Question

The head of maintenance at XYZ Rent-A-Car believes that the mean number of miles between services is 3408 miles, with a variance of 249,001. If he is correct, what is the probability that the mean of a sample of 36 cars would differ from the population mean by less than 198 miles

Answer 1

Answer:

$0.98268$

Step-by-step explanation:

We are given that

Mean, $\mu=3408$miles

Variance, $\sigma^2=249001$

We have to find  the probability that the mean of a sample of 36 cars would differ from the population mean by less than 198 miles.

n=36

$P(|\bar{x}-3408}|<198)=P(3408-198<\bar{x}<198+3408)$

=$P(3210<\bar{x}<3606)$

=$P(\frac{3210-3408}{\sqrt{\frac{249001}{36}}}<\frac{\bar{x}-\mu}{\sqrt{\frac{variance}{n}}}<\frac{3606-3408}{\sqrt{\frac{249001}{36}}}$

$=P(-2.38<Z<2.38)$

$=P(Z<2.38)-P(Z<-2.38)$

$=0.99134-0.00866$

=$0.98268$

Hence,  the probability that the mean of a sample of 36 cars would differ from the population mean by less than 198 miles=$0.98268$