In: Math
Xn are independent random variables and each one of them has a normal distribution with mean 0 and variance 1
what is the distribution of X-bar if n=9
what is the probability of X-bar =<3 if n=9
Solution :
Given that ,
mean =
= 0
standard deviation =
= 1
n = 9
= 0 and = 1
2
= 1
=
/
n = 1 /
9 = 1 / 3
(a)
~ N(
,
)
~ N(0 , 1/3 )
(b)
P( 3) =
P((
-
) /
<
(3 - 0) / 1/3)
= P(z < 9)
Using standard normal table,
Probability = 0