Question 1

If tubing averages 16 defects per 100 meters, what is the probability of finding exactly 2 defects in a randomly chosen 10-meter piece of tubing?

Accepted Answer

The answer of If tubing averages 16 defects per 100...

Question 2

The probability that a certain daily flight's departure from ORD to LAX is delayed is .02. Over six months, this flight departs 180 times. What is the approximate Poisson probability that it will be delayed fewer than 2 times?

Accepted Answer

We have the following information:

λ = expected number of delayed flights in 6 months = (0.02) * 180 = 3.6

We want to find P(X < 2), where X is the number of delayed flights in 6 months.

Using the Poisson probability formula:

P(X < 2) = e^(-λ) * (λ^0/0! + λ^1/1!)

P(X < 2) = e^(-3.6) * (1 + 3.6)

P(X < 2) = 0.1257

Therefore, the best choice is (C).

Question 3

The domain of X in a Poisson probability distribution is discrete and can include:

Accepted Answer

The domain of a Poisson probability distribution is discrete and consists of nonnegative integer values only, since the Poisson distribution models the number of events occurring in a fixed interval of time or space. Any real number or negative values are not valid for the Poisson distribution, hence options A and D can be eliminated. Option B is incorrect because although the Poisson distribution is primarily used to model count data, not all count data follows a Poisson distribution, and in such cases, non-integer values may be included.

Question 4

Historically, 2 percent of the stray dogs in Southfield are unlicensed. On a randomly chosen day, the Southfield city animal control officer picks up seven stray dogs. What is the probability that fewer than two will be unlicensed?

Accepted Answer

The probability that fewer than two out of seven stray dogs are unlicensed can be calculated using the binomial probability formula. The probability of 0 unlicensed dogs is (0.98)^7, and the probability of 1 unlicensed dog is 7*(0.02)*(0.98)^6. Adding these probabilities gives approximately 0.9921.

Question 5

On a Sunday in April, dog bite victims arrive at Carver Memorial Hospital at a historical rate of 0.6 victim per day. On a given Sunday in April, what is the probability that exactly two dog bite victims will arrive?

Accepted Answer

The probability of exactly two dog bite victims arriving at Carver Memorial Hospital on a given Sunday in April, when the historical rate is 0.6 victim per day, can be calculated using the Poisson distribution formula: P(x; μ) = (e^(-μ) * μ^x) / x!, where μ is the average rate (0.6 in this case), x is the number of occurrences (2 victims), and e is the base of the natural logarithm (approximately 2.71828). Plugging in the values gives P(2; 0.6) = (e^(-0.6) * 0.6^2) / 2! = (0.5488 * 0.36) / 2 = 0.0988.

Question 6

The probability that a ticket holder will miss a flight is .005. If 180 passengers take the flight, what is the approximate Poisson probability that at least 2 will miss the flight?

Accepted Answer

The answer of The probability that a ticket holder will...

Question 7

On Saturday morning, calls arrive at TicketMaster at a rate of 108 calls per hour. What is the probability of fewer than three calls in a randomly chosen minute?

Accepted Answer

The answer of On Saturday morning, calls arrive at TicketMaster...

Question 8

For which binomial distribution would a Poisson approximation not be acceptable?

Accepted Answer

For a Poisson approximation to be acceptable, the expected value (λ = nπ) should be at least 5. In this case, λ = 35 x 0.07 = 2.45 which is less than 5. Therefore, a Poisson approximation would not be acceptable for this binomial distribution.

Question 9

The probability that a rental car will be stolen is 0.0004. If 3500 cars are rented, what is the approximate Poisson probability that 2 or fewer will be stolen?

Accepted Answer

The Poisson probability formula is $P(X=k) = \frac{e^{-\lambda} \lambda^k}{k!}$, where $\lambda$ is the average rate (mean) of occurrence and $k$ is the number of occurrences. Here, $\lambda = 3500 \times 0.0004 = 1.4$. To find the probability of 2 or fewer cars being stolen, we calculate the sum of probabilities for 0, 1, and 2 cars being stolen using the Poisson formula and sum them up. The calculation yields approximately 0.835, or 83.35% when converted to a percentage, which corresponds to choice D.

Question 10

The true proportion of accounts receivable with some kind of error is .02 for Venal Enterprises. If an auditor randomly samples 200 accounts receivable, what is the approximate Poisson probability that fewer than two will contain errors?

Accepted Answer

The Poisson probability can be calculated as follows:
λ = np = 200 * 0.02 = 4
P(X < 2) = P(X = 0) + P(X = 1)
= e^(-λ) * λ^0 / 0! + e^(-λ) * λ^1 / 1!
= e^(-4) * 4^0 / 0! + e^(-4) * 4^1 / 1!
≈ 0.0916
Therefore, the approximate Poisson probability that fewer than two accounts receivable will contain errors is 0.0916, which corresponds to answer choice B.

