Question

the summary statistics shows the distribution of the percent of foreign - born residents in the 50 states.

n	mean	sd	min	q1	med	q3	max

montana had 1.9% foreign - born residents. find and interpret the z - score for montana.
o z = 1.12
montanas percent of foreign - born residents is 1.12 standard deviations above the mean percent of foreign - born residents for all states.
o z=-1.12
montanas percent of foreign - born residents is 1.12 standard deviations below the mean percent of foreign - born residents for all states.
o z = 0.47
montanas percent of foreign - born residents is 0.47 standard deviations above the mean percent of foreign - born residents for all states.
o z=-1.12
it is 1.12 standard deviations below the average.

Explanation:

Step1: Recall z - score formula

The formula for the z - score is $z=\frac{x - \mu}{\sigma}$, where $x$ is the data - point, $\mu$ is the mean, and $\sigma$ is the standard deviation.

Step2: Identify values from the table

We are given that $x = 1.9$ (percent of foreign - born residents in Montana), $\mu=8.73$ (mean percent of foreign - born residents), and $\sigma = 6.12$ (standard deviation of percent of foreign - born residents).

Step3: Calculate the z - score

$z=\frac{1.9 - 8.73}{6.12}=\frac{- 6.83}{6.12}\approx - 1.12$

Step4: Interpret the z - score

A negative z - score means the data - point is below the mean. So Montana's percent of foreign - born residents is 1.12 standard deviations below the mean percent of foreign - born residents for all states.

Answer:

$z=-1.12$. Montana's percent of foreign - born residents is 1.12 standard deviations below the mean percent of foreign - born residents for all states.