Question

You may need to use the appropriate appendix table or technology to answer this question.

According to the National Association of Colleges and Employers, the 2015 mean starting salary for new college graduates in health sciences was $51,541. The mean 2015 starting salary for new college graduates in business was $53,901. † Assume that starting salaries are normally distributed and that the standard deviation for starting salaries for new college graduates in health sciences is $11,000. Assume that the standard deviation for starting salaries for new college graduates in business is $17,000.

(a)

What is the probability that a new college graduate in business will earn a starting salary of at least $65,000? (Round your answer to four decimal places.)

(b)

What is the probability that a new college graduate in health sciences will earn a starting salary of at least $65,000? (Round your answer to four decimal places.)

(c)

What is the probability that a new college graduate in health sciences will earn a starting salary less than $46,000? (Round your answer to four decimal places.)

(d)

How much would a new college graduate in business have to earn in dollars in order to have a starting salary higher than 99% of all starting salaries of new college graduates in the health sciences? (Round your answer to the nearest whole number.)

$

Answer 1

a)
Here, μ = 53901, σ = 17000 and x = 65000. We need to compute P(X >= 65000). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z = (65000 - 53901)/17000 = 0.65

Therefore,
P(X >= 65000) = P(z <= (65000 - 53901)/17000)
= P(z >= 0.65)
= 1 - 0.7422
= 0.2578

b)

Here, μ = 51541, σ = 11000 and x = 65000. We need to compute P(X >= 65000). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z = (65000 - 51541)/11000 = 1.22

Therefore,
P(X >= 65000) = P(z <= (65000 - 51541)/11000)
= P(z >= 1.22)
= 1 - 0.8888
= 0.1112

c)

Here, μ = 51541, σ = 11000 and x = 46000. We need to compute P(X <= 46000). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z = (46000 - 51541)/11000 = -0.5

Therefore,
P(X <= 46000) = P(z <= (46000 - 51541)/11000)
= P(z <= -0.5)
= 0.3085

d)

z value at 99% = 2.33

z = (x - mean)/sigma

2.33 = ( x - 51541)/11000

x =11000 * 2.33+51541
x = 77171