Question

Human Resource Consulting (HRC) surveyed a random sample of 53 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 $501 with a standard deviation of $102.

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

Answer)

We need to use standard normal z table to estimate the answers.

Given mean = 501

Standard deviation = 102

A)

Standard error is given by

Standard deviation/√sample size

Sample size = n = 53

Standard error (SE) = 102/√53

SE = 14.010777522766

B)

P(477<x<527) = p(x<527) - p(x<477)

P(x<527)

Z = (x-mean)/(SE)

Z = (527-501)/14.010777522766

Z = 1.86

From z table, p(z<1.86) = 0.9686

P(x<477)

Z = -1.71

From z table, p(z<-1.71) = 0.0436

Required probability is 0.9686 - 0.0436 = 0.925

C)

P(492<x<512)

P(x<512)

Z = (512-501)/(SE)

Z = 0.79

From z table, p(z<0.79) = 0.7852

P(x<492)

Z = -0.64

From z table, p(z<-0.64) = 0.2611

Required answer is 0.7852 - 0.2611

= 0.5241

D)

P(x>530)

Z = (530-501)/(SE)

Z = 2.07

From z table, p(z>2.07) = 0.0192