Question 1

A bank manager estimates that an average of two customers enters the tellers' queue every five minutes. Assume that the number of customers that enters the tellers' queue is Poisson distributed. What is the probability that fewer than two customers enter the queue in a randomly selected five-minute period?

Accepted Answer

The answer of A bank manager estimates that an average...

Question 2

A bank manager estimates that an average of two customers enter the tellers' queue every five minutes. Assume that the number of customers that enter the tellers' queue is Poisson distributed. What is the probability that exactly three customers enter the queue in a randomly selected five-minute period?

Accepted Answer

We are given that the number of customers that enter the tellers' queue is Poisson distributed with an average of two customers every five minutes. This means that the parameter λ of the Poisson distribution is equal to 2. We want to find the probability that exactly three customers enter the queue in a randomly selected five-minute period. We can use the Poisson probability distribution to calculate this probability:

P(X = 3) = e^(-λ) * λ^3 / 3! = e^(-2) * 2^3 / 3! ≈ 0.1804

Therefore, the answer is B.

Question 3

According to geologists, the San Francisco Bay Area experiences five earthquakes with a magnitude of 6.5 or greater every 100 years. What is the standard deviation of the number of earthquakes with a magnitude of 6.5 or greater striking the San Francisco Bay Area in the next 40 years?

Accepted Answer

The answer of According to geologists, the San Francisco Bay...

Question 4

An urn contains 12 balls, five of which are red. Selection of a red ball is desired and is therefore considered to be a success. If three balls are selected, what is the expected value of the distribution of the number of selected red balls?

Accepted Answer

The expected value E(X) for the number of red balls selected is calculated using the formula for the hypergeometric distribution: E(X) = n * (k/N), where n is the number of draws (3), k is the number of successes in the population (5 red balls), and N is the population size (12 balls). Thus, E(X) = 3 * (5/12) = 1.25.

Question 5

The foreclosure crisis has been particularly devastating in housing markets in much of the south and west United States, but even when analysis is restricted to relatively strong housing markets the numbers are staggering. For example, in 2017 an average of three residential properties were auctioned off each weekday in the city of Boston, up from an average of one per week in 2011. What is the probability that at least one foreclosure auction occurred in Boston on a randomly selected weekday of 2017?

Accepted Answer

The probability of at least one foreclosure auction occurring on a randomly selected weekday can be found by subtracting the probability of no auctions occurring from 1. Given that an average of three residential properties were auctioned off each weekday, the probability of no auction occurring is very low. Thus, 1 minus a very low probability gives a high probability, which is 0.9502.

Question 6

The foreclosure crisis has been particularly devastating in housing markets in much of the south and west United States, but even when analysis is restricted to relatively strong housing markets the numbers are staggering. For example, in 2017 an average of three residential properties were auctioned off each weekday in the city of Boston, up from an average of one per week in 2011. What is the probability that no more than two foreclosure auctions occurred on a randomly selected weekday of 2017 in Boston?

Accepted Answer

The average number of auctions per weekday in 2017 is 3. Using the Poisson distribution, the probability of having no more than 2 auctions is the sum of probabilities of having 0, 1, or 2 auctions. This calculation results in approximately 0.4232.

Question 7

According to geologists, the San Francisco Bay Area experiences five earthquakes with a magnitude of 6.5 or greater every 100 years. What is the probability that more than two earthquakes with a magnitude of 6.5 or greater will strike the San Francisco Bay Area in the next 40 years?

Accepted Answer

The probability of more than two earthquakes in the next 40 years can be calculated using the Poisson distribution formula, where the average rate ($\lambda$) of events (earthquakes) in a given time period needs to be determined first. Given that there are 5 earthquakes every 100 years, the rate for 40 years is $5 \times \frac{40}{100} = 2$. The probability of having more than two earthquakes is the complement of the probability of having 0, 1, or 2 earthquakes. Using the Poisson formula $P(X=k) = \frac{e^{-\lambda} \lambda^k}{k!}$, we calculate the probabilities for 0, 1, and 2 earthquakes and subtract their sum from 1. The correct calculation aligns with option C, indicating it accounts for the complement of the probabilities of having 0, 1, or 2 earthquakes in the given period.

Question 8

A bank manager estimates that an average of two customers enters the tellers' queue every five minutes. Assume that the number of customers that enters the tellers' queue is Poisson distributed. What is the probability that exactly seven customers enter the queue in a randomly selected 15-minute period?

Accepted Answer

We know that $\lambda$, the average number of customers that enters the queue in a 5-minute period, is 2.

