...
A watch manufacturer creates watch springs whose properties must be consistent. In particular, the standard deviation in their weights must be no greater than 2.0 grams. Fifteen watch springs are selected from the production line and measured; their weights are 1, 4, 8, 7, 1, 5, 1, 4, 1, 4, 3, 9, 1, 3, and 2 grams. Assume = 0.01.
· State the null and alternate hypotheses
· Calculate the sample standard deviation
· Determine which test statistic is appropriate (chi-square or F), and calculate its value.
· Determine the critical value(s).
· State your decision: Should the null hypothesis be rejected?
Answer by stanbon(25959) (Show Source):
You can put this solution on YOUR website!A watch manufacturer creates watch springs whose properties must be consistent. In particular, the standard deviation in their weights must be no greater than 2.0 grams. Fifteen watch springs are selected from the production line and measured; their weights are 1, 4, 8, 7, 1, 5, 1, 4, 1, 4, 3, 9, 1, 3, and 2 grams. Assume = 0.01.
· State the null and alternate hypotheses
Ho: sigma^2 <= 4 (claim)
Ha: sigma^2 > 4
· Calculate the sample standard deviation
std of the sample = 2.667
-----------------------------------
· Determine which test statistic is appropriate (chi-square or F), and calculate its value.
Chi-Sq = (n-1)*s^2/sigma^2 = 14*2.667^2/4 = 24.895
-------------------------------------
· Determine the critical value(s)
Chi-Sq lower = 4.075 ; Chi-Sq upper = 31.32
--------------------------------------
State your decision: Should the null hypothesis be rejected?
Since the ts is not in either rejection interval, do not
reject Ho based on these test results.
======================================
Cheers,
Stan H.
Question 235828: There is a 50% chance of rain on Saturday and a 50% chance of rain on Sunday. What is the probability of 2 sunny days this weekend?
Answer by stanbon(25959) (Show Source):
You can put this solution on YOUR website!There is a 50% chance of rain on Saturday and a 50% chance of rain on Sunday. What is the probability of 2 sunny days this weekend?
--------
The events are independent, so
P(sunny and sunny) = P(sunny)*P(sunny) = 0.5*0.5 = 0.25
===========================================================
Cheers,
Stan H.
Question 235846: From a bag containing 2 rupee-coins and 3 twenty paise coins, a person is asked to draw two coins at random. Find the values of his expectatioon.
Answer by stanbon(25959) (Show Source):
You can put this solution on YOUR website!From a bag containing 2 rupee-coins and 3 twenty paise coins, a person is asked to draw two coins at random. Find the values of his expectatioon.
-----------
Random variable values are ?
----------------
What are the relative values of rupee and paise coins?
=========================
Cheers,
Stan H.
Question 235847: In a college 30% students fail in Physics, 25% fail in Mathematics and 10% students fail in both. One student is chosen at random. What is the probability that
a) He fails in Mathematics if he has failed in Physics
b) He fails in Physics if he has failed in Mathematics.
c) He fails in Physics or Mathematics.
Answer by stanbon(25959) (Show Source):
You can put this solution on YOUR website!In a college 30% students fail in Physics, 25% fail in Mathematics and 10% students fail in both. One student is chosen at random. What is the probability that
a) He fails in Mathematics if he has failed in Physics
P(m|p) = P(m and p)/P(p) = 0.10/0.30 = 1/3
---------------------------------------------------
b) He fails in Physics if he has failed in Mathematics.
P(p|m) = P(p and m)/P(m) = 0.10/0.25 = 10/25 = 2/5
---------------------------------------------------
c) He fails in Physics or Mathematics.
P(p or m) = P(p) + P(m) - P(p and m)
= 0.30 + 0.25 - 0.10 = 0.45
=========================================
Cheers,
Stan H.
Question 235843: Solve the following system of equations by Gauss-Elimination method
x + y + z = 9
x – 2y + 3z = 8
2x + y – z = 3
Answer by Stitch(34) (Show Source):
You can put this solution on YOUR website!Aghhh so that was not Gauss-Elimination but below is a nother way to do it. Sorry about that....
