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 160000 dollars. Assume the standard deviation is 42000 dollars. Suppose you take a simple random sample of 100 graduates.

Find the probability that a single randomly selected policy has a mean value between 155800 and 157900 dollars. P(155800 < X < 157900) = (Enter your answers as numbers accurate to 4 decimal places.)

Find the probability that a random sample of size n = 100 n=100 has a mean value between 155800 and 157900 dollars. P(155800 < M < 157900) =

(Enter your answers as numbers accurate to 4 decimal places.)

Answer 1

Here, μ = 160000, σ = 42000, x1 = 155800 and x2 = 157900. We need to compute P(155800<= X <= 157900). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z1 = (155800 - 160000)/42000 = -0.1
z2 = (157900 - 160000)/42000 = -0.05

Therefore, we get
P(155800 <= X <= 157900) = P((157900 - 160000)/42000) <= z <= (157900 - 160000)/42000)
= P(-0.1 <= z <= -0.05) = P(z <= -0.05) - P(z <= -0.1)
= 0.4801 - 0.4602
= 0.0199

Here, μ = 160000, σ = 4200, x1 = 155800 and x2 = 157900. We need to compute P(155800<= X <= 157900). The corresponding z-value is calculated using Central Limit Theorem

z = (x - μ)/σ
z1 = (155800 - 160000)/4200 = -1
z2 = (157900 - 160000)/4200 = -0.5

Therefore, we get
P(155800 <= X <= 157900) = P((157900 - 160000)/4200) <= z <= (157900 - 160000)/4200)
= P(-1 <= z <= -0.5) = P(z <= -0.5) - P(z <= -1)
= 0.3085 - 0.1587
= 0.1498