Question 1

The Ace Construction Company has entered into a contract to widen a street in Boston. The possible payoffs for this project have been determined by management. The probabilities for these payoffs could be determined using a binomial distribution.

Accepted Answer

The binomial distribution is used for scenarios with two outcomes (success or failure) in repeated trials, which doesn't typically apply to the payoffs of a construction project. Payoffs would more likely follow a continuous distribution, considering the range of possible outcomes based on costs, delays, and other variables.

Question 2

A direct marketing company believes that the probability of making a sale when a call is made to an individual's home is .02. The probability of making 2 or 3 sales in a sample of 20 calls is .0593.

Accepted Answer

The probability of making 2 or 3 sales in 20 calls can be calculated using the binomial probability formula, and the given probability of .0593 is consistent with such a calculation for a binomial distribution with p = 0.02 and n = 20.

Question 3

Each week American Stores receives a shipment from a supplier. The contract specifies that the maximum allowable percent defective is 5 percent. When the shipment arrives, a sample of 20 parts is randomly selected. If 2 or more of the sampled parts are defective, the shipment is rejected and returned to the supplier. Assume that a shipment arrives that actually has 4 percent defective parts and the distribution of defective parts is described by a binomial distribution. The probability that the shipment is rejected is approximately 0.19.

Accepted Answer

The probability of rejecting a shipment with 4% defective parts is approximately 0.19, which is less than 0.25.

Question 4

Ace Computer Manufacturer buys disk drives in lots of 5,000 units from a supplier in California. The contract calls for, at most, 3 percent of the disk drives to be defective. When a shipment arrives, a sample of n = 15 parts is selected. If zero defects are found in this sample, the shipment is accepted. If 3 or more defects are found, the shipment is rejected and sent back to the supplier. If the number of defects found is 1 or 2, a second sample of 15 parts is selected. If this sample yields 1 or fewer defects, the shipment is accepted; otherwise the shipment is rejected. Based on a binomial distribution, the probability that Ace will reject a shipment of parts that meets the contract requirements is approximately 0.0355.

Accepted Answer

The probability of rejecting a shipment that meets the contract requirements can be calculated using the binomial distribution. Given the contract allows for up to 3% defects, the probability of finding defects in the samples can be computed, and the probability of rejecting a shipment under these conditions is approximately 0.0355.

Question 5

In a Florida town, the probability distribution for the number of legitimate emergency calls per day for the Fire Department is given as follows. Also shown is the probability distribution for the number of false alarms:  Given this information, the expected number of total calls to the fire department is 1.60 calls.

Accepted Answer

The answer of In a Florida town, the probability distribution...

Question 6

The Nationwide Motel Company has determined that 70 percent of all calls for motel reservations request nonsmoking rooms. Recently, the customer service manager for the company randomly selected 25 calls. Assuming that the distribution of calls requesting nonsmoking rooms is described by a binomial distribution, the probability that more than 20 customers in the sample will request nonsmoking rooms is approximately 0.09.

Accepted Answer

Using the binomial distribution formula with n=25, p=0.7, and calculating P(X>20), the probability is approximately 0.09.

Question 7

Each week American Stores receives a shipment from a supplier. The contract specifies that the maximum allowable percent defective is 5 percent. When the shipment arrives, a sample of 20 parts is randomly selected. If 2 or more of the sampled parts are defective, the shipment is rejected and returned to the supplier. Assume that a shipment arrives that actually has 4 percent defective parts and the distribution of defective parts is described by a binomial distribution. The probability that the shipment is accepted is approximately 0.81

Accepted Answer

The probability of accepting the shipment is the probability that 0 or 1 part is defective in the sample of 20. Using the binomial distribution with p = 0.04, the probability of 0 defects is approximately 0.442, and the probability of 1 defect is approximately 0.369. Adding these gives approximately 0.811.

Question 8

A pizza restaurant uses 7 different toppings on its pizzas. At lunch time it has a pizza buffet and makes pizzas with 2 toppings. If it wants to serve every possible combination of 2 toppings, it would need to make 14 different pizzas.

Accepted Answer

The number of different pizzas with 2 toppings that can be made from 7 toppings is calculated using combinations, specifically 7 choose 2, which equals 21, not 14.

Question 9

Holmstead Company owns two small engine repair stores. The expected value of the number of complaints received per month at store 1 is 4.5 complaints. Further, the expected number of complaints per month for store 1 and store 2 combined is 13.6. This means the expected number of complaints per month at store 2 must be 9.1 complaints.

Accepted Answer

The expected number of complaints for both stores combined (13.6) minus the expected number for store 1 (4.5) equals the expected number for store 2, which is 9.1 complaints.

Question 10

The Nationwide Motel Company has determined that 70 percent of all calls for motel reservations request nonsmoking rooms. Recently, the customer service manager for the company randomly selected 25 calls. Assuming that the distribution of calls requesting nonsmoking rooms is described by a binomial distribution, the probability that fewer than 5 customers will request smoking rooms is approximately 0.09.

