Let X be the mean of a random sample of size n 12 from the

Let X be the mean of a random sample of size n = 12 from the uniform distribution on the interval (0, 1).

Approximate P(1/2 X 2/3).

Solution

as its given that n =12 not infinity we can get the sidtribution of x , the sampling distribution of X N(µ, ²/n) as n 12.

by using this we can the sampling distribution

N(µ, ²/n)
= N(1/2, (1/12)/12)
= N(1/2, 1/144)
now what we need to find is that x lines in range of 1/2 and 2/3 and we can do this by by dividing the variance and subtract the mean

Let Z = (X-bar - µ) / (²/n)
= (X-bar - 1/2) / (1/144)
= 144(X-bar - 1/2)

let it be z variable , which is standard normal and variance = 1 and mean =0

now we have to just it in probability expression and look from the table

P(1/2 < X-bar < 2/3)
= P{144(1/2 - 1/2) < Z < 144(2/3 - 1/2)}
= P{0 < Z < 24}
= 0.50000 by normal table