Question 11

Cars are arriving at a toll booth at a rate of four per minute. What is the probability that exactly eight cars will arrive in the next two minutes?

Accepted Answer

The number of arrivals in a two-minute interval follows a Poisson distribution with mean = 2*4 = 8. Therefore, the probability of getting exactly 8 arrivals in the next two minutes is:
P(X=8) = (e^-8)*(8^8)/8! = 0.1396 (rounded to four decimal places) 
Therefore, the best choice is B.

Question 12

For which binomial distribution would a Poisson approximation be acceptable?

Accepted Answer

The Poisson approximation to a binomial distribution is acceptable if n is large and π is small. Option C has the smallest value of π and a reasonably large value of n, making it the best choice for a Poisson approximation.

Question 13

The probability that a customer will use a stolen credit card to make a purchase at a certain Target store is 0.003. If 400 purchases are made in a given day, what is the approximate Poisson probability that 4 or fewer will be with stolen cards?

Accepted Answer

The Poisson probability formula is used here, given by P(X=k) = (λ^k * e^(-λ)) / k!, where λ is the mean number of successes (λ = np), and k is the number of successes we're interested in. For this scenario, λ = 400 * 0.003 = 1.2. We're interested in the probability of 4 or fewer successes, which requires summing the probabilities for 0, 1, 2, 3, and 4 successes. This calculation is cumbersome by hand but is well approximated by Poisson probability tables or a calculator. The result is approximately 0.9923, indicating a high probability of 4 or fewer purchases with stolen cards.

Question 14

On average, a major earthquake (Richter scale 6.0 or above) occurs three times a decade in a certain California county. Find the probability that at least one major earthquake will occur within the next decade.

Accepted Answer

The probability of at least one major earthquake occurring is 1 minus the probability of no earthquakes occurring. The average rate is 3 per decade, so using the Poisson distribution: P(X=0) = e^(-3) * 3^0 / 0! = e^(-3). Therefore, the probability of at least one is 1 - e^(-3) ≈ 0.9502.

Question 15

On average, an IRS auditor discovers 4.7 fraudulent income tax returns per day. On a randomly chosen day, what is the probability that she discovers fewer than two?

Accepted Answer

The scenario describes a Poisson distribution with a mean ($\lambda$) of 4.7 fraudulent returns per day. The probability of observing fewer than two fraudulent returns ($k < 2$) is the sum of the probabilities of observing 0 and 1 fraudulent returns. Using the Poisson probability formula $P(k; \lambda) = \frac{e^{-\lambda} \lambda^k}{k!}$, we calculate $P(0; 4.7) + P(1; 4.7)$. This calculation results in a value that matches choice A, indicating a misunderstanding or a mistake in the calculation process provided in the options.

Question 16

Which statement concerning the binomial distribution is correct?

Accepted Answer

The binomial distribution is defined for a given number of trials, n, and probability of success, π. Its probability density function (PDF) covers all integer values of X from 0 to n, representing the number of successes in n trials. Therefore, statement A is correct. Statement B is not true, as the PDF and CDF of the binomial distribution are different functions. Statement C is partly true, as the CDF of the binomial distribution shows the cumulative probability up to each value of X. Statement D is not true, as the shape of the CDF of the binomial distribution depends on the value of π, but it is not generally skewed right or left.

Question 17

The coefficient of variation for a Poisson distribution with λ = 4 is:

Accepted Answer

The coefficient of variation (CV) for a Poisson distribution is calculated as the standard deviation divided by the mean. For a Poisson distribution, the standard deviation is the square root of λ, and the mean is λ. So, CV = (sqrt(λ) / λ) * 100%. For λ = 4, CV = (sqrt(4) / 4) * 100% = (2 / 4) * 100% = 50.0 percent.

Question 18

Arrival of cars per minute at a toll booth may be characterized by the Poisson distribution if:

Accepted Answer

The Poisson distribution is used to model the number of events occurring within a fixed interval of time or space if these events occur with a known constant mean rate and independently of the time since the last event.

Question 19

For which binomial distribution would a Poisson approximation be unacceptable?

Accepted Answer

The Poisson approximation to the binomial distribution is generally considered acceptable if n is large and π is small, with the product nπ (mean of the binomial distribution) not too large. In option C, n = 200 and π = 0.10, which does not meet the criteria for a small π as well as the product nπ = 20, which can be considered too large for an ideal Poisson approximation.

Question 20

The coefficient of variation for a Poisson distribution with λ = 5 is:

Accepted Answer

The answer of The coefficient of variation for a Poisson...

Quiz 6: Discrete Probability Distributions