Human Resource Consulting (HRC) surveyed a random sample of 60 Twin Cities construction companies to find information on the costs of their health care plans. One of the items being tracked is the annual deductible that employees must pay. The Minnesota Department of Labor reports that historically the mean deductible amount per employee is $502 with a standard deviation of $100.

a. Compute the standard error of the sample mean for HRC.
b. What is the chance HRC finds a sample mean between $477 and $527?
c. Calculate the likelihood that the sample mean is between $492 and $512.
d. What is the probability the sample mean is greater than $550?

a. Compute the standard error of the sample mean for HRC.

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

answer; i believe both technician (a) and (b) are correct; /// good luck