Question 1165215: Human Resource Consulting (HRC) surveyed a random sample of 72 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 $125. (Round your z value to 2 decimal places and final answers to 4 decimal places. Leave no cells—blankbe certain to enter "0" if required.)
Compute the standard error of the sample mean for HRC.
What is the chance HRC finds a sample mean between $477 and $527?
Calculate the likelihood that the sample mean is between $492 and $512.
What is the probability the sample mean is greater than $550?
Answer by Boreal(15235)   (Show Source):
You can put this solution on YOUR website! The SEM is the sd/sqrt(n)=125/sqrt(72)=$14.73
z between (477-501)/14.73 and (527-501)/14.73
that is -1.63 and +1.77
probability is 0.9101
for the second, it would be -0.61 and +0.75 with probability of 0.5024
the last is probability z is > 49/14.73 or 0.0004
checking on the calculator which will be different because not rounded
2nd VARS 2 normalcdf ENTER (477,427,501,14.73) for the second and get 0.9096, so the above is a reasonable answer.
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