Question

a set of statistics exam scores are normally distributed with a mean of 76.55 points and a standard deviation of 5 points. what proportion of exam scores are between 79 and 86.05 points? you may round your answer to four decimal places.

Explanation:

Step1: Calculate z - scores

The z - score formula is $z=\frac{x-\mu}{\sigma}$, where $\mu$ is the mean, $\sigma$ is the standard deviation, and $x$ is the data - point.
For $x = 79$, $z_1=\frac{79 - 76.55}{5}=\frac{2.45}{5}=0.49$.
For $x = 86.05$, $z_2=\frac{86.05 - 76.55}{5}=\frac{9.5}{5}=1.9$.

Step2: Use the standard normal distribution table

We want to find $P(0.49We know that $P(0.49From the standard - normal distribution table, $P(Z < 1.9)=0.9713$ and $P(Z < 0.49)=0.6879$.

Step3: Calculate the probability

$P(0.49

Answer:

$0.2834$