a set of statistics exam scores are normally distributed with a mean of 76.55 points and a standard…

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.
Answer
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 $P(0.49<Z<1.9)$. We know that $P(0.49<Z<1.9)=P(Z < 1.9)-P(Z < 0.49)$. From the standard - normal distribution table, $P(Z < 1.9)=0.9713$ and $P(Z < 0.49)=0.6879$.
Step3: Calculate the probability
$P(0.49<Z<1.9)=0.9713 - 0.6879=0.2834$.
Answer:
$0.2834$