Question 1

Chauncey Billups,a current shooting guard for the Los Angeles Clippers,has a career free-throw percentage of 89.4%.Suppose he shoots six free throws in tonight's game.What is the expected number of free throws that Billups will make?

Accepted Answer

The expected number of free throws Billups will make is calculated by multiplying his free-throw percentage by the number of attempts. So, 0.894 (or 89.4%) times 6 equals approximately 5.364.

Question 2

Chauncey Billups,a current shooting guard for the Los Angeles Clippers,has a career free-throw percentage of 89.4%.Suppose he shoots six free throws in tonight's game.What is the probability that Billups makes all six free throws?

Accepted Answer

The probability of Chauncey Billups making all six free throws is found by raising his free-throw percentage to the sixth power, since each shot is an independent event. Given his free-throw percentage is 89.4% or 0.894, the calculation is $0.894^6 \approx 0.5105$.

Question 3

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 2011 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 2005.What is the probability that exactly 10 foreclosure auctions occurred during a randomly selected five-day week in 2011 in Boston?

Accepted Answer

To calculate the probability of exactly 10 foreclosure auctions occurring during a randomly selected five-day week in 2011 in Boston, we need to use the Poisson distribution formula:
P(X = 10) = (e^(-λ) * λ^x) / x!
where λ is the average number of foreclosure auctions per weekday and x is the number of foreclosure auctions we want to find the probability for.
We know that on average three residential properties were auctioned off each weekday in Boston in 2011, so λ = 3. Therefore:
P(X = 10) = (e^(-3) * 3^10) / 10!
P(X = 10) = 0.0486 (rounded to four decimal places)
Therefore, the probability that exactly 10 foreclosure auctions occurred during a randomly selected five-day week in 2011 in Boston is approximately 0.0486 or 4.86%. The best choice is B.

Question 4

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 2011 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 2005.What is the probability that no more than two foreclosure auctions occurred on a randomly selected weekday of 2011 in Boston?

Accepted Answer

The problem can be modeled using a Poisson distribution with an average rate (λ) of 3 auctions per weekday. The probability of having no more than two auctions is the sum of the probabilities of having 0, 1, or 2 auctions, which is calculated using the Poisson probability formula.

Question 5

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 6

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 7

According to a Department of Labor report,the city of Detroit had a 20% unemployment rate in May of 2011.Eight working-age residents were chosen at random.What was the expected number of unemployed residents,when eight working-age residents were randomly selected?

Accepted Answer

To find the expected number of unemployed residents, we need to multiply the total number of working-age residents (which is assumed to be very large and infinite) by the unemployment rate. 
So, 20% of the working-age population is unemployed, which means that the probability of any individual being unemployed is 0.2. 
If eight working-age residents are randomly selected, we can use the binomial distribution formula to find the expected number of unemployed residents: 
Expected number of unemployed residents = (number of trials) x (probability of success) 
Expected number of unemployed residents = 8 x 0.2 = 1.6 
Therefore, the best choice is B) 1.6.

Question 8

Chauncey Billups,a current shooting guard for the Los Angeles Clippers,has a career free-throw percentage of 89.4%.Suppose he shoots six free throws in tonight's game.What is the probability that Billups makes five or more of his free throws?

Accepted Answer

The answer of Chauncey Billups,a current shooting guard for the...

Question 9

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 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 probability that no earthquakes with a magnitude of 6.5 or greater strike the San Francisco Bay Area in the next 40 years?

Accepted Answer

The average rate of earthquakes is 5 per 100 years, or 0.05 per year. For 40 years, the expected number is 40 * 0.05 = 2. Using the Poisson distribution, the probability of 0 events is e^(-2) = 0.1353.

Question 11

Which of the following statements is the most accurate about a Poisson random variable?

Accepted Answer

A Poisson random variable counts the number of successes in a specified time or space interval, such as the number of cars passing through a toll booth in a given hour or the number of defects in a production line during a certain period of time. It is a discrete random variable, meaning it can only take on whole numbers. It does not have a bell-shaped distribution, as it can have a wide range of values with a lower probability of extreme values.

Question 12

Chauncey Billups,a current shooting guard for the Los Angeles Clippers,has a career free-throw percentage of 89.4%.Suppose he shoots six free throws in tonight's game.What is the standard deviation of the number of free throws that Billups will make?

Accepted Answer

The answer of Chauncey Billups,a current shooting guard for the...

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 given problem can be solved using Poisson distribution formula:
P(X ≥ 2) = 1 - P(X < 2) = 1 - P(X = 0) - P(X = 1)
where X is the number of customers entering the queue in a five-minute period.

Now, given the average rate of customer arrival, we can find the probability of X using Poisson distribution formula:
P(X=k)=[(λ^k * e^(-λ))/k!]
where λ=rate of customer arrival=2/5=0.4 customers per minute
and k=0,1,2,3,⋯

Using the formula, we get:
P(X=0) = [(0.4^0 * e^(-0.4))/0!] = 0.6703
P(X=1) = [(0.4^1 * e^(-0.4))/1!] = 0.2681

Therefore, 
P(X ≥ 2) = 1 - P(X < 2) = 1 - (P(X=0) + P(X=1)) = 1 - (0.6703+0.2681) = 0.0616.

Rounding the result to four decimal places gives 0.5940 as the answer, which is option 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 2011 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 2005.What is the probability that exactly four foreclosure auctions occurred on a randomly selected weekday of 2011 in Boston?

Accepted Answer

The problem can be modeled using a Poisson distribution, where the average rate (λ) is 3 foreclosures per weekday. The probability of exactly k events (k=4) is given by the formula P(X=k) = (e^(-λ) * λ^k) / k!. Plugging in the values, we get P(X=4) = (e^(-3) * 3^4) / 4! = 0.1680.

Question 15

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 16

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 17

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 can use the Poisson distribution formula to solve this problem.
Let lambda = average number of customers in a 5-minute period = 2
Then, mu = lambda * (time period / 5 minutes) = 2 * (15 / 5) = 6
We want to find P(X = 7), where X is the number of customers that enter the queue in a 15-minute period.
Using the Poisson formula, we get:
P(X = 7) = (e^-6)*(6^7)/7! = 0.1377
Therefore, the probability that exactly seven customers enter the queue in a randomly selected 15-minute period is 0.1377, which is option C.

Question 18

According to a Department of Labor report,the city of Detroit had a 20% unemployment rate in May of 2011.Eight working-age residents were chosen at random.What is the probability that exactly four residents were unemployed?

Accepted Answer

The situation can be modeled by a binomial distribution with n=8 and p=0.20 (probability of unemployment). The probability of exactly 4 residents being unemployed is given by: ${8 \choose 4} (0.20)^4 (1-0.20)^4 \approx 0.0459$

Question 19

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 2011 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 2005.What is the probability that at least one foreclosure auction occurred in Boston on a randomly selected weekday of 2011?

Accepted Answer

To find the probability of at least one foreclosure auction occurring on a randomly selected weekday of 2011, we need to find the complement probability of none of the auctions occurring.

The average number of residential properties auctioned off per weekday in 2011 = 3. 
The number of weekdays in 2011 = 365.

Therefore, the expected number of residential properties auctioned off in Boston in 2011 = 3 x 365 = 1095.

The probability of none of the auctions occurring on a randomly selected weekday = (1 - 1/1095)^365 = 0.0498 (rounded to four decimal places).

The probability of at least one auction occurring on a randomly selected weekday = 1 - 0.0498 = 0.9502 (rounded to four decimal places).

Therefore, the best answer is D, with a probability of 0.9502.

Question 20

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...

Quiz 5: Discrete Probability Distributions