Answer:
The range that you would expect 68.26 percent of the grades to fall is between 53 and 83.
Step-by-step explanation:
Problems of normally distributed samples can be 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 question:
![\mu = 68, \sigma = 15](https://tex.z-dn.net/?f=%5Cmu%20%3D%2068%2C%20%5Csigma%20%3D%2015)
Middle 68.26% of the grades:
From the
50 - (68.26/2) = 15.87th percentile
To the
50 + (68.26/2) = 84.13rd percentile.
15.87th percentile:
X when Z has a pvalue of 0.1587. So X when Z = -1.
![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 = \frac{X - 68}{15}](https://tex.z-dn.net/?f=-1%20%3D%20%5Cfrac%7BX%20-%2068%7D%7B15%7D)
![X - 68 = -1*15](https://tex.z-dn.net/?f=X%20-%2068%20%3D%20-1%2A15)
![X = 53](https://tex.z-dn.net/?f=X%20%3D%2053)
84.13rd percentile:
X when Z has a pvalue of 0.8413. So X when Z = 1.
![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 = \frac{X - 68}{15}](https://tex.z-dn.net/?f=1%20%3D%20%5Cfrac%7BX%20-%2068%7D%7B15%7D)
![X - 68 = 1*15](https://tex.z-dn.net/?f=X%20-%2068%20%3D%201%2A15)
![X = 83](https://tex.z-dn.net/?f=X%20%3D%2083)
The range that you would expect 68.26 percent of the grades to fall is between 53 and 83.