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Sociology graduates, upon entering the workforce, earn a mean salary of
$25,000 with a standard deviation of $5,000. Assume that the distribution of salaries is Normal.
a) What is the probability that a randomly chosen salary exceeds $35,000?
b) What is the probability that a randomly chosen salary is less than $22,500?
c) What is the probability that a randomly chosen salary lies between
$25,000 and $37,500?
d) Find the 15th percentile of the sociology salaries.
e) Find the two salaries that are the cutoff values for the middle 60% of
salaries.


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