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Sagot :
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) 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.
Normal Probability Distribution
In a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the z-score of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
- 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 [tex]\mu = 25000[/tex].
- The standard deviation is of [tex]\sigma = 5000[/tex].
Item a:
The probability is 1 subtracted by the p-value of Z when X = 35000, hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{35000 - 25000}{5000}[/tex]
[tex]Z = 2[/tex]
[tex]Z = 2[/tex] 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 p-value of Z when X = 22500, hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{22500 - 25000}{5000}[/tex]
[tex]Z = -0.5[/tex]
[tex]Z = -0.5[/tex] 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 p-value of Z when X = 37500 subtracted by the p-value of Z when X = 25000, hence:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{37500 - 25000}{5000}[/tex]
[tex]Z = 2.5[/tex]
[tex]Z = 2.5[/tex] has a p-value of 0.9938.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{25000 - 25000}{5000}[/tex]
[tex]Z = 0[/tex]
[tex]Z = 0[/tex] 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 X when Z = -1.035.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]-1.035 = \frac{X - 25000}{5000}[/tex]
[tex]X - 25000 = -1.035(5000)[/tex]
[tex]X = 19825[/tex]
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:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]-0.84 = \frac{X - 25000}{5000}[/tex]
[tex]X - 25000 = -0.84(5000)[/tex]
[tex]X = 20800[/tex]
80th percentile:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]0.84 = \frac{X - 25000}{5000}[/tex]
[tex]X - 25000 = 0.84(5000)[/tex]
[tex]X = 29200[/tex]
The salaries are $20,800 and $29,200.
More can be learned about the normal distribution at https://brainly.com/question/24663213
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