A survey reported that the mean starting salary for college graduates in the spr

Question

A survey reported that the mean starting salary for college graduates in the spring of 2011 was $33 176. Assume that the distribution of starting salaries follows the normal distribution with a standard deviation of $3 500. What percent of the graduates have starting salaries:

a. Between $30 000 and $38 000? (Round final answer to 4 decimal places.)

Probability

b. More than $42 000? (Round final answer to 4 decimal places.)

Probability

c. Between $38 000 and $42 000? (Round final answer to 4 decimal places.)

Probability

Explanation / Answer

Given Mean = $33,176 and SD = $3,500

a. Between $30 000 and $38 000?

In order to find the probability we first need to find the Z values.

P(30,000<x<38,000)

When x = 30000

The z value is calculated as follows:

z = (30000 - 33176)/3500 = -0.9074

When x = 38000

The z value is calculated as follows:

z = (38000 - 33176)/3500 = 1.378

From the normal distribution table the values for the above mentioned z values are

For Z = -0.9074 = 0.1660

For Z = 1.378 = 0.9131

P = 0.9131-0.1660

P = 0.7471

b. More than $42 000

P(x>42000)

When x = 42000

The z value is calculated as follows:

z = (42000 - 33176)/3500 =2.521

P (Z>2.251) = [total area] - [area to the left of z = 2.521]

= 1 - 0.9941

P = 0.0059

c. Between $38 000 and $42 000

P(38000<x<42000)

When x = 38000

The z value is calculated as follows:

z = (38000 - 33176)/3500 = 1.378

From the normal distribution table the value for Z is as follows:

Z = 1.378 = 0.9131

When x = 42000

The z value is calculated as follows:

z = (42000 - 33176)/3500 =2.521

Z = 0.9941

Hence the Probability is 0.9941 - 0.9131 = 0.081

P = 0.081

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