STAT Excel Test 2

The proportions of families with various numbers of children age 18 or under in a small town are given in the following table. One family is randomly selected from this town. X 0 1 2 3 4 5 P(X) 0.1 0.4 0.3 0.1 0.05 0.05 Find the probability that the selected family has at least 3 children age 18 or under. 0.20 0.40 0.10 0.30 0.50

0.20

A woman buys 20 one-dollar lottery tickets per month. The probability of any ticket being a winning ticket is 0.10 or 10%. Which of the following shows the correct EXCEL formula to find the probability that in any one month, at least three of the tickers that the woman buys are winning tickets? =1 - BINOM.DIST(2, 20, 0.10, TRUE) =BINOM.DIST(2, 20, 0.10, TRUE) =1 - BINOM.DIST(2, 20, 0.10, FALSE) =BINOM.DIST(2, 20, 0.10, FALSE) =1 - NORM.DIST(3, 20, 0.10, TRUE) =1 - BINOM.DIST(3, 20, 0.10, TRUE)

= 1 - BINOM.DIST(2, 20, 0.10, TRUE)

A city government asks 500 randomly selected people whether or not they are employed. The population percentage of employment is 0.60. Which of the following shows the correct EXCEL formula to calculate the probability that at least 426 of these people are employed? = 1 - BINOM.DIST(426, 500, 0.60, TRUE) =BINOM.DIST(425, 500, 0.60, FALSE) = 1 - BINOM.DIST(425, 500, 0.60, TRUE) = BINOM.DIST(425, 500, 1 - 0.60, TRUE) = NORM.DIST(425, 500, 1 - 0.60, TRUE)

= 1 - BINOM.DIST(425, 500, 0.60, TRUE)

An inspection on 160 parts made from two production lines at a factory yields the following table. A part is randomly selected from these 160 parts. The probability that this part is defective or is made from production line 1 is about =(147+70-65)/160 =(147+90-82)/160 =(13+90-8)/160 =(13+70-5)/160 =(70+90-8)/160

=(13+70-5)/160

At Hopewell Electronics, all 140 employees were asked about their political affiliations: Democrat, Republican or Independent. The employees were grouped by type of work, as executives or production workers. The results with row and column totals are shown in the following table. Suppose an employee is selected at random from the 140 Hopewell employees. Democrat Republican Independent Total Executive 5 34 9 48 Production Worker 63 21 8 92 Total 68 55 17 140 The probability that this employee is a production worker or is a Republican is about _________________. =(92+55-21)/140 =(92+55)/140 =(48+55-34)/140 =(92+55-21)/92 =21

=(92+55-21)/140

Cora is playing a game that involves flipping three coins at once. Let the random variable X be the number of coins that land showing "heads". Here is the probability distribution for X: 0 1 2 5 P(X) 0.125 0.375 0.375 0.125 Which of the following calculates the expected value of the number of coins that land showing "heads"? =AVERAGE(E4:H5) =AVERAGE(E5:H5) =PRODUCTSUM(E4:H4, E5:H5) =(0+1+2+3)*(0.125+0.375+0.375+0.125) =0*0.125 + 1*0.375 + 2*0.375 + 3*0.125

=0*0.125 + 1*0.375 + 2*0.375 + 3*0.125

The number of baseball players unavailable due to injury for a game is summarized by the following probability distribution: x: 0 1 2 3 4 5 P(x): 0.28 0.27 0.17 0.12 0.10 0.06 What is the probability that no more than 2 players will be unavailable for a randomly selected game? =0.17 =0.17 + 0.12 + 0.10 + 0.06 =0.28 + 0.27 + 0.17 =0.28 + 0.27 =0.12 + 0.10 + 0.06

=0.28 + 0.27 + 0.17

Calculate P(z≥1.5), where z is the standard normal random variable. =BINOM.DIST(1.5, 0, 1, TRUE) =1 - NORM.DIST(1.5, 0, 1, TRUE) =1 - BINOM.DIST(1.5, 0, 1, FALSE) =1 - NORM.DIST(1.5, 0, 1, FALSE) =NORM.DIST(1.5, 0, 1, TRUE)

