Question

Use technology to solve the following problem: The mean annual income for people in a certain city (in thousands of dollars) is 44, with a standard deviation of 35. A pollster draws a sample of 59 people to interview. What is the probability that the sample mean income is between 42 and 48 (thousands of dollars)?

Answer 1

Answer:

$P(42 < \bar X$

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 scores for the limits of the interval we got:

$z= \frac{42-44}{\frac{35}{\sqrt{59}}}= -0.4$

$z= \frac{48-44}{\frac{35}{\sqrt{59}}}= 0.878$

And we want to find this probability:

$P(-0.4$

And we can use the foolowing excel command and we got:

=NORM.DIST(0.878;0;1;TRUE)-NORM.DIST(-0.4;0;1;TRUE)

And we got:

$P(-0.4$

Step-by-step explanation:

For this case we know the following parameters:

$\mu = 44 ,\sigma =35$

We select a sample size of n =59. So then the sample size is large enough to use the central limit theorem and the distribution for the sample mean is given by:

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

We want to find the following probability:

$P(42 < \bar X$

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 scores for the limits of the interval we got:

$z= \frac{42-44}{\frac{35}{\sqrt{59}}}= -0.4$

$z= \frac{48-44}{\frac{35}{\sqrt{59}}}= 0.878$

And we want to find this probability:

$P(-0.4$

And we can use the foolowing excel command and we got:

=NORM.DIST(0.878;0;1;TRUE)-NORM.DIST(-0.4;0;1;TRUE)

And we got:

$P(-0.4$