Here is the Gauss-Elimination way:
Equation 1: 
Equation 2: 
Equation 3: 
-*(Equation 1) + Equation 2 = Equation 2
Multiply equation 1 by negative 1 and then add to equation 2
-(X + Y + Z = 9) + (X - 2Y +3Z = 8) Rewrite the equation
(-X - Y - Z = -9) + (X - 2Y +3Z = 8) Combine like terms, notice that the X's cancel out
Equation 2 now equals 
-------
To eliminate X for equation 3, multiply equation 1 by -2
-2*(Equation 1) + Equation 3 = Equation 3
-2*(X + Y + Z = 9) + (2X + Y - Z = 3) Rewrite the equation
(-2X - 2Y -2Z = -18) + (2X + Y - Z = 3) Combine like terms, notice that the X's cancel out
Equation 3 now equals 
-------
Now we have to eliminate another variable
Equation 2 now equals
Equation 3 now equals
-3*(Equation 3) + Equation 2 = Equation 2
-3*(-Y -3Z = -15) + (-3Y + 2Z = -1) Rewrite the equation
(3Y + 9Z = 45) + (-3Y + 2Z = -1) Combine like terms, notice that the Y's cancel out
Equation 2 now equals  Divide both sides by 11
Now we have to go back and do substitution
Lets plug 4 in for Z in the new equation 3
 Simplify
 Add Y to both sides
 Add 15 to both sides
Now lets plug our answers into equation 1
 Combione like terms
 Subtract 7 from both sides
You can see the checks down below
-------------------------------------------------------------------------------
Anothor way
Given:
Equation 1:
Equation 2:
Equation 3:
------------------------------------
Solution:
The first step is to solve for 1 of the variables.
Lets set equations 1 & 3 equal to zero
Equation 1: Subtract 9 from both sides
Equation 3: Subtract 3 from both sides
Now since both equations equal zero, we can set them equal to each other
 Add 3 to both sides
 Subtract X from both sides
 Combine like terms
 Subtract Y from both sides
 Add Z to both sides
 Combine like terms
Lets label  as equation 4
-------------------------------------------
Now plug (2Z - 6) into equation 1 for X
Equation 1:
 Combine like terms
 Add 6 to both sides
 Subtract 3Z from both sides
Lets label as equation 5
-------------------------------------------
Now plug equation 4 in for X, and equation 5 in for Y in Equation 2
Equation 2:
 Combine like terms
 Add 36 to both sides
 Divide both sides by 11
-------------------------------------------
Now plug 4 into equation 5 for Z
Equation 5:
 Simplify
 Combine like terms
-------------------------------------------
Now plug 4 into equation 4 for Z
Equation 4:
 Simplify
 Combine like terms
-------------------------------------------
Its time to check your answers
X = 2, Y = 3, Z = 4
Plug your answers into the three given equations and make sure that they are true
Equation 1:
Equation 2:
Equation 3:
----
Equation 1: 
 Equation 1 checks out
----
Equation 2:
 Equation 2 checks out
----
Equation 3:
 Equation 3 checks out
Question 235413: Two groups of ten sprinters run 100 meters. The times required by sprinters in the first group are as follows:
12.0 12.9 10.1 14.6 11.9 11.3 13.0 12.9 13.3 10.0
The times required by sprinters in the second group are as follows:
11.6 12.5 12.3 10.2 16.3 17.0 19.1 18.7 12.3 16.4
Assuming that = 0.02, test the hypothesis that the means of the two populations are equal.
· State the null and alternate hypotheses
· Calculate the mean and standard deviation for each group
· Calculate the value of the test statistic.
· Determine the critical value(s).
· State your decision: Should the null hypothesis be rejected?
Thank you so very much for helping with this problem. I really do appreciate all of your hard work.
Answer by stanbon(25959) (Show Source):
You can put this solution on YOUR website!Two groups of ten sprinters run 100 meters. The times required by sprinters in the first group are as follows:
12.0 12.9 10.1 14.6 11.9 11.3 13.0 12.9 13.3 10.0
The times required by sprinters in the second group are as follows:
11.6 12.5 12.3 10.2 16.3 17.0 19.1 18.7 12.3 16.4
Assuming that alpha = 0.02, test the hypothesis that the means of the two populations are equal.
· State the null and alternate hypotheses
Ho: u(1)-u(2) = 0
Ha: u(1)-u(2) is not 0
---------------------------------
· Calculate the mean and standard deviation for each group
Grp 1: mean = 12.2 ; std = 1.445
Grp 2: mean = 14.64 ; std = 3.202
----------------------------------
· Calculate the value of the test statistic.
