In the EAI sampling problem, the population mean is $51,900 and the population s
ID: 3359763 • Letter: I
Question
In the EAI sampling problem, the population mean is $51,900 and the population standard deviation is $5,000. When the sample size is n = 30, there is a 0.4161 probability of obtaining a sample mean within +/- $500 of the population mean. Use z-table. What is the probability that the sample mean is within $500 of the population mean if a sample of size 60 is used (to 4 decimals)? What is the probability that the sample mean is within $500 of the population mean if a sample of size 120 is used (to 4 decimals)?
Explanation / Answer
When n=60 is used
To find P(a <= Z <=b) = F(b) - F(a)
P(X < 51400) = (51400-51900)/5000/ Sqrt ( 60 )
= -500/645.49722
= -0.7746
= P ( Z <-0.7746) From Standard Normal Table
= 0.21929
P(X < 52400) = (52400-51900)/5000/ Sqrt ( 60 )
= 500/645.49722 = 0.7746
= P ( Z <0.7746) From Standard Normal Table
= 0.78071
P(51400 < X < 52400) = 0.78071-0.21929 = 0.5614
When n=120 is used
To find P(a <= Z <=b) = F(b) - F(a)
P(X < 51400) = (51400-51900)/5000/ Sqrt ( 120 )
= -500/456.43546
= -1.09545
= P ( Z <-1.09545) From Standard Normal Table
= 0.13666
P(X < 52400) = (52400-51900)/5000/ Sqrt ( 120 )
= 500/456.43546 = 1.09545
= P ( Z <1.09545) From Standard Normal Table
= 0.86334
P(51400 < X < 52400) = 0.86334-0.13666 = 0.7267