Answer:
The probability that at most 10 of the next 150 items produced are unacceptable is 0.8315.
Step-by-step explanation:
Let <em>X</em> = number of items with unacceptable quality.
The probability of an item being unacceptable is, P (X) = <em>p</em> = 0.05.
The sample of items selected is of size, <em>n</em> = 150.
The random variable <em>X</em> follows a Binomial distribution with parameters <em>n</em> = 150 and <em>p</em> = 0.05.
According to the Central limit theorem, if a sample of large size (<em>n</em> > 30) is selected from an unknown population then the sampling distribution of sample mean can be approximated by the Normal distribution.
The mean of this sampling distribution is: ![\mu_{\hat p}= p=0.05](https://tex.z-dn.net/?f=%5Cmu_%7B%5Chat%20p%7D%3D%20p%3D0.05)
The standard deviation of this sampling distribution is: ![\sigma_{\hat p}=\sqrt{\frac{ p(1-p)}{n}}=\sqrt{\frac{0.05(1-.0.05)}{150} }=0.0178](https://tex.z-dn.net/?f=%5Csigma_%7B%5Chat%20p%7D%3D%5Csqrt%7B%5Cfrac%7B%20p%281-p%29%7D%7Bn%7D%7D%3D%5Csqrt%7B%5Cfrac%7B0.05%281-.0.05%29%7D%7B150%7D%20%7D%3D0.0178)
If 10 of the 150 items produced are unacceptable then the probability of this event is:
![\hat p=\frac{10}{150}=0.067](https://tex.z-dn.net/?f=%5Chat%20p%3D%5Cfrac%7B10%7D%7B150%7D%3D0.067)
Compute the value of
as follows:
![P(\hat p\leq 0.067)=P(\frac{\hat p-\mu_{p}}{\sigma_{p}} \leq\frac{0.067-0.05}{0.0178})=P(Z\leq 0.96)=0.8315](https://tex.z-dn.net/?f=P%28%5Chat%20p%5Cleq%200.067%29%3DP%28%5Cfrac%7B%5Chat%20p-%5Cmu_%7Bp%7D%7D%7B%5Csigma_%7Bp%7D%7D%20%5Cleq%5Cfrac%7B0.067-0.05%7D%7B0.0178%7D%29%3DP%28Z%5Cleq%200.96%29%3D0.8315)
*Use a <em>z</em>-table for the probability.
Thus, the probability that at most 10 of the next 150 items produced are unacceptable is 0.8315.