Question

texes mathematics 4-8 (115)

1. given a normal distribution with a mean of 60 and the standard deviation is 5, what percent of the data lies between the values of 65 and 75?

a. 13.59%
b. 15.7%
c. 21.50%
d. 15%

Explanation:

⚡ Using what you learned: normal distribution

Step 1: Calculate the z-scores

Find the z-scores for the boundary values \( x_1 = 65 \) and \( x_2 = 75 \) using the formula:

$$ z = \frac{x - \mu}{\sigma} $$

Given:

• Mean (\( \mu \)) = \( 60 \)
• Standard deviation (\( \sigma \)) = \( 5 \)

For \( x_1 = 65 \):

$$ z_1 = \frac{65 - 60}{5} = 1 $$

For \( x_2 = 75 \):

$$ z_2 = \frac{75 - 60}{5} = 3 $$

Step 2: Find the percentage between the z-scores

Using the Empirical Rule (68-95-99.7 rule) for a normal distribution:

• The area within \( 1 \) standard deviation of the mean (\( z = -1 \) to \( z = 1 \)) is approximately \( 68.2\% \).
• The area within \( 2 \) standard deviations of the mean (\( z = -2 \) to \( z = 2 \)) is approximately \( 95.4\% \).
• The area within \( 3 \) standard deviations of the mean (\( z = -3 \) to \( z = 3 \)) is approximately \( 99.7\% \).

To find the percentage of data between \( z = 1 \) and \( z = 3 \):

$$ P(1 < Z < 3) = P(0 < Z < 3) - P(0 < Z < 1) $$

Using the symmetry of the normal distribution:

$$ P(0 < Z < 3) = \frac{99.7\%}{2} = 49.85\% $$

$$ P(0 < Z < 1) = \frac{68.2\%}{2} = 34.1\% $$

Subtract the two values:

$$ P(1 < Z < 3) = 49.85\% - 34.1\% = 15.75\% $$

This matches option B (\( 15.7\% \)).

Answer:

B. 15.7%