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

Population mean, for heath science µ = 51541

Population standard deviation, σ = 11000

Population mean for business, µ = 53901

Population standard deviation, σ = 17000

a) Probability that a new college graduate in business will earn a starting salary of at least $65000 =

= P(X > 65000)

= P( (X-µ)/σ > (65000-53901)/17000)

= P(z > 0.6529)

= 1 - P(z < 0.6529)

Using excel function:

= 1 - NORM.S.DIST(0.6529, 1)

= 0.2569

b) Probability that a new college graduate in health sciences will earn a starting salary of at least 65000 =

= P(X > 65000)

= P( (X-µ)/σ > (65000-51541)/11000)

= P(z > 1.2235)

= 1 - P(z < 1.2235)

Using excel function:

= 1 - NORM.S.DIST(1.2235, 1)

= 0.1106

c) Probability that a new college graduate in health sciences will earn a starting salary less than 46000 =

= P(X < 46000)

= P( (X-µ)/σ < (46000-51541)/11000 )

= P(z < -0.5037)

Using excel function:

= NORM.S.DIST(-0.5037, 1)

= 0.3072

d) P(z > x) = 0.99

= 1 - P(z < x) = 0.99

= P(z < x) = 0.01

Z score at p = 0.01 using excel = ABS(NORM.S.INV(0.01)) = 2.33

Value of X = µ + z*σ = 51541 + (2.33)*11000 = $ 77171