I ran a 2-Sample Ttest and got the following:
ts = t = -2.1965
--------------------------
· Determine the critical value(s).
invT(0.01, with df=12.517) = -2.3266
----------------------------------------
· State your decision: Should the null hypothesis be rejected?
Since the test statistic is not in the reject interval, do not
reject Ho.
----------------------------------------
Cheers,
Stan H.
Question 235411: I submitted this question on yesterday and did not get a response. I would like to thank the tutor for helping me on this problem.
A telephone survey gives 670 consumers two choices: Do they prefer Coke or Pepsi? Exactly 158 of those surveyed state that they prefer Coke. Assuming that = 0.10, test the hypothesis that the proportion of the population that prefers Coke is 50%.
· State the null and alternate hypotheses
· Calculate the sample proportion
· Calculate the value of the test statistic.
· Determine the critical value(s).
· State your decision: Should the null hypothesis be rejected?
Answer by stanbon(25959) (Show Source):
You can put this solution on YOUR website!A telephone survey gives 670 consumers two choices: Do they prefer Coke or Pepsi?
Exactly 158 of those surveyed state that they prefer Coke.
Assuming that alpha = 0.10, test the hypothesis that the proportion of the population that prefers Coke is 50%.
· State the null and alternate hypotheses
Ho: p = 0.5
Ha: p is not 0.5
-----------------------
· Calculate the sample proportion
p-hat = 158/670 = 0.2358
-----------------------------------
· Calculate the value of the test statistic.
z(0.2358) = (0.2358-0.5000)/sqrt[.5*.5/(670)] = -13.6773
-------------
· Determine the critical value(s).
invNorm(0.05) = +/-1.645
--------------
Decision: Should Ho be rejected.
Yes because the test statistic is in the rejection interval
to the left of -1.645.
=====================================
Cheers,
Stan H.
Question 235513: i have a question in the grouped data in statistics , they asking the probability that an emloyee . taken at random , earns more than 20000 pounds , total workers are 68 and they have group which interval is between 15000-24.999 . plzzzzzz help meeeeee
Answer by stanbon(25959) (Show Source):
You can put this solution on YOUR website!i have a question in the grouped data in statistics , they asking the probability that an emloyee . taken at random , earns more than 20000 pounds , total workers are 68 and they have group which interval is between 15000-24.999 .
------------------
If the interval is 15000 to 24999
the range is 9999
-----------------------------
Use the median of the interval as the mean:
mean = (15000+24999)/2 = 19999.5
Rounding that up the mean is 20,000
-----------
If 20,000 is the mean, the probability an employee
picked at random earns more that 20,000 is 50%.
=====================================================
Cheers,
Stan H.
Question 235541: A car battery manufacturer introduces a new technique to increase the life of the battery. 25 batteries were tested and got a mean life of 4.6 years with a standard deviation of 0.70 years. Construct a 90% confidence interval for the mean life of the new battery.
THX
Answer by stanbon(25959) (Show Source):
You can put this solution on YOUR website!A car battery manufacturer introduces a new technique to increase the life of the battery. 25 batteries were tested and got a mean life of 4.6 years with a standard deviation of 0.70 years. Construct a 90% confidence interval for the mean life of the new battery.
------------------
x-bar = 4.6 yrs.
standard error = E = 1.645*0.7/sqrt(25) = 0.2303
Comment: Since you want 90% confidence you build 45% on each
side of the mean. That leaves 5% in each of the tails. The
z-value border for that 5% tail is 1.645 as you can get using
your z-chart or calculator. I used a TI calculator to get
invNorm(0.95) = 1.645.
-------------------
90% CI: 4.6-0.2303 < u < 4.6+0.2303
90% CI: 4.3697 < u < 4.8303
==================================
Cheers,
Stan H.
Question 235546: a manufacture of tv tubes has for many years used a process giving a mean tube life of 4900 hrs and a standard deviation of 1200 hours.
A new process is tried to see if it will increase the life significantly. A sample of 100 new tubes gave a mean life of 5000 hrs. Is the new process better than the old process at 1% level of significant? Construct hypotheses test it and interpret the result.
Answer by stanbon(25959) (Show Source):
You can put this solution on YOUR website!a manufacture of tv tubes has for many years used a process giving a mean tube life of 4900 hrs and a standard deviation of 1200 hours.
--------------------------------------------------------------------
A new process is tried to see if it will increase the life significantly.
