Human Resource Consulting (HRC) is surveying a sample of 60 Twin cities in order

Question

Human Resource Consulting (HRC) is surveying a sample of 60 Twin cities in order to study health care costs for a client. One of the items being tracked is the annual deductible that employees must pay. The Minnesota Department of Labor reports the mean of this distribution is $502 with a standard deviation of $100. (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. Round your 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. Round your 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.) Probability previous attempt

Explanation / Answer

Calculate a z score for each (add .5 an subtract .5 to account for the continuity correction) Z=(527.5-502)/100 = .255 Z= (476.5-502)/100 = -.255 Normalcdf(-.255, .255) = .2013

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