Question

suppose a normal distribution of scores has a mean of 48 and a standard deviation of 2 for #15 - 16. what percentage of scores fall within 42 - 54?
o 68%
o 95%
o 99.7%
o 100%

Explanation:

Step1: Calculate z - scores

The formula for the z - score is $z=\frac{x-\mu}{\sigma}$, where $\mu$ is the mean, $\sigma$ is the standard deviation, and $x$ is the data - point. Given $\mu = 48$ and $\sigma=2$. For $x = 42$, $z_1=\frac{42 - 48}{2}=\frac{-6}{2}=-3$. For $x = 54$, $z_2=\frac{54 - 48}{2}=\frac{6}{2}=3$.

Step2: Use the empirical rule

In a normal distribution, approximately 99.7% of the data lies within $z=-3$ and $z = 3$.

Answer:

99.7%