A sample of 100 new tubes gave a mean life of 5000 hrs.
--------------------------------------------------------------------
Is the new process better than the old process at 1% level of significant? Construct hypotheses test it and interpret the result.
---------------------
Ho: u = 4900
Ha: u > 4900 (claim)
-------------------------
Test statistic: t(5000) = (5000-4900)/[1200/Sqrt(100)] =1000/1200= 0.8333..
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p-value: P(t>0.8333 with df=99) = 0.2033
---------------------------------
Since the p-value is greater than 1% do not reject Ho.
The new process is not an improvement over the old process.
============================================================
Cheers,
Stan H.
Question 235577: a miniature golf course offers a free game to golfers who make a hole-in-one on the last hole.Last week, 56 out of 270 golfers made a hole-in-one on the last hole.Find the experimental probability that a golfer makes a hole-in-one on the last hole.please round your answer to two decmial places.
Answer by stanbon(25959) (Show Source):
You can put this solution on YOUR website!a miniature golf course offers a free game to golfers who make a hole-in-one on the last hole.Last week, 56 out of 270 golfers made a hole-in-one on the last hole.Find the experimental probability that a golfer makes a hole-in-one on the last hole.please round your answer to two decmial places.
--------------------
Based on one week's data the probability is 56/270 = 0.2074..
Cheers,
Stan H.
Question 235550: The apollo space craft program lasted from 1967 to 1972 and incl 13 missions. The missions lasted from as little as 7 hrs to as long as 301 hrs. The duration of each flight is listed below:
9-195-241-301-216-260-7-244-192-147-10-295-142-
Is this dataset a sample or a population?
find the mmean and median of the flight time?
Find the range and quartile range of the flight times.
Answer by stanbon(25959) (Show Source):
You can put this solution on YOUR website!The apollo space craft program lasted from 1967 to 1972 and incl 13 missions. The missions lasted from as little as 7 hrs to as long as 301 hrs. The duration of each flight is listed below:
9-195-241-301-216-260-7-244-192-147-10-295-142-
Is this dataset a sample or a population?
It is the population of the set of flights described.
It is a sample from the set of all space flights.
---------------------------------------------------------
find the mmean and median of the flight time?
mean: add the numbers and divide by 13 to get 166.08
median is the middle number after putting the numbers in order: 195
Find the range and quartile range of the flight times.
range = max-min = 301-7 = 294
-----------------
I'll leave quartile range to you.
====================================
Cheers,
Stan H.
Question 235547: Tthe mean amount purchased by each customer at a casino is 23.50 euros. The population is positively skewed and the standard deviation is not known. A MARKETING REseARCHER TOOK A sample of 50 customers. If the standard deviation of the sample is 5 euros, what is the likelihood that the sample mean is greater than 22.5 but less then 25 euros
Answer by stanbon(25959) (Show Source):
You can put this solution on YOUR website!The mean amount purchased by each customer at a casino is 23.50 euros.
The population is positively skewed and the standard deviation is not known.
A MARKETING REseARCHER TOOK A sample of 50 customers.
If the standard deviation of the sample is 5 euros, what is the likelihood that the sample mean is greater than 22.5 but less then 25 euros
--------------
Although the population is not normally distributed, the sample means
will tend toward normality as n gets larger.
-----------------
Mean of the sample means: 23.50
If 5 euros is the std of the population, 5/Sqrt(50) is std of sample means:
t(25) = (25-23.5)/[5/sqrt(50)] = 2.1213
t(22.5) = (22.5-23.5)/[5/sqrt(50)] = -1.4142
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P(22.5 < x < 25) = P(-1.4142 < t < 2.1213 with df=49) = 0.8987 ===================================================================
Cheers,
Stan H.
Question 235379: the fm radio stations with high signal strength in new orleans breaks down into the following categories:
hip hop 2
religious 6
country 2
rock 3
talk radio 2
other 5
if someone driving in new orleans tuned to a fm station at random, what would be the chance he would get a relirious or talk radio station? give in fraction at lowest terms.
Answer by checkley77(6671) (Show Source):
Question 235316: An oil drilling company knows that it costs $25,000 to sink a test well. If oil is hit, the income for the drilling company will bve $425,000. If only natural gas is hit, the income will be $125,000. If nothing is hit, there will be no income. If the probability of hitting oil is 1/40 and the probability of hitting gas is 1/20, what is the expectation for the drilling company? Should the company sink the test well.
