Question

A factory makes processors for use in computers. it is known that about 90% of the processors meet specifications. every week a sample of 18 processors is selected at random for testing. assume a binomial distribution is valid. suppose we collect a large number of these samples of 18 processors and determine the number meeting specifications in each sample. what is the mean number of processors meeting specifications?

Answer 1

First recall the binomial distribution:
$B(k,n,p)=C_n^kp^k(1-p)^{n-k}\\\text{Here n=18,p=0.9, therefore:}\\B(k,n,p)=C_{18}^k0.9^k\times0.1^{18-k}\\\text{In order to find the mean, we have to evaluate the following formula:}\\\sum P(X=k)\times k=\sum_{k=0}^{k=18} \frac{18!}{k!(18-k)!}0.9^k\times0.1^{18-k}k$

Using a calculator we get the mean equal approximatively 91.

Answer 2

The mean number of processors meeting specifications is 16.2In order to get the mean number of the precessors meeting specification we will multiply the 90% to 18 which will give us 16.2, it is the basic or the most easy way to get the mean number. The answer is 16.2