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Chapter 7: Problem 103

Flashlight bulbs manufactured by a certain company are sometimes defective. a. If \(5 \%\) of all such bulbs are defective, could the techniques of this section be used to approximate the probability that at least five of the bulbs in a random sample of size 50 are defective? If so, calculate this probability; if not, explain why not. b. Reconsider the question posed in Part (a) for the probability that at least 20 bulbs in a random sample of size 500 are defective.

Short Answer

Expert verified
In the first part of the problem, the normal approximation cannot be used because the threshold \(np=5\) is not reached. Thus, the exact probability has to be calculated using the binomial distribution. In the second part, the normal approximation can be used, so the probability can be calculated more easily.

Step by step solution

01

Determine parameters of Binomial Distribution

For both parts of the problem the parameters are clear: The probability of success \(p\) (i.e., a bulb is defective) is given as \(0.05\) or \(5\%\) and the number \(n\) of experiments (i.e., the sample size) is given as \(50\) for the first task and \(500\) for the second task. From these parameters we could now define the expected value \(E(X)=np\) and the standard deviation \(\sqrt{np(1-p)}\) of the binomial distribution.
02

Check if Normal Approximation can be used

The Normal approximation to the binomial distribution applies if \(np \geq 5\), \(n(1-p) \geq 5\) and the number of events \(n\) is large. For the first part of the problem, \(np = 50*0.05 = 2.5\), which does not meet the threshold of 5, meaning normal approximation cannot be used here. For the second part of the problem, \(np = 500*0.05 = 25\), which is larger than 5, so we can use the normal approximation for this case.
03

Calculate probability using Binomial Distribution

Since we can't use the normal approximation in the first part, we have to use the binomial distribution. We want to find the probability that the number \(x\) of defective lightbulbs is at least 5, which corresponds to \(P(X \geq 5)\). Expressing this as the complement of the event with fewer than 5 defectives, i.e. \(P(X \geq 5) = 1 - P(X < 5)\), we can calculate the probability using the binomial probability formula.
04

Calculate probability using Normal Approximation

In the second part of the problem, we want to find the probability that at least 20 lightbulbs are defective in a sample of 500. That is \(P(X \geq 20)\) where \(X\) is Normal with mean \(E(X) = np\) and standard deviation \(\sqrt{np(1-p)}\). To bring \(X\) into the standard normal form, we subtract the mean and divide by the standard deviation, i.e.\(Z = (X - np) / \sqrt{np(1-p)}\). Then we look up \(P(Z \geq 20/ \sqrt{np(1-p)})\) in the standard normal table.

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Most popular questions from this chapter

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