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 82 and 90? (remember to refer to the normal distribution located on page 8 of the lesson).

a. 47.5%

b. 68%

c. 34%

d. 13.5%

Explanation:

Step1: Calculate the z - score

The z - score formula is $z=\frac{x-\mu}{\sigma}$, where $x$ is the value, $\mu$ is the mean, and $\sigma$ is the standard deviation. Here, $\mu = 82$, $\sigma=4$, and $x = 90$. So $z=\frac{90 - 82}{4}=\frac{8}{4}=2$.
The mean has a z - score of $z_1=\frac{82 - 82}{4}=0$.

Step2: Use the standard normal distribution

The area under the standard - normal curve between $z = 0$ and $z = 2$ can be found using the properties of the standard normal distribution. The total area under the standard - normal curve is 1. The area to the left of $z = 0$ is 0.5, and the area to the left of $z = 2$ is approximately 0.9772.
The area between $z = 0$ and $z = 2$ is $0.9772-0.5 = 0.4772\approx47.5\%$.

Answer:

A. 47.5%