Given a normal distribution with mean = 100 and standard deviation = 10, if you select a sample of n = 25, what is the probability that x-bar is above 102.2?

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Answer 1 · 2017-05-07T15:52:38

#0.1357#

#X~N(100,10^2)#

for samples of #n=25#

the sampling distribution of the mean is

#barX~N(100,10^2/25)#

we want

#P(barX>102.2)#

#=1-P(barX<102.2)#

#1-P(Z<(102.2-100)/(sqrt(100/25)))#

#=1-P(Z<1.1)#

From tables we get

#=1-0.8643#

#=0.1357#