Answer:
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
and standard deviation
, the z-score of a measure X is given by:
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
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](https://tex.z-dn.net/?f=%5Cmu%20%3D%2075%2C%20%5Csigma%20%3D%2015)
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}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![Z = \frac{75 - 75}{15}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B75%20-%2075%7D%7B15%7D)
![Z = 0](https://tex.z-dn.net/?f=Z%20%3D%200)
has a p-value of 0.5
X = 48
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![Z = \frac{48 - 75}{15}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B48%20-%2075%7D%7B15%7D)
![Z = -1.8](https://tex.z-dn.net/?f=Z%20%3D%20-1.8)
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.