To find the probability that exactly 7 customers enter the queue in a 15-minute period, we need to use the Poisson distribution formula:

$$P(X=7) = \frac{e^{-\lambda}\lambda^x}{x!}$$

where x is the number of customers entering the queue, and $\lambda$ is the average rate of customers entering the queue.

Substituting the values, we get:

$$P(X=7) = \frac{e^{-2}\cdot 2^7}{7!} = 0.1377$$

Therefore, the answer is C.

Question 9

A roll of steel is manufactured on a processing line. The anticipated number of defects in a 10-foot segment of this roll is two. What is the probability of no defects in 10 feet of steel?

Accepted Answer

The situation described follows a Poisson distribution, where the average number of successes (defects, in this case) in a given interval is 2. The formula for the probability of observing exactly k successes is $P(k; \lambda) = \frac{e^{-\lambda} \lambda^k}{k!}$, where $\lambda$ is the average number of successes (2 defects per 10 feet), and $k$ is the number of successes we're interested in (0 defects). Plugging in the values, we get $P(0; 2) = \frac{e^{-2} 2^0}{0!} = e^{-2} \approx 0.1353$, indicating the probability of observing no defects in 10 feet of steel.

Question 10

According to geologists, the San Francisco Bay Area experiences five earthquakes with a magnitude of 6.5 or greater every 100 years. What is the expected value of the number of earthquakes with a magnitude of 6.5 or greater striking the San Francisco Bay Area in the next 40 years?

Accepted Answer

The expected value is calculated by taking the rate of earthquakes per year and multiplying it by the number of years in question. Given that there are 5 earthquakes every 100 years, the rate is 5/100 = 0.05 earthquakes per year. For 40 years, the expected number of earthquakes is 0.05 * 40 = 2.

Question 11

According to geologists, the San Francisco Bay Area experiences five earthquakes with a magnitude of 6.5 or greater every 100 years. What is the probability that one or more earthquakes with a magnitude of 6.5 or greater will strike the San Francisco Bay Area in the next year?

Accepted Answer

The probability of one or more earthquakes occurring in a given year is found by subtracting the probability of no earthquakes occurring from 1. If there are five earthquakes every 100 years, the probability of no earthquake in a given year is $ (1 - \frac{5}{100}) = 0.95 $. Thus, the probability of at least one earthquake is $ 1 - 0.95 = 0.05 $ or 0.0488 when calculated precisely.

Question 12

Suppose a baseball team has 14 players on the roster who are not members of the pitching staff. Of those 14 players, assume that three have recently taken a performance-enhancing drug. Suppose the league decides to randomly test five members of the team. What is the probability that exactly two of the tested players are found to have taken a performance-enhancing drug?

Accepted Answer

The probability that exactly two out of the five tested players have taken a performance-enhancing drug can be calculated using the binomial probability formula:
P(X = 2) = (C(3,2) * C(11,3)) / C(14,5)
where C(n,r) represents the number of combinations of n items taken r at a time.

Simplifying this expression gives:
P(X = 2) = (3 * 165) / 2002
P(X = 2) = 495 / 2002
P(X = 2) ≈ 0.2473

Therefore, the best choice is B, with a probability of 0.2473.

Question 13

A bank manager estimates that an average of two customers enters the tellers' queue every five minutes. Assume that the number of customers that enters the tellers' queue is Poisson distributed. What is the probability that at least two customers enter the queue in a randomly selected five-minute period?

Accepted Answer

The parameter for the Poisson distribution is λ = 2 (since on average, 2 customers enter the queue in a five-minute period). To find the probability that at least two customers enter the queue, we need to find P(X ≥ 2), where X is the number of customers entering the queue in a five-minute period. 
Using the Poisson distribution, we have: 
P(X ≥ 2) = 1 - P(X < 2) 
P(X < 2) = P(X = 0) + P(X = 1) 
P(X = 0) = e^-2 * (2^0 / 0!) = 0.1353 
P(X = 1) = e^-2 * (2^1 / 1!) = 0.2707 
P(X < 2) = 0.1353 + 0.2707 = 0.406 
P(X ≥ 2) = 1 - 0.406 = 0.594. Therefore, the answer is D.

Question 14

The foreclosure crisis has been particularly devastating in housing markets in much of the south and west United States, but even when analysis is restricted to relatively strong housing markets the numbers are staggering. For example, in 2017 an average of three residential properties were auctioned off each weekday in the city of Boston, up from an average of one per week in 2011. What is the probability that exactly 10 foreclosure auctions occurred during a randomly selected five-day week in 2017 in Boston?

Accepted Answer

We can use the Poisson distribution to solve this problem, since the number of foreclosure auctions per day follows a Poisson distribution with a mean of 3.

