Question

The quality control manager at a computer manufacturing company believes that the mean life of a computer is 8181 months, with a variance of 6464. If he is correct, what is the probability that the mean of a sample of 6060 computers would differ from the population mean by less than 2.942.94 months? Round your answer to four decimal places.

Answer 1

`P(78.36 \leq \bar X \leq 83.64)=P(z<2.56)-P(z<-2.56)=0.9895`

Explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

The Z-score is "a numerical measurement used in statistics of a value's relationship to the mean (average) of a group of values, measured in terms of standard deviations from the mean".

Let X the random variable that represent interest on this case, and for this case we know the distribution for X is given by:

`X \sim N(\mu=81,\sigma=8)`

And let
`\bar X` represent the sample mean, the distribution for the sample mean is given by:

`\bar X \sim N(\mu,(\sigma)/(√(n)))`

On this case
`\bar X \sim N(81,(8)/(√(60))=1.03)`

For this case we can use the z score formula to solve the problem.The z score on this case is given by this formula:

`z=(\bar x-\mu)/((\sigma)/(√(n)))`

We want to find this probability:

`P(81-2.64 \leq \bar X \leq 81+2.64)=P(78.36 \leq \bar X \leq 83.64)`

And if we replace we got:

`z=(83.64-81)/((8)/(√(60)))=2.56`

`z=(78.36-81)/((8)/(√(60)))=-2.56`

And we can find the probability on this way:

`P(78.36 \leq \bar X \leq 83.64)=P(z<2.56)-P(z<-2.56)=0.9895`