91Ó°ÊÓ

York Steel Corporation produces a special bearing that must meet rigid specifications. When the production process is running properly, \(10 \%\) of the bearings fail to meet the required specifications. Sometimes problems develop with the production process that cause the rejection rate to exceed \(10 \%\). To guard against this higher rejection rate, samples of 15 bearings are taken periodically and carefully inspected. If more than 2 bearings in a sample of 15 fail to meet the required specifications, production is suspended for necessary adjustments. a. If the true rate of rejection is \(10 \%\) (that is, the production process is working properly), what is the probability that the production will be suspended based on a sample of 15 bearings? b. What assumptions did you make in part a?

Step by step solution

01

Understanding Binomial Distribution

The checking of each bearing can be considered as a binomial trial where it can either pass or fail. Hence, we can use the formula for binomial distribution probability which is \(P(x; n, p) = C(n,x) * p^x * q^{n-x}\) where \(P(x; n, p)\) is the probability of \(x\) successes in \(n\) trials when the probability of success in any given trial is \(p\). \(C(n,x)\) refers to combination of \(n\) items taken \(x\) at a time. \(p\) is the probability of one event and \(q\) is the probability of its complement.

02

Using the Binomial Probability Formula

The question asks for the probability of finding more than 2 defective bearings. This is equivalent to 1 subtracted by the probability of finding 2 or fewer defective bearings. Hence, we need to calculate \(P(X > 2) = 1 - P(X \leq 2)\). This involves summing up probabilities for \(\text{X}=0\), \(\text{X}=1\), and \(\text{X}=2\).

03

Calculation of Individual Probabilities

Using the given rejection rate of \(10 \%\), we first calculate each probability separately for x being 0, 1, and 2. For example, for \(x=0\), it would be \(P(X=0)=C(15,0)*(0.10)^0*(1-0.10)^{15} = (0.90)^{15}\).

04

Summation of Probabilities and Calculation of the Desired Probability

We then add these probabilities and then subtract the sum from 1 to get the desired probability which is the probability of more than 2 defective bearings. Hence, \(P(X>2) = 1 - [P(X = 0) + P(X = 1) + P(X = 2)]\). The resultant Value provides the probability that the production will be suspended.

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