PLEASE SHOW WORK he Economic Policy Institute reported that the annual salary of

Question

PLEASE SHOW WORK

he Economic Policy Institute reported that the annual salary of all MBA graduates, 5 years after graduation, was $150,000. Assume the standard deviation of annual salary of all MBA graduates is $35,000. a. What is the probability that the mean salary of a sample of 34 MBA graduates is withirn above the population mean)? Show your steps and manual (hand and calculator) calculations in 4 decimal points, where possible. b. What is the probability that the mean salary of a sample of 34 MBA graduates is at most $150,000? Show your steps and manual (hand and calculator) calculations in 4 decimal points, where possible. c, what is the average salary of the top 8% of the MBA graduates? Show your steps and manual calculations.

Explanation / Answer

given mean = 150000 sd = 35000 , please keep the z tables ready

a)
we know the z score for sample is given ass

Z = (X-Mean)/(sd/sqrt(n)), where n is sample size
below mean = 150,000 - 12000 = 138000

Z = (138000-150000)/(35000/sqrt(35)) = -2.02

above mean = 150,000 + 12000 = 1162000

Z = (162000-150000)/(35000/sqrt(35)) = 2.02
To find the probability of P (2.02<Z<2.02), we use the following formula:

P (2.02<Z<2.02 )=P ( Z<2.02 )P (Z<2.02 )
We see that P ( Z<2.02 )=0.9783.
P ( Z<2.02 ) can be found by using the following fomula.
P ( Z<a)=1P ( Z<a )
After substituting a=2.02 we have:
P ( Z<2.02)=1P ( Z<2.02 )
We see that P ( Z<2.02 )=0.9783 so,
P ( Z<2.02)=1P ( Z<2.02 )=10.9783=0.0217
At the end we have:
P (2.02<Z<2.02 )=0.9566

b) at most 150000
Z = (150000-150000)/(35000/sqrt(35))
Z = 0
We conclude that:
P ( Z<0 )=0.5

c)

for the tope 8% we first calculate the z score from the z table for p = 0.92 , which is 1.405

now using the formula
Z = (X-Mean)/(sd)

X = 1.405*35000 +150000 = 164175

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