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