Accepted Answer

Using the binomial distribution with n=25 and p=0.7 for nonsmoking rooms, the probability of fewer than 5 customers requesting smoking rooms (which means 21 or more requesting nonsmoking) is approximately 0.09.

Question 11

A construction company has found it has a probability of 0.10 of winning each time it bids on a project. The probability of winning a given number of projects out of 12 bids could be determined with a binomial distribution.

Accepted Answer

The scenario meets the criteria for a binomial distribution: a fixed number of trials (12 bids), each trial has only two outcomes (win or lose), the probability of success (winning) is constant (0.10), and the trials are independent.

Question 12

The number of defects discovered in a random sample of 100 products produced at the Berdan Manufacturing Company is binomially distributed with p = .03. Based on this, the standard deviation of the number of defects per sample of size 100 is 2.91.

Accepted Answer

The standard deviation of a binomial distribution is given by the formula $\sqrt{n \cdot p \cdot (1-p)}$, where $n$ is the sample size and $p$ is the probability of success (in this case, finding a defect). For $n = 100$ and $p = .03$, the standard deviation is $\sqrt{100 \cdot .03 \cdot (1-.03)} = \sqrt{2.91} \approx 1.706$.

Question 13

The Nationwide Motel Company has determined that 70 percent of all calls for motel reservations request nonsmoking rooms. Recently, the customer service manager for the company randomly selected 25 calls. Assuming that the distribution of calls requesting nonsmoking rooms is described by a binomial distribution, the expected number of requests for nonsmoking rooms is 14.

Accepted Answer

The expected number of requests for nonsmoking rooms is calculated by multiplying the probability of a nonsmoking room request (0.70) by the total number of calls (25), which equals 17.5, not 14.

Question 14

When using the binomial distribution, the maximum possible number of success is the number of trials.

Accepted Answer

In a binomial distribution, the number of successes can range from 0 to the number of trials, making the number of trials the maximum possible number of successes.

Question 15

The distribution for the number of emergency calls to a city's 911 emergency number in a one-hour time period is likely to be described by a binomial distribution.

Accepted Answer

The number of emergency calls to a city's 911 emergency number in a one-hour time period is better described by a Poisson distribution, as it deals with the number of events happening in a fixed interval of time and space where these events occur independently.

Question 16

One difference between the binomial distribution and Poisson distribution is that the binomial's upper bound is the number of trials while the Poisson has no particular upper bound.

Accepted Answer

The binomial distribution is used for a fixed number of trials, so its upper bound is the number of trials (n). In contrast, the Poisson distribution models the number of events in a fixed interval and theoretically has no upper limit to the count of events, thus no particular upper bound.

Question 17

Bill Price is a sales rep in northern California representing a line of athletic socks. Each day, he makes 10 sales calls. The chance of making a sale on each call is thought to be 0.30. The probability that he will make exactly two sales is approximately 0.2335.

Accepted Answer

This scenario can be modeled using the binomial probability formula, which is P(X = k) = (n choose k) * p^k * (1-p)^(n-k), where n is the number of trials (in this case, sales calls, which is 10), k is the number of successes (sales made, which is 2), and p is the probability of success on each trial (0.30). Plugging these values into the formula gives a probability of approximately 0.2335 for making exactly two sales, confirming the statement is true.

Question 18

One of the characteristics of the binomial distribution is that the probability of success for each trial depends on whether the previous trial was a success or not.

Accepted Answer

In a binomial distribution, the probability of success on each trial is constant and does not depend on the outcomes of previous trials.

Question 19

The Nationwide Motel Company has determined that 70 percent of all calls for motel reservations request nonsmoking rooms. Recently, the customer service manager for the company randomly selected 25 calls. Assuming that the distribution of calls requesting nonsmoking rooms is described by a binomial distribution, the standard deviation of requests for nonsmoking rooms is 5.25 customers.

Accepted Answer

The standard deviation for a binomial distribution is calculated using the formula $\sigma = \sqrt{n \cdot p \cdot (1-p)}$, where $n$ is the number of trials (in this case, 25 calls), and $p$ is the probability of success (in this case, 0.70 for requesting nonsmoking rooms). Plugging in these values, we get $\sigma = \sqrt{25 \cdot 0.70 \cdot (1-0.70)} = \sqrt{25 \cdot 0.70 \cdot 0.30} = \sqrt{5.25} \approx 2.29$, not 5.25.

Question 20

Six managers at a company all enjoy golf. Each Saturday, four of the six get together for 18 holes of golf. They have decided to set up a schedule so that the same foursome does not play twice before all possible foursomes have played. The number of weekends that will pass before the same group would play twice is 15.

Accepted Answer

The number of unique combinations of 4 managers out of 6 is calculated using the combination formula C(n, k) = n! / (k!(n-k)!), where n is the total number of managers and k is the number of managers per group. C(6, 4) = 15, so it will take 15 weekends for all possible foursomes to play.

Quiz 5: Discrete Probability Distributions