Question

mr. emmer gave a test in his chemistry class. the scores were normally distributed with a mean of 82 and a standard deviation of 4. what percent of students would you expect to score between 74 and 78? (remember to refer to the normal - distribution located on page 3 of the lesson).
○ a. 47.5%
○ b. 68%
○ c. 34%
○ d. 13.5%

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.
For $x = 74$, $z_1=\frac{74 - 82}{4}=\frac{- 8}{4}=-2$.
For $x = 78$, $z_2=\frac{78 - 82}{4}=\frac{-4}{4}=-1$.

Step2: Use the standard normal distribution table

The area to the left of $z_1=-2$ from the standard - normal table is $0.0228$.
The area to the left of $z_2=-1$ from the standard - normal table is $0.1587$.

Step3: Find the area between the two z - scores

The area between $z=-2$ and $z = - 1$ is $P(-2Converting to a percentage, we get $13.59\%$.

Answer:

D. 13.59%