Respuesta :
Answer:
[tex]z = \frac{4-4.25}{\frac{1.5}{\sqrt{40}}}= -1.054[/tex]
And we can find the following probability:
[tex] P(z<-1.054) = 0.146[/tex]
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:
[tex]\bar X = 4.25[/tex] represent the mean
[tex]\sigma =18/12=1.5[/tex] represent the deviation in years
They select a sample size of n=40>30. And they want to find this probability:
[tex] P(\bar X<40)[/tex]
Since the sample size is large enough we can use the central limit theorem and the distribution for the sample mean would be:
[tex]\bar X \sim N(\mu, \frac{\sigma}{\sqrt{n}}) [/tex]
We can use the z score formula given by:
[tex] z = \frac{\bar X -\mu}{\frac{\sigma}{\sqrt{n}}}[/tex]
And if we find the z score for 4 we got:
[tex]z = \frac{4-4.25}{\frac{1.5}{\sqrt{40}}}= -1.054[/tex]
And we can find the following probability:
[tex] P(z<-1.054) = 0.146[/tex]
And the last probability can be founded using the normal standard distribution or excel.
