a)
for normal distribution z score =(X-μ)/σ | |
here mean= μ= | 185 |
std deviation =σ= | 110.0000 |
probability = | P(X<190) | = | P(Z<0.05)= | 0.5199 |
b)
sample size =n= | 130 |
std error=σx̅=σ/√n= | 9.6476 |
probability = | P(Xbar<190) | = | P(Z<0.52)= | 0.6985 |
c)
as sample size is greater then 30 therefore sample mean can be apprximated as normally distributed
hence
probability = | P(Ybar<190) | = | P(Z<0.52)= | 0.6985 |
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