Your Teacher WageWeb i earn a mean annual salary of $87,100. Assume that annual

Question

Your Teacher WageWeb i earn a mean annual salary of $87,100. Assume that annual salaries s a service of HRPDt and provides compensation information ion on mare than 170 benchmark positions in human resources. The October 2003 posting indicated that labor relation managers more salares are normally distributed and have a standard deviation of $9125.(Give your answers correct to four decimal places) ) what is the probabnty that randomiy selected iabor relation manager earned more than $107,200 in 2003? (b) A sample of 21 labor relation managers is taken, and annual salaries are reported. What is the probability that the sample mean annual salary fails between $85,200 and $88, 500 is the

Explanation / Answer

the PDF of normal distribution is = 1/ * 2 * e ^ -(x-u)^2/ 2^2
standard normal distribution is a normal distribution with a,
mean of 0,
standard deviation of 1
equation of the normal curve is ( Z )= x - u / sd ~ N(0,1)
mean ( u ) = 87100
standard Deviation ( sd )= 9125
a.
GREATER THAN
P(X > 107200) = (107200-87100)/9125
= 20100/9125 = 2.2027
= P ( Z >2.2027) From Standard Normal Table
= 0.0138

b.
mean of the sampling distribution ( x ) = 87100
standard Deviation ( sd )= 9125/ Sqrt ( 21 ) =1991.2382
sample size (n) = 21
BETWEEN THEM
To find P(a <= Z <=b) = F(b) - F(a)
P(X < 85200) = (85200-87100)/9125/ Sqrt ( 21 )
= -1900/1991.23825
= -0.95418
= P ( Z <-0.95418) From Standard Normal Table
= 0.17
P(X < 88500) = (88500-87100)/9125/ Sqrt ( 21 )
= 1400/1991.23825 = 0.70308
= P ( Z <0.70308) From Standard Normal Table
= 0.759
P(85200 < X < 88500) = 0.759-0.17 = 0.589

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