Question 1

According to the IRS, 1.1% of tax returns will be audited next year. A random sample of 40 tax returns was selected. Use the Poisson distribution to approximate the binomial distribution to determine the probability that none of these tax returns will be audited.

Accepted Answer

The Poisson distribution can be used to approximate the binomial distribution for a large number of trials (n) and a small probability of success (p), with λ = np. Here, λ = 40 * 0.011 = 0.44. The probability that none of these tax returns will be audited (k=0) is given by P(X=0) = e^(-λ) * λ^0 / 0! = e^(-0.44) ≈ 0.6440.

Question 2

The --------- of a discrete probability distribution measures the dispersion of each outcome of the random variable from the mean of the distribution.

Accepted Answer

The standard deviation of a discrete probability distribution measures how spread out the values of a random variable are from the mean, indicating the dispersion of outcomes.

Question 3

Dwight Howard plays center in the National Basketball Association and averaged 1.8 blocked shots per game during a recent season. Assume that the number of blocked shots per game follows the Poisson distribution. What is the probability that Dwight Howard will block two or three shots during the next game?

Accepted Answer

The probability of Dwight Howard blocking two or three shots in a game, given that his average is 1.8 blocked shots per game and assuming a Poisson distribution, can be calculated using the Poisson probability formula: P(X=k) = (λ^k * e^(-λ)) / k!, where λ is the average number of events (1.8 in this case), and k is the number of events for which we're calculating the probability (2 or 3). The probabilities for 2 and 3 blocks are calculated separately and then added together. The calculation yields a combined probability of approximately 0.4285, making option B the correct answer.

Question 4

Which of the following is not a condition of a discrete probability distribution?

Accepted Answer

The condition that the probability of a success must exceed the probability of a failure is not a requirement for a discrete probability distribution. The key requirements are that the probabilities sum to 1, each probability is between 0 and 1, and outcomes are mutually exclusive.

Question 5

According to the Mortgage Bankers Association, 11% of U.S. mortgages were delinquent last year. A delinquent mortgage is one that has missed at least one payment but has not yet gone to foreclosure. A random sample of eight mortgages was selected. What is the probability that one or more of these mortgages is delinquent?

Accepted Answer

The probability of at least one delinquent mortgage is 1 minus the probability of none being delinquent. The probability of a mortgage not being delinquent is 0.89. For eight mortgages, this is 0.89^8. Subtracting this from 1 gives the probability of one or more being delinquent.

Question 6

According to the Mortgage Bankers Association, 11% of U.S. mortgages were delinquent last year. A delinquent mortgage is one that has missed at least one payment but has not yet gone to foreclosure. A random sample of eight mortgages was selected. What is the probability that less than three of these mortgages are delinquent?

Accepted Answer

This problem can be solved using the binomial probability formula, which is P(X = k) = (nCk) * p^k * (1-p)^(n-k), where n is the number of trials, k is the number of successful trials, p is the probability of success on a single trial, and nCk is the combination of n items taken k at a time. Here, n = 8 (the number of mortgages), p = 0.11 (the probability of a mortgage being delinquent), and we want the probability of k being less than 3, which means we need to calculate the probability for k = 0, 1, and 2, and then sum those probabilities.Calculating for k = 0, 1, and 2 and summing those probabilities gives a result close to 0.9794, indicating that there's a 97.94% chance that less than three of the selected mortgages are delinquent.

Question 7

According to LIMRA, 77% of husband-wife families with kids under 18 years old have life insurance. A random sample of six husband-wife families was selected. What is the probability that five or more families have life insurance?

Accepted Answer

The answer of According to LIMRA, 77% of husband-wife families...

Question 8

According to the Deloitte EV Survey, 5% of respondents indicated that they would be willing to pay an extra $500 or more for an electric vehicle versus one with a gasoline engine. A random sample of 25 adults has been selected. Use the Poisson distribution to approximate the binomial distribution to determine the probability that exactly one adult from this sample would be willing to pay an extra $500 or more for an electric vehicle versus one with a gasoline engine.

Accepted Answer

The answer of According to the Deloitte EV Survey, 5%...

Question 9

A company that bakes chocolate chip cookies averages 5.2 chocolate chips per cookie. Assume that the number of chocolate chips per cookie follows the Poisson distribution. What is the probability that a randomly selected cookie will contain two or more chocolate chips?

Accepted Answer

The probability that a randomly selected cookie will contain two or more chocolate chips can be found by subtracting the probability of finding 0 or 1 chocolate chips from 1, since the Poisson distribution's probabilities sum to 1. The formula for the Poisson probability is P(X=k) = (λ^k * e^(-λ)) / k!, where λ is the average rate (5.2 in this case), k is the number of occurrences (0 or 1), and e is the base of the natural logarithm. Calculating P(X=0) and P(X=1) and subtracting their sum from 1 gives the probability of having two or more chocolate chips.

