On a standard coordinate

464 students will score between 48 and 75. Using the z-distribution, we measure how many standard deviations each score is from the mean, then find the p-value associated with each score to find the proportion, and from the proportion, we find how many out of 1000.

Step-by-step explanation:

Normal Probability Distribution:

Problems of normal distributions can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

It assumes the scores are normally distributed with a mean score of 75 and a standard deviation of 15.

This means that $\mu = 75, \sigma = 15$

How many students will score between 48 and 75?

First we find the proportion, which is the pvalue of Z when X = 75 subtracted by the pvalue of Z when X = 48. So

X = 75

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{75 - 75}{15}$

$Z = 0$

$Z = 0$ has a p-value of 0.5

X = 48

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{48 - 75}{15}$

$Z = -1.8$

$Z = -1.8$ has a p-value of 0.0359

1 - 0.0359 = 0.4641

Out of 1000:

0.4641*1000 = 464

464 students will score between 48 and 75. Using the z-distribution, we measure how many standard deviations each score is from the mean, then find the p-value associated with each score to find the proportion, and from the proportion, we find how many out of 1000.

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