=1 - NORM.DIST(1.5, 0, 1, TRUE)

Let Z be the standard normal random variable. Find P(Z>−0.50) =1 - NORM.DIST(-0.50, 0, 1, FALSE) =1 - NORM.DIST(0.50, 0, 1, TRUE) =1 - NORM.DIST(-0.50, 0, 1, TRUE) =NORM.DIST(-0.50, 0, 1, FALSE) =NORM.DIST(-0.50, 0, 1, TRUE)

=1-NORM.DIST(-0.50, 0, 1, FALSE)

Assume that children's IQs (Age 6-12) follow a normal distribution with mean 100 and standard deviation of 12. Find the probability that a randomly selected child has IQ above 118. =1-NORM.INV(118, 100, 12, TRUE) =1-NORM.DIST(118, 100, 100/SQRT(12), TRUE) =1-NORM.DIST(118, 100, 12, TRUE) =NORM.DIST(118, 100, 12, TRUE) =1-T.DIST(118, 100, 12, TRUE)

=1-NORM.DIST(118, 100, 12, TRUE)

The actual amount of hot water dispensed by a coffee brewer, when it is set to dispense 8.0 ounces, is normally distributed with mean 8.0 ounce and standard deviation 0.2 oz. The probability that a randomly selected nominal 8.0 ounce cup of coffee will be brewed with more than 8.125 ounces is about: =NORM.DIST(8.125, 8.0, 0.2, TRUE) =NORM.DIST(8.125, 8.0, 0.2/SQRT(8.0), TRUE) =1-NORM.DIST(8.125, 8.0, 0.2, TRUE) =1-NORM.DIST(8.125, 8.0, 0.2, FALSE) =NORM.INV(8.125, 8.0, 0.2, TRUE)

=1-NORM.DIST(8.125, 8.0, 0.2, TRUE)

The table shows the results of a survey in which 90 dog owners were asked how much they have spent in the last year for their dog's health care, and whether their dogs were purebred or mixed breeds. Find the probability that a randomly selected dog owner spent at least $100 on health care and the dog was a mixed breed. =71/90 =15/36 =15/50 =15/90 =86/90

=15/90

The following two-way contingency table gives the breakdown of the voters in a particular locale according to gender and political party preference. A person is selected at random from this population. Democratic Republican total Male 240 280 520 Female 290 190 480 total 530 470 1000 Given that a randomly selected voter is a male, find the probability that he is a Democratic. =240/520 =240/1000 =240/530 =280/520 =280/470

=240/250

The table shows the results of a survey in which 90 dog owners were asked how much they have spent in the last year for their dog's health care, and whether their dogs were purebred or mixed breeds. Find the probability that less than $100 was spent on a randomly selected dog's health care in the last year. =40/50 =50/90 =21/40 =40/90 =19/40

=40/90

The manager of the dairy section of a large supermarket chose a random sample of 550 egg cartons and found that 80 cartons had at least one broken egg. let p denote the proportion of all cartons which have at least one broken egg. Which of the following calculates the point estimate for p and also the critical value for the 96% confidence interval for p? =80/550 and =NORM.INV(1-0.04/2, 0, 1) =550/80 and =NORM.INV(1-0.04/2, 0, 1) =80/550 and =NORM.DIST(1-0.04/2, 0, 1, TRUE) =470 and =NORM.INV(1-0.96/2, 0, 1) =80 and =NORM.INV(1-0.40/2, 0, 1) =470/550 and =NORM.INV(1-0.04/2, 80, 550)

=80/550 and =NORM.INV(1-0.04/2, 0, 1)

