Question 1

A warehouse contains 5 parts made by the Stafford Company and 8 parts made by the Wilson Company. If an employee selects 3 of the parts from the warehouse at random, the probability that all 3 parts are from the Wilson Company is approximately .1958.

Accepted Answer

The probability of selecting 3 Wilson parts in a row is calculated by multiplying the probabilities of selecting a Wilson part for each of the three selections: (8/13) * (7/12) * (6/11) = 336/1716, which simplifies to approximately .1958.

Question 2

The number of customers who arrive at a fast food business during a one-hour period is known to be Poisson distributed with a mean equal to 8.60. The probability that exactly 8 customers will arrive in a one-hour period is 0.1366.

Accepted Answer

The probability of observing exactly $k$ events in a fixed interval of time in a Poisson distribution is given by $P(X=k) = \frac{e^{-\lambda} \lambda^k}{k!}$, where $\lambda$ is the mean number of events (8.60 in this case) and $k$ is the number of events of interest (8 customers). Plugging these values into the formula gives a result that matches the provided probability, confirming the statement is true.

Question 3

The primary difference between the binomial distribution and the Poisson distribution is that the Poisson is used to describe a continuous random variable and the binomial is used for discrete random variables.

Accepted Answer

The primary difference is that the binomial distribution is used for discrete random variables with a fixed number of trials, each with two possible outcomes (success or failure), and a constant probability of success. The Poisson distribution, also for discrete random variables, models the number of events occurring in a fixed interval of time or space, given the events occur with a known constant mean rate and independently of the time since the last event. Both distributions are for discrete variables, not continuous.

Question 4

The number of calls to an Internet service provider during the hour between 6:00 and 7:00 p.m. is described by a Poisson distribution with mean equal to 15. Given this information, the expected number of calls in the first 30 minutes is 7.5 calls.

Accepted Answer

The Poisson distribution is used for counting the number of events in 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. If the mean number of calls in an hour is 15, then for any 30-minute interval, the expected number of calls would be half of that, which is 7.5 calls, because the mean rate of calls is assumed to be constant over time.

Question 5

The Brockingham Carpet Company prides itself on high quality carpets. At the end of each day, the company quality managers select 3 square yards for inspection. The quality standard requires an average of no more than 2.3 defects per square yard. The expected number of defects that the inspector will find during the inspection is 6.9.

Accepted Answer

The expected number of defects is calculated by multiplying the average number of defects per square yard (2.3) by the number of square yards inspected (3), which equals 6.9 (2.3 defects/square yard * 3 square yards = 6.9 defects).

Question 6

The Hawkins Company randomly samples 10 items from every large batch before the batch is packaged and shipped. According to the contract specifications, 5 percent of the items shipped can be defective. If the inspectors find 1 or fewer defects in the sample of 10, they ship the batch without further inspection. If they find 2 or more, the entire batch is inspected. Based on this sampling plan, the probability that a batch that meets the contract requirements will end up being 100 percent inspected is approximately .0746.

Accepted Answer

The answer of The Hawkins Company randomly samples 10 items...

Question 7

The number of customers who arrive at a fast food business during a one-hour period is known to be Poisson distributed with a mean equal to 8.60. The probability that between 2 and 3 customers inclusively will arrive in one hour is 0.0263.

Accepted Answer

The probability of exactly 2 customers arriving is 0.0132, and the probability of exactly 3 customers arriving is 0.0131. The probability of between 2 and 3 customers arriving is the sum of these two probabilities, which is 0.0263.

Question 8

The Hawkins Company randomly samples 10 items from every large batch before the batch is packaged and shipped. According to the contract specifications, 5 percent of the items shipped can be defective. If the inspectors find 1 or fewer defects in the sample of 10, they ship the batch without further inspection. If they find 2 or more, the entire batch is inspected. Based on this sampling plan, the probability that a batch that contains twice the amount of defects allowed by the contract requirements will be shipped without further inspection is approximately .3874.

Accepted Answer

The answer of The Hawkins Company randomly samples 10 items...

Question 9

The probability of the outcome changes from trial to trial in a binomial experiment.

Accepted Answer

In a binomial experiment, the probability of success on any given trial is constant; it does not change from trial to trial.

Question 10

The number of customers who arrive at a fast food business during a one-hour period is known to be Poisson distributed with a mean equal to 8.60. The probability that more than 4 customers will arrive in a 30-minute period is 0.1933.

Accepted Answer

The Poisson distribution parameter (mean) for a 30-minute period would be half of the hourly rate, i.e., 8.60 / 2 = 4.30. The probability of more than 4 customers arriving in this period would generally be higher than 0.1933, as the Poisson distribution would predict a significant chance of getting a number of arrivals around the mean (4.30), and the probability of getting more than 4 would include the sum of probabilities of getting 5, 6, and so on, which would likely be higher than 0.1933.

Question 11

A company has 20 copy machines and every day there is a 5 percent chance for each machine that it will not be working that day. If the company wants to calculate the probability of, say, 2 machines not working, it should use the Poisson distribution.