Answer by stanbon(25959) (Show Source):
You can put this solution on YOUR website!An oil drilling company knows that it costs $25,000 to sink a test well.
If oil is hit, the income for the drilling company will be $425,000.
If only natural gas is hit, the income will be $125,000.
If nothing is hit, there will be no income.
If the probability of hitting oil is 1/40 and the probability of hitting gas is 1/20, what is the expectation for the drilling company?
Should the company sink the test well.
------------
Let the random variable, x, represent company profit.
Values of "x": 400,000 , 100,000, -25000
Matched probabilities are (1/40) , 1/20, 37/40
----------------------------------------------------
Expected profit = (1/40)*400,000 + (2/40)100,000 - (35/40)(25000)
E(x) = [400,000+200,000-875000)/40
E(x) = -$6875
They can expect to lose that amount every time they drill.
================================================================
Cheers,
Stan H.
Question 235129: The pretzel pack company advertized that their pretzels weigh 2.5 ounces. The weights are normally distributed with mean 2.5 ounces and standard deviation 0.15 ounces. For a pretzel pack selected at random find the probabilty that it weighs less than 2.2 ounces and find the probability that it weighs more than 2.8 ounces.
Answer by stanbon(25959) (Show Source):
You can put this solution on YOUR website!The pretzel pack company advertized that their pretzels weigh 2.5 ounces.
The weights are normally distributed with mean 2.5 ounces and standard deviation 0.15 ounces.
For a pretzel pack selected at random find the probabilty that it weighs less than 2.2 ounces and find the probability that it weighs more than 2.8 ounces.
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z(2.2) = (2.2-2.5)/0.15 =-2
z(2.8) = (2.8-2.5)/0.15 = 2
------------------------
P(2.2 < x < 2.8) = P(-2 < z < 2) = 0.9545
=============================================
Cheers,
Stan H.
Question 235235: How many different ways can a couple have 3 daughters and 2 sons in five distinct pregnancies (no multiple births). list all gender string (GGGBB etc) and compute the probability this happens.
Answer by jim_thompson5910(13679) (Show Source):
You can put this solution on YOUR website!Here are all of the possible combinations of 5 children:
G,G,G,G,G
G,G,G,G,B
G,G,G,B,G
G,G,G,B,B
G,G,B,G,G
G,G,B,G,B
G,G,B,B,G
G,G,B,B,B
G,B,G,G,G
G,B,G,G,B
G,B,G,B,G
G,B,G,B,B
G,B,B,G,G
G,B,B,G,B
G,B,B,B,G
G,B,B,B,B
B,G,G,G,G
B,G,G,G,B
B,G,G,B,G
B,G,G,B,B
B,G,B,G,G
B,G,B,G,B
B,G,B,B,G
B,G,B,B,B
B,B,G,G,G
B,B,G,G,B
B,B,G,B,G
B,B,G,B,B
B,B,B,G,G
B,B,B,G,B
B,B,B,B,G
B,B,B,B,B
There are a total of 32 possible combinations.
From this list, we have the following combinations for 3 girls and 2 boys:
G,G,G,B,B
G,G,B,G,B
G,B,G,G,B
B,G,G,G,B
B,G,G,B,G
B,G,B,G,G
B,B,G,G,G
G,B,B,G,G
G,B,G,B,G
G,G,B,B,G
In this list, there are 10 different possible combinations.
So the probability of having 3 girls and 2 boys is:
P(3 girls & 2 boys) = 10/32 = 5/16 = 0.3125
which is a 31.25% chance
Question 235276: A factory worker places 86 newly created circuits on a shelf to be checked for quality. Of these, 6 will not work correctly. Suppose that she is asked to randomly select two circuits, without replacement, from the shelf. What is the chance that both circuits she selects will be defective? Approximate the soulution to the nearest ten thousandth.
Answer by stanbon(25959) (Show Source):
You can put this solution on YOUR website!A factory worker places 86 newly created circuits on a shelf to be checked for quality.
Of these, 6 will not work correctly.
Suppose that she is asked to randomly select two circuits, without replacement, from the shelf.
What is the chance that both circuits she selects will be defective? Approximate the soulution to the nearest ten thousandth.
---
Number of ways to select two defective from 6: 6C2 = 15
Nomber of ways to select two with no restriction: 86C2 = (86*85)/(1*2)
43*85 = 3655
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Ans: 15/3655 = 0.0041
============================
Cheers,
Stan H.