The manager of the dairy section of a large supermarket chose a random sample of 550 egg cartons and found that 80 cartons had at least one broken egg. let p denote the proportion of all cartons which have at least one broken egg. Which of the following calculates the point estimate for p and also the critical value for the 96% confidence interval for p? =550/80 and =NORM.INV(1-0.04/2, 0, 1) =80/550 and =NORM.DIST(1-0.04/2, 0, 1, TRUE) =80/550 and =NORM.INV(1-0.04/2, 0, 1) =470/550 and =NORM.INV(1-0.04/2, 80, 550) =80 and =NORM.INV(1-0.40/2, 0, 1) =470 and =NORM.INV(1-0.96/2, 0, 1)

=80/550 and =NORM.INV(1-0.04/2,0,1)

You are taking a multiple-choice quiz that consists of five questions. Each question had four possible answers, only one of which is correct. To complete the quiz, you randomly guess the answer to each question. Which of the following shows the correct EXCEL formula to compute the probability of guessing less than three answers correctly. =1 - BINOM.DIST(2, 5, 0.25, FALSE) =NORM.DIST(2, 5, 0.25, TRUE) =BINOM.DIST(3, 5, 0.25, FALSE) =BINOM.DIST(2, 5, 0.25, TRUE) =1 - NORM.DIST(2, 5, 0.25, TRUE)

=BINOM.DIST(2, 5, 0.25, TRUE)

A city government asks 500 randomly selected people whether or not they are employed. The population percentage of employment is 0.60. Which equation would calculate the probability that exactly 250 of these people are employed? BINOM.DIST(250, 500, 0.60, 0.50) =BINOM.DIST(250, 500, 0.60, FALSE) =250/500 =BINOM.DIST(250, 500, 0.60, TRUE) =BINOM.DIST(250, 500, 0.50, FALSE)

=BINOM.DIST(250, 500, 0.60, FALSE)

You are taking a multiple-choice quiz that consists of twenty questions. Each question had six possible answers, only one of which is correct. To complete the quiz, you randomly guess the answer to each question. Which of the following shows the correct EXCEL formula to compute the probability of guessing less than five answers correctly. =BINOM.DIST(4, 20, 1/6, TRUE) =1 - BINOM.DIST(4, 20, 1/6, FALSE) =BINOM.DIST(5, 20, 1/6, FALSE) =NORM.DIST(4, 20, 1/6, TRUE) =1 - NORM.DIST(4, 20, 1/6, TRUE)

=BINOM.DIST(4, 20, 1/6, TRUE)

An insurance company study shows that 60% of the auto insurance claims submitted for property damage were submitted by males under 25 years of age. Suppose 8 property damage claims involving automobiles are selected at random from that region. Let X be the number of claims (among the 8 selected) that were made by males under the age of 25. To find the probability that exactly 6 of the 8 claims are made by males under 25, we would use the excel function =BINOM.DIST(6,8,0.6,TRUE) =NORM.DIST(6,8,0.6,FALSE) =NORM.DIST(6,8,0.6,TRUE) =T.DIST(6,8,0.6,TRUE) =BINOM.DIST(6,8,0.6,FALSE)

=BINOM.DIST(6,8,0.6,FALSE)

A physician wanted to estimate the mean length of time that a patient had to wait to see him after arriving at the office. A random sample of 100 patients showed a mean waiting time of 30 minutes and a standard deviation of 8 minutes. Which of the following calculates the margin of error (EBM) for the 86% confidence interval for the mean waiting time? =CONFIDENCE.T(0.14/2, 8, 100) =CONFIDENCE.T(0.14, 8, 100) =CONFIDENCE.NORM(0.14/2, 30, 8, 100) =CONFIDENCE.NORM(0.86, 8, 100) =CONFIDENCE.NORM(0.14, 30, 8/SQRT(100)) =CONFIDENCE.NORM(0.14, 8, 100)

=CONFIDENCE.NORM(0.14, 8, 100)

