Question

A survey reported that the mean starting salary for college graduates after a three-year program was $35,710.Assume that the distribution of starting salaries follows the normal distribution with a standard deviation of $3320. What percentage of the graduates have starting salaries: (Round z-score computation to 2 decimal places and the final answers to 4 decimal places.) a. Between $31,800 and $39,200? Probability b. More than $43,600? Probability c. Between $39,200 and $43,600? Probability

Answer 1

Mean = = 35710

Standard deviation = = 3320

a)

We have to find P( 31800 < X < 39200)

For finding this probability we have to find z score.

That is we have to find P( - 1.18 < Z < 1.05)

P( - 1.18 < Z < 1.05) = P(Z < 1.05) - P(Z < - 1.18) = 0.8531 - 0.1190 = 0.7341 ( From z table)

73.41%

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b)

We have to find P(X > 43600)

For finding this probability we have to find z score.

That is we have to find P(Z > 2.38)

P(Z > 2.38) = 1 - P(Z < 2.38) = 1 - 0.9913 = 0.0087 ( From z table)

0.87%

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c)

We have to find P( 39200 < X < 43600)

For finding this probability we have to find z score.

That is we have to find P( 1.05 < Z < 2.38)

P( 1.05 < Z < 2.38) = P(Z < 2.38) - P(Z < 1.05) = 0.9913 - 0.8531 = 0.1382 ( From z table)

13.82%

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