Question 235267: A bag contains 16 yellow marbles, 6 green marbles, and 7 red marbles. What is the chance of drawing a yellow marble? If a yellow marble is drawn the first time and then a second marble is drawn without replacement, what is the probability of drawing a second yellow marble. Give solution in reduced fraction form, separated by a comma.
Answer by stanbon(25959) (Show Source):
You can put this solution on YOUR website!A bag contains 16 yellow marbles, 6 green marbles, and 7 red marbles.
What is the chance of drawing a yellow marble?
Ans: P(yellow) = 16/29
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If a yellow marble is drawn the first time and then a second marble is drawn without replacement, what is the probability of drawing a second yellow marble.
P(2nd yellow | 1st yellow) = 15/28
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Cheers,
Stan H.
Question 235035: What is the probability that a 89% foul shooter will make at least 2 shots in a basketball game?
I know it is some sort of conditional probability, but I have now idea on what formula I need to use to solve it and my book is not very helpful.
Answer by stanbon(25959) (Show Source):
You can put this solution on YOUR website!What is the probability that a 89% foul shooter will make at least 2 shots in a basketball game?
--------------------------------------
P(2<= x <=89) = 1 - P(0<= x <= 1)
= 1 - [2C0(0.89)^0(0.11)2 + 2C1(0.89)(0.11)]
---
= 1 -[0.2079]
---
= 0.7921
====================
Cheers,
Stan H.
Question 235126: Ed earned 88 on his math midterm and 84 on his english midterm. In his math class the mean score was 75 with standard deviation 8. In his english class the mean score was 73 with standard deviation 5. Convert each score to a standard score, then find which was the higher score relative to the rest of the class.
Answer by stanbon(25959) (Show Source):
You can put this solution on YOUR website!Ed earned 88 on his math midterm and 84 on his english midterm. In his math class the mean score was 75 with standard deviation 8. In his english class the mean score was 73 with standard deviation 5. Convert each score to a standard score, then find which was the higher score relative to the rest of the class.
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Math z-score: (88-75)/8 = 1.625
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English z-score: (84-73)/5 = 2.2
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His English score is further above average and is a higher relative score.
==============================================================
Cheers,
Stan H.
Question 235136: The life span of one species of a maple tree is normally distributed with mean 20 years and standard deviation 2.3 years. If one tree is choosen at random, what is the probability it will live 21 years or longer? What is the probability that the average lifetime of 9 such trees will be 21 years or more?
Answer by stanbon(25959) (Show Source):
You can put this solution on YOUR website!The life span of one species of a maple tree is normally distributed with mean 20 years and standard deviation 2.3 years. If one tree is choosen at random, what is the probability it will live 21 years or longer?
---
z(21) = (21-20)/2.3 = 0.4348
P(x >= 21) = P(z >= 0.4348) = 0.3319
----------------------------------------------
What is the probability that the average lifetime of 9 such trees will be 21 years or more?
---
z(21) = (21-20)/[2.3/Sqrt(9)] = 1/[2.3/3] = 3/2.3 = 1.305
P(x-bar >= 21) = P(z >= 1.305) = 0.09606
=============================================
Cheers,
Stan H.
Question 235051: Evaluate the expression using a calculator
17!
Answer by Alan3354(5863)  (Show Source):
Question 235049: n! where n=12 and r=4
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r!(n-r)!
Answer by Alan3354(5863) (Show Source):
Question 235072: what is 5 to the 10th power?
Answer by Stitch(34) (Show Source):
Question 235056: five cards are dealt, what is the probability that the fifth one is a king
Answer by solver91311(4818) (Show Source):
You can put this solution on YOUR website!
Depends on whether you can see the value of the first four or not.
If you don't know the value of the first four cards, then there are 4 kings in 52 cards, or  .
On the other hand, if you can see the first four cards, then the denominator of the fraction changes from 52 to 48, that is 52 minus the 4 cards you can see leaving you 48 cards you don't know about. The numerator is a little more complicated.
The numerator of the probability depends on the number of kings that appear in the first 4 cards. Quite obviously, if the first 4 cards are themselves kings, then the probability that the fifth card will be a king is zero -- because there aren't any left.
So, let  represent the number of kings that appear in the first four cards, then the probability that the fifth card is a king is calculated by the following:
 ,
and you can say:
John
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