Let X be the number of foreclosure auctions during a five-day week in Boston. Then X follows a Poisson distribution with parameter λ = 3 * 5 = 15 (since the mean number of auctions over 5 days is just 3 * the number of days).

Using the Poisson probability formula, we find:

P(X = 10) = (e^(-15) * 15^10) / 10! ≈ 0.0486

Therefore, the probability of exactly 10 foreclosure auctions occurring during a randomly selected five-day week in 2017 in Boston is 0.0486 or approximately 4.86%.

The best choice is B because it correctly uses the Poisson distribution to calculate the probability.

Question 15

Which of the following is true about the hypergeometric distribution?

Accepted Answer

The hypergeometric distribution is used in situations where the trials are not independent because the probability of success depends on the outcome of previous trials. However, the probability of success may change from trial to trial because the population being sampled from is changing. Therefore, choice C is the correct answer.

Question 16

Suppose a baseball team has 14 players on the roster who are not members of the pitching staff. Of those 14 players, assume that three have recently taken a performance-enhancing drug. Suppose the league decides to randomly test five members of the team. What is the probability that at least one of the tested players is found to have taken a performance-enhancing drug?

Accepted Answer

The probability of at least one player being found to have taken a performance-enhancing drug is easier to calculate by finding the complement, which is the probability that none of the tested players have taken the drug, and then subtracting that from 1. There are 11 players who haven't taken the drug, so the probability of selecting 5 of these from the total 14 without selecting any of the 3 who did take the drug is (11 choose 5) / (14 choose 5). This gives 1 - (462/600) = 1 - 0.77 = 0.2308 for the complement, so the probability of at least one being found is 1 - 0.2308 = 0.7692.

Question 17

Cars arrive randomly at a tollbooth at a rate of 20 cars per 10 minutes during rush hour. What is the probability that exactly five cars will arrive over a five-minute interval during rush hour?

Accepted Answer

Since 20 cars arrive per 10 minutes, the arrival rate is 2 cars per minute during rush hour. The number of arrivals in a five-minute interval follows a Poisson distribution with a mean of $\lambda = (2	ext{ cars/minute})(5	ext{ minutes}) = 10	ext{ cars}$.
The probability of exactly 5 cars arriving in this interval is:
$P(X=5) = \dfrac{e^{-10}10^5}{5!} \approx 0.0378$
Therefore, the answer is choice A.

Question 18

According to geologists, the San Francisco Bay Area experiences five earthquakes with a magnitude of 6.5 or greater every 100 years. What is the probability that no earthquakes with a magnitude of 6.5 or greater will strike the San Francisco Bay Area in the next 40 years?

Accepted Answer

The probability of an earthquake with a magnitude of 6.5 or greater occurring in any given year is 5/100 = 0.05. Therefore, the probability of no such earthquake occurring in a given year is 1 - 0.05 = 0.95. Over 40 years, the probability of no such earthquakes occurring is $0.95^{40}$, which approximately equals 0.1353.

Question 19

An urn contains 12 balls, five of which are red. The selection of a red ball is desired and is therefore considered to be a success. If a person draws three balls from the urn, what is the probability of two successes?

Accepted Answer

The probability of a success (i.e. selecting a red ball) is P(S) = 5/12. The probability of a failure (i.e. selecting a non-red ball) is P(F) = 7/12. 
To find the probability of two successes, we need to use the binomial probability formula:
P(2 successes) = (3 choose 2) * (5/12)^2 * (7/12)^1
= 3 * (25/144) * (7/12)
= 0.3182
Therefore, the best choice is B.

Question 20

There are currently 18 pit bulls at the pound. Of the 18 pit bulls, four have attacked another dog in the last year. Joe, a member of the staff, randomly selects six of the pit bulls for his group. What is the probability that exactly one of the pit bulls in Joe's group attacked another dog last year?

Accepted Answer

To find the probability that exactly one of the pit bulls in Joe's group attacked another dog last year, we need to use the binomial distribution. Letting $p$ be the probability that any one of the pit bulls in the pound attacked another dog last year, we have $p=\frac{4}{18}=\frac{2}{9}$. The probability of exactly one success (i.e. one pit bull that attacked another dog) in six trials (i.e. six pit bulls selected for Joe's group) is given by the binomial probability:
$$P(X=1) = \binom{6}{1}p^1(1-p)^5 = 6\left(\frac{2}{9}ight)\left(\frac{7}{9}ight)^5 \approx 0.4314$$
Therefore, the best choice is $\boxed{	extbf{(C)}}$.

Quiz 5: Discrete Probability Distributions