Need Statistics

label Mathematics
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Assume mu - 9.80

P(|Z| > Z_0.01) =       CP(|Z| > Z_0.01)+P(|Z| < -Z_0.01)

=          P(sample mean - 9.80)*sqrt(4)/sigma > Z_0.01 + (9.81 -9.80)*sqrt(4)/sigma

            P(sample mean - 9.80)*sqrt(4)/sigma < Z_0.01 + (9.81 -9.80)*sqrt(4)/sigma

that is the algebra in the deviation of the formula.  

P(|Z| > Z_0.01) = o.372088166 (ecxel formula) =1-NORM.S.DIST(NORM.S.INV(0.99)+(0.01*2)/0.01,TRUE)+NORM.S.DIST(-NORM.S.INV(0.99)+(0.01*2)/0.01,TRUE)

you need to use the morm.s.dist and norm.inv functions


Dec 11th, 2015

Thank you for the opportunity to help you with your question!

P { |Z| > 1}  =  P { Z < -0.01 }  + P { Z < 0.01}  =  0.4960  +  0.4960  =  0.992 = 99.2%. 

Please let me know if you need any clarification. I'm always happy to answer your questions.
Dec 11th, 2015

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