Question

The quality control manager at a computer manufacturing company believes that the mean life of a computer is 120 months, with a variance of 64. If he is correct, what is the probability that the mean of a sample of 89 computers would be less than 118.81 months? Round your answer to four decimal places.

Answer 1

Answer:

P(X<118.81)=0.0803

Step-by-step explanation:

Assuming the distribution for the mean life is approximately normal, with mean 120 months and variance 64 months^2, we can calculate the parameters for a sampling distribution with sample size = 89 computers.

The sampling distribution mean will be equal to the mean for a single computer:

$\mu_{89}=\mu=120$

The standard deviation will be adjusted by the sample size as:

$\sigma_{89}=\sqrt{\sigma^2/n}=\sqrt{64/89}=\sqrt{0.719}=0.848$

With these parameters, we can calculate the z-score for X=118.81.

$z=\dfrac{X-\mu}{\sigma}=\dfrac{118.81-120}{0.848}=\dfrac{-1.19}{0.848}=-1.403$

Then, the probability that the mean of a sample of 89 computers is less than 118.81 months is:

$P(X<118.81)=P(z<-1.403)=0.0803$