Foreign-born residents Refer to Exercise $11$ Here are summary statistics for

the percent of foreign-born residents in the $50$ states:

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.

The table is

Value, $x=1.9$

Mean, $\overline{x}=8.73$

Standard deviation, $SD=6.12$

The formula used:$z=\frac{value-mean}{s\tan darddeviation}$

The amount of standard deviations a value deviates from the mean is represented by the $z-$score.

A positive$z-$score implies a value that is higher than the mean.

Whereas,

When the $z-$score is negative, it means the value is below the mean.

Calculate the $z-$score:

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

Therefore,

Montana's percentage of foreign-born residents is $1.12$standard deviations lower than the national average of foreign-born residents.

The amount of standard deviations a value deviates from the mean is represented by the $z-$score.

A positive$z-$score implies a value that is higher than the mean.

Whereas,

When the $z-$score is negative, it means the value is below the mean.

Calculate the$z-$score:

$z=\frac{x-\overline{x}}{SD}$

Substitute the values:

$2.10=\frac{x-8.73}{6.12}$

Multiply both sides by $6.12$:

12.852 = x − 8.73

Add $8.73$to both sides:

$12.852+8.73=x-8.73+8.73$

That becomes

$x=21.582$

Therefore,

New York has approx. $21.582\%$of foreign $-$based residents.