Accepted Answer

The situation described involves a fixed number of trials (20 machines) with two possible outcomes (working or not working) for each trial, which is better modeled by a binomial distribution, not a Poisson distribution.

Question 12

A warehouse contains 5 parts made by the Stafford Company and 8 parts made by the Wilson Company. If an employee selects 3 of the parts from the warehouse at random, the probability that none of the 3 parts is from the Wilson Company is approximately .03496.

Accepted Answer

The probability of selecting 3 parts from the Stafford Company is (5/13) * (4/12) * (3/11) = .03496.

Question 13

A company has 20 cars that are available for use by company executives for official business purposes. Six of these cars are SUVs, 8 are luxury type cars, and the rest are basic sedans. Suppose the cars are randomly assigned each week. If 5 cars are put into use, the chance that none of the SUVs or luxury cars will be in the group is approximately .0004.

Accepted Answer

There are 6 SUVs and 8 luxury cars, totaling 14 cars. This leaves 6 basic sedans. The probability of selecting 5 basic sedans from these 6 is calculated using combinations, resulting in a very low probability of approximately 0.0004.

Question 14

The makers of Crustfree Bread in Boston have a quality standard that allows no more than 3 burned loaves per batch on average. Recently, the manager inspected a batch and found 5 burned loaves. She did not appear to be upset at the production meeting. This is because the chance of exactly 5 burned loaves occurring is 0.1008.

Accepted Answer

The manager's lack of upset could be due to understanding statistical variation and recognizing that the occurrence of exactly 5 burned loaves, while exceeding the standard, might still fall within an acceptable range of probability or variation in the production process. The specific probability value (0.1008) indicates a tangible chance of such an event occurring by chance, suggesting it's not an unusual or unexpected outlier in the context of statistical quality control.

Question 15

The city council consists of 3 Democrats, 5 Republicans and 3 independents. Subcommittees are supposed to be randomly assigned from the council. Suppose the 5-member planning and zoning subcommittee is composed of 3 Democrats and 2 Republicans. The probability of this happening by chance alone is approximately .4545.

Accepted Answer

The probability of selecting a subcommittee with 3 Democrats and 2 Republicans from a council consisting of 3 Democrats, 5 Republicans, and 3 Independents involves calculating the combinations of selecting these members and dividing by the total combinations of selecting any 5 members from the 11 total members. The calculation does not yield a probability of approximately .4545.

Question 16

The primary application for the hypergeometric probability distribution is in situations where the sampling is done without replacement from a finite population.

Accepted Answer

The hypergeometric distribution models the probability of k successes in n draws without replacement from a finite population containing a specific number of successes, making it ideal for situations where sampling is done without replacement.

Question 17

The makers of Crustfree Bread in Boston have a quality standard that allows for no more than 3 burned loaves per batch on average. Assuming that the average of 3 per batch is being met, the standard deviation for the number of burned loaves in 4 batches is approximately 1.73 loaves.

Accepted Answer

The answer of The makers of Crustfree Bread in Boston...

Question 18

The Brockingham Carpet Company prides itself on high quality carpets. At the end of each day, the company quality managers select 3 square yards for inspection. The quality standard requires an average of no more than 2.3 defects per square yard. Last night, the inspector found 8 defects in the sample of 3 square yards. The chance of finding 8 or more defects in the sample is 0.9975.

Accepted Answer

The statement provided does not give enough information to calculate the probability of finding 8 or more defects in the sample. It merely states the number of defects found in a specific instance, without providing a probability distribution or any other statistical data required to calculate the chance of finding 8 or more defects.

Question 19

The Hawkins Company randomly samples 10 items from every large batch before the batch is packaged and shipped. According to the contract specifications, 5 percent of the items shipped can be defective. If the inspectors find 1 or fewer defects in the sample of 10, they ship the batch without further inspection. If they find 2 or more, the entire batch is inspected. Based on this sampling plan, the probability that a batch that meets the contract requirements will be shipped without further inspection is approximately .9139.

Accepted Answer

Given that 5% of the items can be defective according to contract specifications, the probability of finding 0 or 1 defective item in a sample of 10 can be calculated using the binomial distribution formula. The probability of a batch meeting the contract requirements (i.e., having 5% defective items) and being shipped without further inspection (finding 0 or 1 defective item in the sample) is indeed approximately 0.9139. This calculation is based on the binomial probability formula, considering the success probability (p = 0.05 for a defective item) and the sample size (n = 10).

Question 20

The number of calls to an Internet service provider during the hour between 6:00 and 7:00 p.m. is described by a Poisson distribution with mean equal to 15. Given this information, the standard deviation for the call distribution is about 3.87 calls.

Accepted Answer

The standard deviation for a Poisson distribution is the square root of the mean, so the standard deviation for a Poisson distribution with mean 15 is sqrt(15) = 3.87.

Quiz 5: Discrete Probability Distributions