Question

a set of chemistry exam scores are normally distributed with a mean of 70 points and a standard deviation of 5 points. what proportion of exam scores are between 68 and 73 points? you may round your answer to four decimal places.

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 value from the data set.
For $x = 68$, $\mu=70$, and $\sigma = 5$, the z - score $z_1=\frac{68 - 70}{5}=\frac{- 2}{5}=-0.4$.
For $x = 73$, $\mu = 70$, and $\sigma=5$, the z - score $z_2=\frac{73 - 70}{5}=\frac{3}{5}=0.6$.

Step2: Use the standard normal distribution table

We want to find $P(-0.4We know that $P(-0.4From the standard - normal distribution table, $P(Z < 0.6)=0.7257$ and $P(Z<-0.4)=0.3446$.
So $P(-0.4

Answer:

$0.3811$