Household credit card debt in a certain region has a mean of $16,890 and a standard deviation of $4,220. Calculate the probability that the mean credit card debt of a random sample of 35 households would be $16,000 or less. =NORM.DIST(16000, 16890, 16890/SQRT(35), TRUE) =NORM.DIST(16000, 16890, 4220, TRUE) =NORM.DIST(16890, 16000, 4220/SQRT(35), TRUE) =NORM.DIST(16000, 16890, 4220/SQRT(35), TRUE) =1-NORM.DIST(16000, 16890, 4220/SQRT(35), TRUE)

=NORM.DIST(16000, 16890, 4220/SQRT(35), TRUE)

The length of a wild lemur's tail has a normal distribution with a mean of 1.95 feet with a standard deviation of 0.2 feet. A random sample of 64 lemurs is selected. Calculate the probability that the average of their tail lengths is between 1.9 and 2.11 feet. =NORM.DIST(2.11, 1.95, 0.2, TRUE) - NORM.DIST(1.9, 1.95, 0.2, TRUE) =NORM.INV(2.11, 1.95, 0.2/SQRT(64), TRUE) - NORM.INV(1.9, 1.95, 0.2/SQRT(64), TRUE) =NORM.DIST(2.11, 1.95, 0.2/SQRT(64), FALSE) - NORM.DIST(1.9, 1.95, 0.2/SQRT(64), FALSE) =NORM.DIST(1.9, 1.95, 0.2/SQRT(64), TRUE) - NORM.DIST(2.11, 1.95, 0.2/SQRT(64), TRUE) =NORM.DIST(2.11, 1.95, 0.2/SQRT(64), TRUE) - NORM.DIST(1.9, 1.95, 0.2/SQRT(64), TRUE)

=NORM.DIST(2.11, 1.95, 0.2/SQRT(64), TRUE) - NORM.DIST(1.9, 1.95, 0.2/SQRT(64), TRUE)

Scores on the common final exam in the Elementary Statistics course are normally distributed with a mean of 75 and a standard deviation of 10. The department has the rule that in order to receive an A in the course his score must be in the top 25% of all exam scores. Which of the following shows the correct EXCEL formula to calculate the minimum IQ score required by this program? =1 - NORM.INV(0.25, 75, 10) =NORM.INV(75, 75, 10) =NORM.INV(0.25, 75, 10) =NORM.INV(0.75, 75, 10) =T.INV(0.75, 75, 10)

=NORM.INV(0.75, 75, 10)

In a lottery game, players pay $2 for a ticket. Out of each batch of 1000 tickets, 1 ticket wins $500, 10 win $100, and 89 of them win $10. The remaining tickets have no prize awarded. Let the random variable X be the profits earned from the lottery. Here is the probability distribution for X: $500 $100 $10 $0 Profit x: $498 $98 $8 -$2 Probability: 0.001 0.01 0.098 0.9 Which of the following excel function calculates the expected value to the player that purchases one of these tickets? =SUMOFTHEPRODUCT(B2:E2, B3:E3) =SUM(B2:E2)*(B3:E3) =AVERAGE(B2:E3) =SUMPRODUCT(B2:E2, B3:E3) =SUM(B2:E2)*SUM(B3:E3)

=SUMPRODUCT(B2:E2, B3:E3)

The mean monthly salary of a random sample of 20 college graduates under the age of 30 was found to be $1320 with a standard deviation of $677. Assume that the distribution of salaries for all college graduates under the age of 30 is normally distributed. All other information remaining unchanged, which of the following would produce a narrower interval than the 80% confidence interval constructed based on this sample? A sample with a standard deviation of 725 instead of 677. A sample with a standard deviation of 800 instead of 677. A 90% confidence interval rather than a 80% confidence interval. A sample of size 15 instead of 20. A sample of size 50 instead of 20.

A sample of size 50 instead of 20.

