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Given a population with a mean of µ = 100 and a variance σ2 = 13,...

Given a population with a mean of µ = 100 and a variance σ2 = 13, assume the central limit theorem applies when the sample size is n ≥ 25. A random sample of size n = 28 is obtained. What is the probability that   98.02 < x⎯⎯{"version":"1.1","math":"<math xmlns="http://www.w3.org/1998/Math/MathML"><mover><mi>x</mi><mo>&#175;</mo></mover></math>"} < 99.08?

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Answer #1
for normal distribution z score =(X-μ)/σx
here mean=       μ= 100
std deviation   =σ= 3.6056
sample size       =n= 28
std error=σx̅=σ/√n= 0.6814
probability = P(98.02<X<99.08) = P(-2.91<Z<-1.35)= 0.0885-0.0018= 0.0867

(please try 0.0868 if this comes wrong and revert)

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