Question

Human Resource Consulting (HRC) surveyed a random sample of 66 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 $499 with a standard deviation of $100.

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

2. What is the chance HRC finds a sample mean between $477 and $527?

3. Calculate the likelihood that the sample mean is between $492 and $512.

4. What is the probability the sample mean is greater than $530?

Answer 1

(1)

Standard error is given by:
SE = s/

= 100/ = 12.3091

So,

Answer is:

12.3091

(2)

To find P(477 < < 527)

Case1: for from 477 to mid value:

Z = (477 - 499)/12.3091 = - 1.7873

Table of Area Under Standard Normal Curve gives area = 0.4633

Case2: for from mid value to 527:

Z = (527 - 499)/12.3091 = 2.2747

Table gives area = 0.4884

So,

P(477 < 499) = 0.4633 + 0.4884 = 0.9517

So,

Answer is:

0.9547

(3)

To find P(492 < < 512)

Case1: for from 477 to mid value:

Z = (492 - 499)/12.3091 = - 0.5687

Table of Area Under Standard Normal Curve gives area = 0.2157

Case2: for from mid value to 512:

Z = (512 - 499)/12.3091 = 1.0561

Table gives area = 0.3554

So,

P(477 < 499) = 0.2157 + 0.3554 = 0.5711

So,

Answer is:

0.5711

(4)

To find P(>530):
Z = (530 - 499)/12.3091 = 2.5185

Table gives area = 0.4941

So,

P(>530) = 0.5 - 0.4941 = 0.0059

So,

Answer is:

0.0059