Using the normal distribution, it is found that:
a) There is a 0.0228 = 2.28% probability that a randomly chosen salary exceeds $35,000.
b)<em> </em>There is a 0.3085 = 30.85% probability that a randomly chosen salary is less than $22,500.
c) There is a 0.4938 = 49.38% probability that a randomly chosen salary lies between $25,000 and $37,500.
d) The 15th percentile of the sociology salaries is of $19,825.
e) The salaries are $20,800 and $29,200.
<h3>Normal Probability Distribution</h3>
In a normal distribution with mean and standard deviation , the z-score of a measure X is given by:
- It 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, which is the percentile of X.
In this problem:
- The mean is of .
- The standard deviation is of .
Item a:
The probability is <u>1 subtracted by the p-value of Z when X = 35000</u>, hence:
has a p-value of 0.9772.
1 - 0.9772 = 0.0228.
There is a 0.0228 = 2.28% probability that a randomly chosen salary exceeds $35,000.
Item b:
The probability is the <u>p-value of Z when X = 22500</u>, hence:
has a p-value of 0.3085.
There is a 0.3085 = 30.85% probability that a randomly chosen salary is less than $22,500.
Item c:
This probability is the <u>p-value of Z when X = 37500 subtracted by the p-value of Z when X = 25000</u>, hence:
has a p-value of 0.9938.
has a p-value of 0.5.
0.9938 - 0.5 = 0.4938
There is a 0.4938 = 49.38% probability that a randomly chosen salary lies between $25,000 and $37,500.
Item d:
This is X when Z has a p-value of 0.15, so <u>X when Z = -1.035</u>.
The 15th percentile of the sociology salaries is of $19,825.
Item e:
Due to the symmetry of the normal distribution, it is the 20th percentile and the 80th percentile, that is, X when Z = -0.84 and X when Z = 0.84.
20th percentile:
80th percentile:
The salaries are $20,800 and $29,200.
More can be learned about the normal distribution at brainly.com/question/24663213