Question

The head of maintenance at XYZ Rent-A-Car believes that the mean number of miles between services is 2643 miles, with a standard deviation of 368 miles. If he is correct, what is the probability that the mean of a sample of 44 cars would differ from the population mean by less than 51 miles

Answer 1

Answer:

$P(2643-51< \bar X < 2643+51)= P(2592< \bar X$

And we can use the z scoe formula given by:

$z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

And if we find the z score for the limits we got:

$z = \frac{2592-2643}{\frac{368}{\sqrt{44}}}= -0.919$

$z = \frac{2694-2643}{\frac{368}{\sqrt{44}}}= 0.919$

And this probability is equivalent to:

$P(-0.919$

Step-by-step explanation:

For this case we can define the random variable X as "number of miles between services" and we know the following info given:

$\mu = 2643 , \sigma = 368$

The central limit theorem states that "if we have a population with mean μ and standard deviation σ and take sufficiently large random samples from the population with replacement, then the distribution of the sample means will be approximately normally distributed. This will hold true regardless of whether the source population is normal or skewed, provided the sample size is sufficiently large".

From the central limit theorem we know that the distribution for the sample mean $\bar X$ is given by:

$\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}})$

We select a random sample size of n =44. And we want to find this probability:

$P(2643-51< \bar X < 2643+51)= P(2592< \bar X$

And we can use the z scoe formula given by:

$z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}$

And if we find the z score for the limits we got:

$z = \frac{2592-2643}{\frac{368}{\sqrt{44}}}= -0.919$

$z = \frac{2694-2643}{\frac{368}{\sqrt{44}}}= 0.919$

And this probability is equivalent to:

$P(-0.919$