Question 74581
Given a normal distribution with µ = 100 and sigma = 10, if you select a sample of n = 25, what is the probability that X bar is
a.	less than 95?
z(95) = (95-100)/10/sqrt25) = -5/2 = -2.5
P(xbar<95)=P(z<-2.5)=0.006
-------------
b.	Between 95 amd 97.5
z(95)=-2.5 ; z(97.5) = (97.5-100)/10/5=-2.5/2=-1.25
P(95<xbar<97.5)=P(-2.5<z<-1.25)=0.0605975...
--------------
c.	Above 102.2?
z(102.2)=(102.2-100)/2=1.1
P(xbar>102.2)=P(z>1.1)=0.135666...
---------------
d.	There is a 65% chance that X bar is above what value?
The corresponding z-value is z=-0.38532...
Converting to raw score: -0.38532 = (xbar-100)/2;
xbar=2*(-0.38532)+100
xbar=99.23
-------------
======================
Cheers,
Stan H.