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Chapter 6: Problem 79

Hurbert Corporation makes font cartridges for laser printers that it sells to Alpha Electronics Inc. The cartridges are shipped to Alpha Electronics in large volumes. The quality control department at Alpha Electronics randomly selects 100 cartridges from each shipment and inspects them for being good or defective. If this sample contains 7 or more defective cartridges, the entire shipment is rejected. Hurbert Corporation promises that of all the cartridges, only \(5 \%\) are defective. a. Find the probability that a given shipment of cartridges received by Alpha Electronics will be accepted. b. Find the probability that a given shipment of cartridges received by Alpha Electronics will not be accepted.

Short Answer

Expert verified
The calculated probability P(Acceptance) and P(Non-Acceptance) from the previous steps are the required solutions. After performing the calculations, insert the obtained numerical values for P(Acceptance) and P(Non-Acceptance) respectively in the final answer.

Step by step solution

01

Define the Parameters of the Binomial Distribution

Define the parameters of the binomial distribution, n (number of trials: 100 cartridges), p (probability of success: probability of a cartridge being good, which is 1 minus the probability of a cartridge being defective i.e., 1 - 0.05 = 0.95), and x (number of successes sought: less than 7 defective means we are looking for shipments where the number of defective cartridges range from 0 to 6.
02

Calculate the Probability for Acceptance

Use the binomial probability formula to calculate the probability for each possible number of defective cartridges from 0 to 6 and sum them up. The binomial probability formula is defined as: \[P(x; n, p) = C(n, x) \cdot p^x \cdot (1 - p)^{n-x}\] where the binomial coefficient C(n, x) = n! / [x! (n - x)!]. After calculating the sum of probabilities for x from 0 to 6, refer to this as P(Acceptance).
03

Calculate the Probability for Non-Acceptance

The probability that a shipment will not be accepted is simply the complement of the probability that the shipment will be accepted. This can be calculated as: P(Non-Acceptance) = 1 - P(Acceptance).

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Most popular questions from this chapter

The management of a supermarket wants to adopt a new promotional policy of giving a free gift to every customer who spends more than a certain amount per visit at this supermarket. The expectation of the management is that after this promotional policy is advertised, the expenditures for all customers at this supermarket will be normally distributed with a mean of \(\$ 95\) and a standard deviation of \(\$ 20 .\) If the management wants to give free gifts to at most \(10 \%\) of the customers, what should the amount be above which a customer would receive a free gift?

A charter bus company is advertising a singles outing on a bus that holds 60 passengers. The company has found that, on average, \(10 \%\) of ticket holders do not show up for such trips; hence, the company routinely overbooks such trips. Assume that passengers act independently of one another. a. If the company sells 65 tickets, what is the probability that the bus can hold all the ticket holders who actually show up? In other words, find the probability that 60 or fewer passengers show up. b. What is the largest number of tickets the company can sell and still be at least \(95 \%\) sure that the bus can hold all the ticket holders who actually show up?

The amount of time taken by a bank teller to serve a randomly selected customer has a normal distribution with a mean of 2 minutes and a standard deviation of \(.5\) minute. a. What is the probability that both of two randomly selected customers will take less than I minute each to be served? b. What is the probability that at least one of four randomly selected customers will need more than \(2.25\) minutes to be served?

Find the following areas under a normal distribution curve with \(\mu=12\) and \(\sigma=2\). a. Area between \(x=7.76\) and \(x=12\) b. Area between \(x=14.48\) and \(x=16.54\) c. Area from \(x=8.22\) to \(x=10.06\)

For a binomial probability distribution, \(n=80\) and \(p=.50 .\) Let \(x\) be the number of successes in 80 trials. a. Find the mean and standard deviation of this binomial distribution. b. Find \(P(x \geq 42)\) using the normal approximation. c. Find \(P(41 \leq x \leq 48)\) using the normal approximation.
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