Answer:
a. Compute the standard error of the sample mean for HRC.
- mean = 502
- standard deviation = 100
- sample size = 60
- standard error = 100 / √60 = 12.9
b. What is the chance HRC finds a sample mean between $477 and $527?
P(477 ≤ X ≤ 527) = P(477 ≤ X - 502 ≤ 527 - 502)
= [(477 - 502) / 12.9] ≤ [(X - 502) / 12.9] ≤ [(527 - 502) / 12.9]
since (X - 502) / 12.9 = z, then
= -1.938 ≤ z ≤ 1.938
so P(477 ≤ X ≤ 527) = P(-1.938 ≤ z ≤ 0) + P(0 ≤ z ≤ 1.938)
z = 1.4662
P(477 ≤ X ≤ 527) = 0.4718 + 0.4718 = 0.9436
c. Calculate the likelihood that the sample mean is between $492 and $512.
P(492 ≤ X ≤ 512) = P(492 ≤ X - 502 ≤ 512 - 502)
= [(492 - 502) / 12.9] ≤ [(X - 502) / 12.9] ≤ [(512 - 502) / 12.9]
since (X - 502) / 12.9 = z, then
= -0.775 ≤ z ≤ 0.775
so P(477 ≤ X ≤ 527) = P(-0.775 ≤ z ≤ 0) + P(0 ≤ z ≤ 0.775)
z = 0.4906
P(477 ≤ X ≤ 527) = 0.2844 + 0.2844 = 0.5688
d. What is the probability the sample mean is greater than $550?
P(550 ≤ X) = P(550 - 502 ≤ X - 502)
= P(48/12.9 ≤ z)
= P(3.72 ≤ z)
= 0.5 - P(0 ≤ 3.72 ≤ z)
= 0.5 - 0.5 = 0
