Question

Business Weekly conducted a survey of graduates from 30 top MBA programs. On the basis of the survey, assume the mean annual salary for graduates 10 years after graduation is 121000 dollars. Assume the standard deviation is 41000 dollars. Suppose you take a simple random sample of 79 graduates.

Find the probability that a single randomly selected salary exceeds 117000 dollars.
P(X > 117000) =

Find the probability that a sample of size n=79 is randomly selected with a mean that exceeds 117000 dollars.
P(M > 117000) =

Enter your answers as numbers accurate to 4 decimal places.

Answer 1

X : annual salary for a graduate 10 years after graduation

X follows normal distribution with mean : 121000 dollars and  standard deviation : : 41000 dollars

Probability that a single randomly selected salary exceeds 117000 dollars : P(X > 117000) = 1-P(X 117000)

Z : (X-)/ : (X-121000)/41000 follows standard normal with mean=0 and standard deviation = 1

Z-score for 117000 = (117000-)/ = (117000-121000)/41000 = -4000/41000 = -0.10

P(X 117000) = P(Z-0.10)

From standard normal tables P(Z-0.10) = 0.4602

P(X 117000) = P(Z-0.10) = 0.4602

P(X > 117000) = 1-P(X 117000) = 1- 0.4602= 0.5398

Probability that a single randomly selected salary exceeds 117000 dollars :

P(X > 117000) = 0.5398

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When a random sample of n, then sample mean follows a normal distribution with mean : and standard deviation

a simple random sample of 79 graduates then sample mean: M follows a normal distribution mean : 121000 and standard deviation: = 4612.86

probability that a sample of size n=79 is randomly selected with a mean that exceeds 117000 dollars :

P(M > 117000) =1-P(M 117000)

P(M 117000)

Z-score for 117000 =  (117000-121000)/4612.86 = -4000/4612.86 = -0.87

From standard normal tables P(Z-0.87) = 0.1922

P(M 117000) = P(Z-0.87) = 0.1922

P(M > 117000) =1-P(M 117000) = 1 - 0.1922=0.8078

probability that a sample of size n=79 is randomly selected with a mean that exceeds 117000 dollars:

P(M > 117000) = 0.8078