Question

Given a mean score of 1150, standard deviation of 90, and 500 participants, solve the following problem. Using this data and the z-score distribution provided in class. Be sure to give your answer in the units requested. Only place your answer in the box.
1. What is the score for someone in the 15th percentile?
2. What is the percentile rank of someone with a score of 1100?
3. How many students have scores of 1060 or greater?
4. How many students scored between 1200 and 1250?

Answer 1

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 $\mu$ and standard deviation $\sigma$, the z-score of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

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 $\mu = 1150, \sigma = 90$

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.

$Z = \frac{X - \mu}{\sigma}$

$-1.037 = \frac{X - 1150}{90}$

$X - 1150 = -1.037*90$

$X = 1056.67$

2. What is the percentile rank of someone with a score of 1100?

This is the p-value of Z when X = 1100. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{1100 - 1150}{90}$

$Z = -0.555$

$Z = -0.555$ 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

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{1060 - 1150}{90}$

$Z = -1$

$Z = -1$ 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

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{1250 - 1150}{90}$

$Z = 1.1$

$Z = 1.1$ has a p-value of 0.8643.

X = 1200

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{1200 - 1150}{90}$

$Z = 0.555$

$Z = 0.555$ has a p-value of 0.7106

0.8643 - 0.7106 = 0.1537

Out of 500:

0.1537*500 = 77

77 is the answer.