In: Statistics and Probability
Let X have Normal distribution with mean 45 and variance 81. If a random sample of size 25 is taken, which of the following is the probability that the sample average is between 41.40 and 45.63?
Solution :
Given that ,
mean =
= 45
varianc= 81 SO
standard deviation =
= 9
n = 25
= 45
=
/
n= 9/
25=1.8
P(41.40< <45.63
) = P[(41.40 - 45) / 1.8< (
-
) /
< (45.63-45) /1.8 )]
= P(-2 < Z <0.35 )
= P(Z < 0.35) - P(Z <-2 )
Using z table
=0.6368-0.0228
=0.6140
probability= 0.6140