Question

Suppose median family income is normally distributed with mean M = $44000 and standard deviation O = $15000. Let 2 represent median family income. What is the probability that for a family chosen at random 2' lies between $29600 and $ 30500?

Answer 1

First, let's calculate the critical value z for each of the given end values of the interval, that is, 29600 and 30500.

To do so, we can use the formula below:

$\begin{gathered} z=\frac{x-\mu}{\sigma}\\ \\ z_1=\frac{29600-44000}{15000}=-\frac{14400}{15000}=-0.96\\ \\ z_2=\frac{30500-44000}{15000}=-\frac{13500}{15000}=-0.9 \end{gathered}$

Now, we need to find the probabilities associated to each of these critical values.

Looking at the z-table, for z = -0.96 we have p = 0.1685 and for z = -0.9 we have p = 0.1841.

Since we want the values inside this interval, let's calculate the difference of the probabilities:

$p=0.1841-0.1685=0.0156$

Therefore the probability is 0.0156.