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

`z = (4-4.25)/((1.5)/(√(40)))= -1.054`

And we can find the following probability:

`P(z<-1.054) = 0.146`

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

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<40)`

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, (\sigma)/(√(n)))`

We can use the z score formula given by:

`z = (\bar X -\mu)/((\sigma)/(√(n)))`

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

`z = (4-4.25)/((1.5)/(√(40)))= -1.054`

And we can find the following probability:

`P(z<-1.054) = 0.146`

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