Answer:
a) 68.26% probability that a student scores between 350 and 550
b) A score of 638(or higher).
c) The 60th percentile of test scores is 475.3.
d) The middle 30% of the test scores is between 411.5 and 488.5.
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 = 450, \sigma = 100](https://tex.z-dn.net/?f=%5Cmu%20%3D%20450%2C%20%5Csigma%20%3D%20100)
a. What is the probability that a student scores between 350 and 550?
This is the pvalue of Z when X = 550 subtracted by the pvalue of Z when X = 350. So
X = 550
![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{550 - 450}{100}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B550%20-%20450%7D%7B100%7D)
![Z = 1](https://tex.z-dn.net/?f=Z%20%3D%201)
has a pvalue of 0.8413
X = 350
![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{350 - 450}{100}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B350%20-%20450%7D%7B100%7D)
![Z = -1](https://tex.z-dn.net/?f=Z%20%3D%20-1)
has a pvalue of 0.1587
0.8413 - 0.1587 = 0.6826
68.26% probability that a student scores between 350 and 550
b. If the upper 3% scholarship, what score must a student receive to get a scholarship?
100 - 3 = 97th percentile, which is X when Z has a pvalue of 0.97. So it is X when Z = 1.88
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![1.88 = \frac{X - 450}{100}](https://tex.z-dn.net/?f=1.88%20%3D%20%5Cfrac%7BX%20-%20450%7D%7B100%7D)
![X - 450 = 1.88*100](https://tex.z-dn.net/?f=X%20-%20450%20%3D%201.88%2A100)
![X = 638](https://tex.z-dn.net/?f=X%20%3D%20638)
A score of 638(or higher).
c. Find the 60th percentile of the test scores.
X when Z has a pvalue of 0.60. So it is X when Z = 0.253
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![0.253 = \frac{X - 450}{100}](https://tex.z-dn.net/?f=0.253%20%3D%20%5Cfrac%7BX%20-%20450%7D%7B100%7D)
![X - 450 = 0.253*100](https://tex.z-dn.net/?f=X%20-%20450%20%3D%200.253%2A100)
![X = 475.3](https://tex.z-dn.net/?f=X%20%3D%20475.3)
The 60th percentile of test scores is 475.3.
d. Find the middle 30% of the test scores.
50 - (30/2) = 35th percentile
50 + (30/2) = 65th percentile.
35th percentile:
X when Z has a pvalue of 0.35. So X when Z = -0.385.
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![-0.385 = \frac{X - 450}{100}](https://tex.z-dn.net/?f=-0.385%20%3D%20%5Cfrac%7BX%20-%20450%7D%7B100%7D)
![X - 450 = -0.385*100](https://tex.z-dn.net/?f=X%20-%20450%20%3D%20-0.385%2A100)
![X = 411.5](https://tex.z-dn.net/?f=X%20%3D%20411.5)
65th percentile:
X when Z has a pvalue of 0.35. So X when Z = 0.385.
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![0.385 = \frac{X - 450}{100}](https://tex.z-dn.net/?f=0.385%20%3D%20%5Cfrac%7BX%20-%20450%7D%7B100%7D)
![X - 450 = 0.385*100](https://tex.z-dn.net/?f=X%20-%20450%20%3D%200.385%2A100)
![X = 488.5](https://tex.z-dn.net/?f=X%20%3D%20488.5)
The middle 30% of the test scores is between 411.5 and 488.5.