Question

Certain transportation company has a fleet of 210 vehicles. The average age of the vehicles is 4.25 years, with a standard deviation of 18 months. In a random sample of 40 vehicles, what is the probability that the average age of vehicles in the sample will be less than 4 years

Answer 1

Answer:

$z = \frac{4-4.25}{\frac{1.5}{\sqrt{40}}}= -1.054$

And we can find the following probability:

$P(z$

And the last probability can be founded using  the normal standard distribution or excel.

Step-by-step explanation:

For this case we define the random variable X as the ages of vehicles. We know the following info for this variable:

$\bar X = 4.25$ represent the mean

$\sigma =18/12=1.5$ represent the deviation in years

They select a sample size of n=40>30. And they want to find this probability:

$P(\bar X$

Since the sample size is large enough we can use the central limit theorem and the distribution for the sample mean would be:

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

We can use the z score formula given by:

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

And if we find the z score for 4 we got:

$z = \frac{4-4.25}{\frac{1.5}{\sqrt{40}}}= -1.054$

And we can find the following probability:

$P(z$

And the last probability can be founded using  the normal standard distribution or excel.