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 190000 dollars. Assume the standard deviation is 39000 dollars. Suppose you take a simple random sample of 96 graduates. Find the probability that a single randomly selected salary has a mean value between 184029.4 and 194776.5 dollars.

P(184029.4 < X < 194776.5) = (Enter your answers as numbers accurate to 4 decimal places.) Find the probability that a random sample of size n = 96 has a mean value between 184029.4 and 194776.5 dollars.

P(184029.4 < ¯ x < 194776.5) = (Enter your answers as numbers accurate to 4 decimal places.)

Answer 1

Solution:

We are given

Mean = 190000

SD = 39000

n = 96

P(184029.4 < X < 194776.5) = P(X<194776.5) – P(X<184029.4)

Find P(X<194776.5)

Z = (X – mean) / SD

Z = (194776.5 - 190000) / 39000

Z = 0.122474

P(Z<0.122474) = P(X<194776.5) = 0.548738

(by using z-table)

Now, find P(X<184029.4)

Z = (X – mean) / SD

Z = (184029.4 - 190000) / 39000

Z = -0.15309

P(Z<-0.15309) = 0.439163

(by using z-table)

P(X<184029.4) = 0.439163

P(184029.4 < X < 194776.5) = P(X<194776.5) – P(X<184029.4)

P(184029.4 < X < 194776.5) = 0.548738 - 0.439163

P(184029.4 < X < 194776.5) = 0.109576

Required probability = 0.1096

Now, we have to find P(184029.4 < Xbar < 194776.5)

We are given

Mean = 190000

SD = 39000

n = 96

P(184029.4 < Xbar < 194776.5) = P(Xbar<194776.5) – P(Xbar<184029.4)

Find P(Xbar<194776.5)

Z = (Xbar – µ) / [σ/sqrt(n)]

Z = (194776.5 - 190000) / [39000/sqrt(96)]

Z = (194776.5 - 190000) / 3980.421

Z = 1.2

P(Z<1.2) = P(Xbar<194776.5) = 0.88493

(by using z-table)

Now, find P(Xbar<184029.4)

Z = (Xbar – µ) / [σ/sqrt(n)]

Z = (184029.4 - 190000) / [39000/sqrt(96)]

Z = (184029.4 - 190000) / 3980.421

Z = -1.5

P(Z<-1.5) = 0.066808

(by using z-table)

P(Xbar<184029.4) = 0.066808

P(184029.4 < Xbar < 194776.5) = P(Xbar<194776.5) – P(Xbar<184029.4)

P(184029.4 < Xbar < 194776.5) = 0.88493 - 0.066808

P(184029.4 < Xbar < 194776.5) = 0.818122

Required probability = 0.8181