Answer:
1. 1056.67
2. 29th percentile.
3. 79
4. 77
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:
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.
Mean score of 1150, standard deviation of 90
This means that ![\mu = 1150, \sigma = 90](https://tex.z-dn.net/?f=%5Cmu%20%3D%201150%2C%20%5Csigma%20%3D%2090)
1. What is the score for someone in the 15th percentile?
This is X when Z has a p-value of 0.15, so X when Z = -1.037.
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![-1.037 = \frac{X - 1150}{90}](https://tex.z-dn.net/?f=-1.037%20%3D%20%5Cfrac%7BX%20-%201150%7D%7B90%7D)
![X - 1150 = -1.037*90](https://tex.z-dn.net/?f=X%20-%201150%20%3D%20-1.037%2A90)
![X = 1056.67](https://tex.z-dn.net/?f=X%20%3D%201056.67)
2. What is the percentile rank of someone with a score of 1100?
This is the p-value of Z when X = 1100. 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{1100 - 1150}{90}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B1100%20-%201150%7D%7B90%7D)
![Z = -0.555](https://tex.z-dn.net/?f=Z%20%3D%20-0.555)
has a p-value of 0.29, so 29th percentile.
3. How many students have scores of 1060 or greater?
The proportion is 1 subtracted by the p-value of Z when X = 1060. 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{1060 - 1150}{90}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B1060%20-%201150%7D%7B90%7D)
![Z = -1](https://tex.z-dn.net/?f=Z%20%3D%20-1)
has a p-value of 0.1587.
Out of 500:
0.1587*500 = 79
79 is the answer.
4. How many students scored between 1200 and 1250?
The proportion is the p-value of Z when X = 1250 subtracted by the p-value of Z when X = 1200. So
X = 1250
![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{1250 - 1150}{90}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B1250%20-%201150%7D%7B90%7D)
![Z = 1.1](https://tex.z-dn.net/?f=Z%20%3D%201.1)
has a p-value of 0.8643.
X = 1200
![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{1200 - 1150}{90}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B1200%20-%201150%7D%7B90%7D)
![Z = 0.555](https://tex.z-dn.net/?f=Z%20%3D%200.555)
has a p-value of 0.7106
0.8643 - 0.7106 = 0.1537
Out of 500:
0.1537*500 = 77
77 is the answer.