Question 20 (1 point)
Suppose a population is known to be normally distributed with a mean, $\mu$, equal to 116 and a standard deviation, $\sigma$, equal to 14 . Approximately what percent of the population would be between 88 and 102 ?
$13.5 \%$
$34 \%$
$95 \%$
$81.5 \%$
$47.5 \%$
$68 \%$

Step 1 ：First, we need to calculate the z-scores for 88 and 102. The z-score for 88 is calculated as \(z_1 = \frac{x_1 - \mu}{\sigma} = \frac{88 - 116}{14} = -2\), and the z-score for 102 is calculated as \(z_2 = \frac{x_2 - \mu}{\sigma} = \frac{102 - 116}{14} = -1\).

Step 2 ：Next, we need to look up these z-scores in the z-table to find the corresponding probabilities. The probability for a z-score of -2 is approximately 0.0228, and the probability for a z-score of -1 is approximately 0.1587.

Step 3 ：To find the percentage of the population between 88 and 102, we subtract the probability for 88 from the probability for 102. This gives us \(percentage = (p_2 - p_1) \times 100 = (0.1587 - 0.0228) \times 100 = 13.59\%\).

Step 4 ：Final Answer: The percentage of the population that would be between 88 and 102 is approximately \(\boxed{13.5\%}\).