Question

Human Resource Consulting (HRC) is surveying a sample of 66 firms in order to study health care costs for a client. One of the items being tracked is the annualdeductible that employees must pay. The state Bureau of Labor reports the mean of this distribution is $498 with a standard deviation of $85.

(a) Compute the standard error of the sample mean for HRC. (Round your answer to 2 decimal places.)

Standard error

(b) What is the chance HRC finds a sample mean between $477 and $527? (Round z value to 2 decimal places and final answer to 4 decimal places.)


Probability

(c) Calculate the likelihood that the sample mean is between $492 and $512. (Round z value to 2 decimal places and final answer to 4 decimal places.)


Probability

(d) What is the probability the sample mean is greater than $550? (Round z value to 2 decimal places and final answer to 4 decimal places.)


Probability

Answer 1

Let Xbar be the sample mean.
Since the distribution from which the sample is taken has a mean $502 and standard deviation $100, the sample mean Xbar has a mean $502 and standard error$100/Sqrt(60) = $12.91.
Thus Z =(Xbar - 502)/12.91 follows a Standard Normal distribution.

a) The standard error of the sample mean for HRC = s/vn = $502/v60 = $12.91.

b) The chance HRC finds a sample mean between $477 and $527 is given by
P[$477 < Xbar < $527] = P[(477 - 502)/12.91 < (Xbar - 502)/12.91 <(527 - 502)/12.91]
= P[-1.9365 < Z < 1.9365]
= P[Z < 1.9365] - P[Z < -1.9365]
= 0.9736 - 0.0264 = 0.9742
c) The likelihood that the sample mean is between $492 and $512 is given by
P[$492 < Xbar < $512] = P[(492 - 502)/12.91 < (Xbar - 502)/12.91 <(512 - 502)/12.91]
= P[-0.7746 < Z < 0.7746]
= P[Z < 0.7746] - P[Z < -0.7746]
= 0.7807 - 0.2193 = 0.5614
d) The probability the sample mean is greater than $550 is given by,
P[xbar > $550] = P[(Xbar - 502)/12.91 > (550 - 502)/12.91]
= P[Z > 3.7181]
= 1 - P[Z < 3.7181]
= 1 - 0.9999 = 0.0001