Question

Assume that the probability mass function of X is given by P(X = 1) = P(X = 2) = P(X = 3) = 1/3

A random sample of n = 36 is selected from this population. Find the probability that the sample mean is

greater than 2.1 but less than 2.5, assuming that the sample mean would be measured to the nearest tenth.

Answer 1

Answer:

Given,

Let us consider,

X be Uniform distribution with U(1 , 3)

mean = (1 + 3)/2

= 4/2

= 2

Standard deviation = sqrt((b-a+1)^2-1) / 12)

= sqrt(((3-1+1)^2 - 1)/12)

= sqrt(8/12)

= 0.8165

Now consider,

P(2.1 <= xbar <= 2.5) = P((2.1 - 2)/(0.8165/sqrt(36)) <= (xbar-mu)/(s/sqrt(n)) <= (2.5 - 2)/(0.8162/sqrt(36)))

= P(0.74 <= z <= 3.68)

= P(z <= 3.68) - P(z <= 0.74)

= 0.9999 - 0.7704

= 0.2295

Hence the required probability = 0.2295