Question

Given the discrete uniform population 32 f(x) = x = 2, 4, 6, 10, elsewhere, find the probability that a random sample of size 54, selected with replacement, will yield a sample mean greater than 4.1 but less than 4.4. Assume the means are measured to the nearest tenth.

Answer 1

Solution: - 6x) + 4x - 3 6x 3 l= Exp(x) = 2x1 2X1 / 3 4 2. 2 - Elx² Explx) = 2 x 12 56 - 4² ** +4 56 3 X 2/3 +6x) r= 1.633 V 56 - 4 Now, Х Approximatly normal lit XI Eli - 4 0.2 2 2 N 8+ 319 4.4-9) So 0.222 1.633 V54 4.1-4 cza P( 4.1<$< 4.4) = P1 7=7 Mb P/0.45 0.45 CZC 1.80) 0 - 2 2 2 7203 0.2905