Question 10

According to the Insurance Research Council, 14% of U.S. drivers are uninsured. A random sample of seven drivers was selected. What is the probability that three or more of these drivers are uninsured?

Accepted Answer

The probability of selecting three or more uninsured drivers is 0.0620.

Question 11

A statistics class at Wilmington College has 25 students of which 15 are math majors and 10 are business majors. The professor randomly selects five students to work together on a group project. What is the probability that less than three of the students in this group are business majors?

Accepted Answer

The probability that less than three of the students are business majors can be calculated using the hypergeometric distribution. The probability of selecting 0, 1, or 2 business majors from the group of 5 students is calculated and summed to get the final probability.

Question 12

A statistics class at Wilmington College has 25 students of which 15 are math majors and 10 are business majors. The professor randomly selects five students to work together on a group project. What is the standard deviation of this probability distribution if selecting a business major is defined as a success?

Accepted Answer

The answer of A statistics class at Wilmington College has...

Question 13

According to the National Health Interview Survey, 4.7% of U.S. adults 50 and older have had a total knee replacement. A random sample of 20 adults 50 and older has been selected. Use the Poisson distribution to approximate the binomial distribution to determine the probability that three or four adults from this sample have had a total knee replacement.

Accepted Answer

The Poisson distribution is a good approximation to the binomial distribution when the probability of success is small and the sample size is large. In this case, the probability of success is 4.7% and the sample size is 20, so the Poisson distribution is a good approximation. The probability of three or four adults from this sample having had a total knee replacement is given by the Poisson distribution with parameter lambda = 20 * 0.047 = 0.94. This probability is 0.0668.

Question 14

The mean and the variance are equal for a ------- probability distribution.

Accepted Answer

The Poisson distribution is characterized by the property that its mean and variance are equal. This distribution is often used to model the number of events occurring within a fixed interval of time or space under certain conditions.

Question 15

According to the Centers for Medicare & Medicaid Services, 21% of nursing homes in the United States received five stars in overall ratings on a scale of 1 to 5. A random sample of seven nursing homes was selected. What is the standard deviation of this distribution?

Accepted Answer

The standard deviation of a binomial distribution is the square root of np(1-p), where n is the sample size and p is the probability of success. In this case, n = 7 and p = 0.21, so the standard deviation is sqrt(7 * 0.21 * 0.79) = 1.078.

Question 16

According to the Centers for Medicare & Medicaid Services, 21% of nursing homes in the United States received five stars in overall ratings on a scale of 1 to 5. A random sample of seven nursing homes was selected. What is the probability that more than one of them received five stars?

Accepted Answer

The probability that more than one nursing home received five stars can be found by subtracting the probability of none or only one receiving five stars from 1. Let p = 0.21 (probability of a nursing home receiving five stars). The probability of exactly 0 or 1 nursing homes receiving five stars is calculated using the binomial probability formula. The complement of these probabilities gives the probability of more than one receiving five stars.$$ P(\text{more than one}) = 1 - [P(\text{exactly 0}) + P(\text{exactly 1})] $$$$ P(\text{exactly 0}) = \binom{7}{0} p^0 (1-p)^7 $$$$ P(\text{exactly 1}) = \binom{7}{1} p^1 (1-p)^6 $$$$ P(\text{more than one}) = 1 - [(1-0.21)^7 + 7 \times 0.21 \times (1-0.21)^6] $$Calculating these gives a value that matches the option for the probability of more than one nursing home receiving five stars, which is approximately 0.4506.

Question 17

According to LIMRA, 77% of husband-wife families with kids under 18 years old have life insurance. A random sample of six husband-wife families was selected. What is the probability that exactly four families have life insurance?

Accepted Answer

The probability can be calculated using the binomial probability formula: P(X=k) = C(n, k) * p^k * (1-p)^(n-k), where n=6, k=4, and p=0.77.

Question 18

Assume that the number of pieces of junk mail per day that a person receives in their mail box follows the Poisson distribution and averages 4.3 pieces per day. What is the probability that this person will receive exactly four pieces of junk mail over the next two days?

Accepted Answer

The answer of Assume that the number of pieces of...

Question 19

A camera manufacturer just shipped a batch of 36 digital cameras to local retailers and seven of the cameras have a defective sensor. Cutler Electronics received eight of these digital cameras. What is the standard deviation of this probability distribution?

Accepted Answer

The answer of A camera manufacturer just shipped a batch...

Question 20

According to LIMRA, 77% of husband-wife families with kids under 18 years old have life insurance. A random sample of six husband-wife families was selected. What is the mean of this distribution?

Accepted Answer

The mean of a binomial distribution is given by $n \times p$, where $n$ is the number of trials (in this case, 6 families) and $p$ is the probability of success (in this case, 0.77). Thus, the mean is $6 \times 0.77 = 4.62$.

Quiz 5: Reproductive Choices