Human Resource Consulting (HRC) surveyed a random sample of 60 Twin Cities const

Question

Human Resource Consulting (HRC) surveyed a random sample of 60 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 $502 with a standard deviation of $100

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

the solution says probability is 0.9476, found by 0.4738+0.4738. Where do we find 0.4738?

Human Resource Consulting (HRC) surveyed a random sample of 60 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 $502 with a standard deviation of $100

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

the solution says probability is 0.9476, found by 0.4738+0.4738. Where do we find 0.4738?

Explanation / Answer

Normal Distribution
Mean ( u ) =502
Standard Deviation ( sd )=100
Normal Distribution = Z= X- u / sd ~ N(0,1)                  
To find P(a <= Z <=b) = F(b) - F(a)
P(X < 477) = (477-502)/100/ Sqrt ( 60 )
= -25/12.9099
= -1.9365
= P ( Z <-1.9365) From Standard Normal Table
= 0.0264
P(X < 527) = (527-502)/100/ Sqrt ( 60 )
= 25/12.9099 = 1.9365
= P ( Z <1.9365) From Standard Normal Table
= 0.9736
P(477 < X < 527) = 0.9736-0.0264 = 0.9476

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