Answer:
Grades between 62 and 64 result in a D grade.
Step-by-step explanation:
Problems of normally distributed samples are solved using the z-score formula.
In a set with mean
and standard deviation
, the zscore 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 pvalue, we get the probability that the value of the measure is greater than X.
In this problem, we have that:
![\mu = 71.9, \sigma = 7.8](https://tex.z-dn.net/?f=%5Cmu%20%3D%2071.9%2C%20%5Csigma%20%3D%207.8)
Find the numerical limits for a D grade.
D: Scores below the top 84% and above the bottom 10%
So below the 100-84 = 16th percentile and above the 10th percentile.
16th percentile:
This is the value of X when Z has a pvalue of 0.16. So X when Z = -0.995.
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![-0.995 = \frac{X - 71.9}{7.8}](https://tex.z-dn.net/?f=-0.995%20%3D%20%5Cfrac%7BX%20-%2071.9%7D%7B7.8%7D)
![X - 71.9 = -0.995*7.8](https://tex.z-dn.net/?f=X%20-%2071.9%20%3D%20-0.995%2A7.8)
![X = 64](https://tex.z-dn.net/?f=X%20%3D%2064)
10th percentile:
This is the value of X when Z has a pvalue of 0.1. So X when Z = -1.28.
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![-1.28 = \frac{X - 71.9}{7.8}](https://tex.z-dn.net/?f=-1.28%20%3D%20%5Cfrac%7BX%20-%2071.9%7D%7B7.8%7D)
![X - 71.9 = -1.28*7.8](https://tex.z-dn.net/?f=X%20-%2071.9%20%3D%20-1.28%2A7.8)
![X = 62](https://tex.z-dn.net/?f=X%20%3D%2062)
Grades between 62 and 64 result in a D grade.