A survey is conducted to estimate the average household income in a large metropolitan area. A random sample of 150 households in this area yielded an average household income of $55,000 with a standard deviation of $13,200. All other information remaining unchanged, which of the following would produce a wider interval than the 90% confidence interval constructed from this sample? A sample with a standard deviation of $10,000 instead of $13,200. An 80% confidence interval rather than a 90% confidence interval. A sample with a standard deviation of $11,670 instead of $13,200. A sample size of 100 instead of 150. A sample size of 175 instead of 150.

A sample size of 100 instead of 150.

In a survey of 1000 people, 700 people said that they voted in the last presidential election. Let p denote the proportion of all people who voted. Which of the following actions would result in a confidence interval narrower than the 90% confidence interval computed from this sample? Computing a 95% confidence interval rather than a 90% confidence interval Decreasing the sample size Computing a 99% confidence interval rather than a 90% confidence interval None of the answer choices Computing a 80% confidence interval rather than a 90% confidence interval

Computing a 80% confidence interval rather than a 90% confidence interval

A random variable X has a normal distribution. The distributions shown below are the distributions of the sample averages of X (i.e. x̄). One of the distributions is for the averages of a sample of size 30 (n = 30), and the other distribution is for the averages of samples of size 60 (n = 60). Which distribution is for the sample of size 60? A is taller and narrower. B is wider and short bell shaped curves Distribution A Distribution B It is impossible to say

Distribution A

A random variable X has a normal distribution. The distributions shown below are the distributions of the sample averages of X (i.e. x̄). One of the distributions is for the averages of a sample of size 30 (n = 30), and the other distribution is for the averages of samples of size 60 (n = 60). Which distribution is for the sample of size 30? A is taller and narrower. B is wider and short bell shaped curves Distribution A Distribution B It is impossible to say

Distribution B

In a survey of 1002 ​adults, a polling agency​ asked, "When you​ retire, do you think you will have enough money to live comfortably or not?". Of the 1002 surveyed, 525 stated that they were worried about having enough money to live comfortably in retirement. An economist calculated the 90​% confidence interval for the proportion of adults who are worried about having enough money to live comfortably in retirement and he got 0.498 for the lower bound, and 0.550 for the upper bound. Interpret the interval. There is a 90​% probability that the true proportion of worried adults in the population is between 49.8% and 55%. There is 90​% confidence that the true proportion of worried adults in the population is 525/1002. There is 90​% confidence that the true proportion of worried adults in the population is between 49.8% and 55%. This tells us nothing about the true proportion of worried adults in the population. 90% of the population lies in the interval between 49.8% and 55%.

There is 90​% confidence that the true proportion of worried adults in the population is between 49.8% and 55%.

A physician wanted to estimate the mean length of time that a patient had to wait to see him after arriving at the office. A random sample of 64 patients showed a mean waiting time of 30 minutes and a standard deviation of 10 minutes. Which of the following calculates the margin of error (EBM) for the 87% confidence interval for the mean waiting time? =CONFIDENCE.T(0.13, 10, 64) =CONFIDENCE.NORM(0.87, 10, 64) =CONFIDENCE.T(0.13, 30, 10/SQRT(64)) =CONFIDENCE.NORM(0.13, 10, 64) =CONFIDENCE.NORM(0.13/2, 10, 64) =CONFIDENCE.NORM(0.13, 30, 10/SQRT(64))

VIEW=CONFIDENCE.NORM(0.13, 10,64)

Last Wednesday, a random sample of 16 students was surveyed to find how long it takes to walk from the Fretwell building to the College of Education building. The survey team found a sample mean of 10 minutes with a standard deviation of 1.6 minutes. Assuming walking times from Fretwell to the College of Education are normally distributed, which of the following calculates the margin of error (EBM) for the 89% confidence interval for the population mean of walking times? =CONFIDENCE.NORM(0.89, 1.6, 10) =CONFIDENCE.T(0.11, 1.6, 10) =CONFIDENCE.NORM(0.11, 1.6, 16) =CONFIDENCE.NORM(0.11, 1.6, 10) =CONFIDENCE.T(0.11/2, 1.6, 10/SQRT(16)) =CONFIDENCE.T(0.11, 1.6, 16)

