Question

A random sample of n=36 flights were taken and the time recorded to board the flight. Previous studies had determined that mean time to be 40 minutes with standard deviation of 24 minutes.


a) What is the sampling distribution of the sample mean?

b) What is the probability that the average time exceeds 45 minutes?

Answer 1

Answer:

a) $\bar X \sim N(40,\frac{24}{\sqrt{36}}=4)$

b)$P(\bar X >45)=0.106$

Step-by-step explanation:

Previous concepts

Normal distribution, is a "probability distribution that is symmetric about the mean, showing that data near the mean are more frequent in occurrence than data far from the mean".

Part a

The sampling distribution for the sample mean is given by:

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

The deduction is explained below we have this:

$E(\bar X)= E(\sum_{i=1}^{n}\frac{x_i}{n})= \sum_{i=1}^n \frac{E(x_i)}{n}= \frac{n\mu}{n}=\mu$

$Var(\bar X)=Var(\sum_{i=1}^{n}\frac{x_i}{n})= \frac{1}{n^2}\sum_{i=1}^n Var(x_i)$

Since the variance for each individual observation is $Var(x_i)=\sigma^2$ then:

$Var(\bar X)=\frac{n \sigma^2}{n^2}=\frac{\sigma}{n}$

And then for this special case:

$\bar X \sim N(40,\frac{24}{\sqrt{36}}=4)$

Part b

We are interested on this probability:

$P(\bar X >45)$

And we have already found the probability distribution for the sample mean on part a. So on this case we can use the z score formula given by:

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

Applying this we have the following result:

$P(\bar X >45)=P(Z>\frac{45-40}{\frac{24}{\sqrt{36}}})=P(Z>1.25)$

And using the normal standard distribution, Excel or a calculator we find this:

$P(Z>1.25)=1-P(z<1.25)=1-0.894=0.106$