Question

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a set of history exam scores are normally distributed with a mean of 83.2 points and a standard deviation of 8 points.
what proportion of exam scores are between 82 and 89.2 points?
you may round your answer to four decimal places.
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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 $x = 82$, $z_1=\frac{82 - 83.2}{8}=\frac{- 1.2}{8}=-0.15$.
For $x = 89.2$, $z_2=\frac{89.2 - 83.2}{8}=\frac{6}{8}=0.75$.

Step2: Use the standard normal distribution table

We want to find $P(-0.15We know that $P(-0.15From the standard - normal distribution table, $P(Z < 0.75)=0.7734$ and $P(Z < - 0.15)=0.4404$.
So $P(-0.15

Answer:

$0.3330$