VIEW=CONFIDENCE.T(0.11, 1.6, 16)

Last Wednesday, a random sample of 12 students was surveyed to find how long it takes to walk from the Fretwell building to the College of Education building. The survey team found a sample mean of 10 minutes with a standard deviation of 2.5 minutes. Assuming walking times from Fretwell to the College of Education are normally distributed, which of the following calculates the margin of error (EBM) for the 87% confidence interval for the population mean of walking times? =CONFIDENCE.T(1-0.87/2, 2.5, 12) =CONFIDENCE.NORM(0.13, 2.5, 12) =CONFIDENCE.NORM(1-0.13/2, 2.5, 12) =CONFIDENCE.T(0.13, 2.5, 10/SQRT(12)) =CONFIDENCE.T(0.13, 2.5, 10) =CONFIDENCE.T(0.13, 2.5, 12)

VIEW=CONFIDENCE.T(0.13, 2.5, 12)

In a random sample of 32 criminals convicted of a certain​ crime, it was determined that the mean length of sentencing was 57 ​months, with a standard deviation of 12 months. A district attorney calculated the 95​% confidence interval for the mean length of sentencing for this crime and he got 52.7 for the lower bound and 61.3 for the upper bound. Interpret the interval. We can be 95​% confident that the mean length of sentencing for the crime is between 52.7 and 61.3 months. There is 95​% confidence that the mean length of sentencing time for this crime is about 57 months. 95​% of the sentences for the crime are between 52.7 and 61.3 months. There is a 95​% probability that the mean length of sentencing for the crime is between 52.7 and 61.3 months. This tells us nothing about the mean length of sentencing time for this crime.

We can be 95​% confident that the mean length of sentencing for the crime is between 52.7 and 61.3 months.

A manufacturer claims that the mean weight of its ice cream cartons is 10ounces with a standard deviation of 0.6 ounce. Assuming 36 cartons are selected. Let xbar represent the mean sales price of the sample. Find the mean and standard deviation of xbar mean: 10, st.dev: 36 mean: 10, st.dev: 0.6/sqrt(36) mean: 36, st.dev: 0.6/sqrt/(36) mean: 36, st.dev: 0.6 mean: 10, st.dev: 0.6

mean: 10, st.dev: 0.6/sqrt(36)

The average sales price of single-family houses in Mooresville is $345,000 with a standard deviation of $15,000. A random sample of 80 single-family houses in Mooresville is selected. Let xbar represent the mean sales price of the sample. Find the mean and standard deviation of xbar. mean=345000, st.dev=15000/sqrt(80) mean=345000, st.dev=345000/sqrt(80) mean=345000/sqrt(80), st.dev=15000 mean=345000, st.dev=15000 mean=345000, st.dev=15000/80

mean=345000, st.dev=15000/sqrt(80)

A survey in a community states that 660 out of 800 people smoke on a regular basis. Using the information from this survey, a researcher wishes to estimate the required sample size. He wants to be 95% confident and wants the sample proportion to be within 1% of the population proportion. Which of the following is correct? p' = 660; q' = 140; EBP = 0.01 p' = 140/800; q' = 660/800; EBP = NORM.INV(1-0.05/2, 0, 1) p' = 140; q' = 660; EBP =NORM.INV(1-0.01/2, 0, 1) p' = 660/800; q' = 140/800; EBP = 0.05 p' = 660/800; q' = 140/800; EBP = 0.01

p' = 660/800; q' = 140/800; EBP = 0.01

Which of the following random variables is continuous? the number of heads tossed on four distinct coins the height of an elephant to the nearest inch the weight of a baby giraffe The number of passengers in a ca

the weight of a baby giraffe

Which of the following random variables is continuous? the weight of a chicken the number of tacos a teenager can eat in one sitting the number of quarters in a cash register the total rolled on two fair 6-sided dice

the weight of a chicken