Question

The mean annual income for people in a certain city (in thousands of dollars) is 42, with a standard deviation of 30. A pollster draws a sample of 90 people to interview. a. What is the probability that the sample mean income is less than 38?

Answer 1

Answer:

$P(\bar X <38)$

And 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 38 we got:

$z = \frac{38-42}{\frac{30}{\sqrt{90}}}= -1.265$

So then we want to find this probability:

$P(z<-1.265)$

And we can use the normal standard distribution or excel and we got:

$P(z<-1.265)=0.103$

Step-by-step explanation:

For this case we define the random variable X as the annual income for people at certain city. And we know the following properties:

$E(X) = 42, Sd(X) =42$

They select a sample size of n = 90>30. So then we can assume that the central limit can be applied.

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 want to calculate this probability:

$P(\bar X <38)$

And 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 38 we got:

$z = \frac{38-42}{\frac{30}{\sqrt{90}}}= -1.265$

So then we want to find this probability:

$P(z<-1.265)$

And we can use the normal standard distribution or excel and we got:

$P(z<-1.265)=0.103$