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 between 95 and 97.5?

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Answer 1 · 2017-09-19T11:05:24

#=0.0994#

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

then the sampling distribution of the mean with sample size #n#

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

in this case

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

#=>barX~N(100,4)#

we want

#P( 95 < barX<97.5)#

standardising

#P(( 95-100)/2 < Z <(97.5-100)/2)#

where #Z~N(0,1)#

#P(-2.5 < Z <-1.25)#

by symmetry of the Normal distribution

#=P( 2.5< Z<1.25)#

#=P(z<2.5)-P(Z<1.25)#

from tables

#=0.9938-0.8944

#=0.0994#