Question

1. foreign - born residents refer to exercise 11. here are summary statistics for the percent of foreign - born residents in the 50 states: n mean sd min q1 med q3 max 50 8.73 6.12 1.3 4.1 6.2 13.4 27.1 (a) find and interpret the z - score for montana, which had 1.9% foreign - born residents. (b) new york had a standardized score of 2.10. find the percent of foreign - born residents in new york at that time.

Explanation:

Step1: Recall z - score formula

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

Step2: Calculate z - score for Montana

Given $\mu = 8.73$, $\sigma=6.12$, and $x = 1.9$. Substitute these values into the formula: $z=\frac{1.9 - 8.73}{6.12}=\frac{- 6.83}{6.12}\approx - 1.12$. The interpretation is that the percentage of foreign - born residents in Montana is approximately 1.12 standard deviations below the mean percentage of foreign - born residents across the 50 states.

Step3: Use z - score formula to find $x$ for New York

We know $z = 2.10$, $\mu = 8.73$, and $\sigma=6.12$. Rearranging the z - score formula $z=\frac{x-\mu}{\sigma}$ to solve for $x$ gives $x=\mu+z\sigma$. Substitute the values: $x = 8.73+2.10\times6.12=8.73 + 12.852=21.582$.

Answer:

(a) The z - score for Montana is approximately $-1.12$. It means the percentage of foreign - born residents in Montana is about 1.12 standard deviations below the mean.
(b) The percent of foreign - born residents in New York is approximately $21.58\%$.