Answer:
a) 0.84% of students scored over 32
b) 29.12% of students scored under 17
c) 70.04% of students scored between 17 and 32.
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 = 19.8, \sigma = 5.1](https://tex.z-dn.net/?f=%5Cmu%20%3D%2019.8%2C%20%5Csigma%20%3D%205.1)
a) About what percent of students scored over 32?
This is 1 subtracted by the pvalue of Z when X = 32. So
![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{32 - 19.8}{5.1}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B32%20-%2019.8%7D%7B5.1%7D)
![Z = 2.39](https://tex.z-dn.net/?f=Z%20%3D%202.39)
has a pvalue of 0.9916
1 - 0.9916 = 0.0084
So 0.84% of students scored over 32
b) About what percent of students scored under 17?
This is the pvalue of Z when X = 17. So
![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{17 - 19.8}{5.1}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B17%20-%2019.8%7D%7B5.1%7D)
![Z = -0.55](https://tex.z-dn.net/?f=Z%20%3D%20-0.55)
has a pvalue of 0.2912
So 29.12% of students scored under 17
c) About what percent of students scored between 17 and 32?
This is the pvalue of Z when X = 32 subtracted by the pvalue of Z when X = 12. So
X = 32
![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{32 - 19.8}{5.1}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B32%20-%2019.8%7D%7B5.1%7D)
![Z = 2.39](https://tex.z-dn.net/?f=Z%20%3D%202.39)
has a pvalue of 0.9916
X = 17
![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{17 - 19.8}{5.1}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B17%20-%2019.8%7D%7B5.1%7D)
![Z = -0.55](https://tex.z-dn.net/?f=Z%20%3D%20-0.55)
has a pvalue of 0.2912
0.9916 - 0.2912 = 0.7004
70.04% of students scored between 17 and 32.