Question

for a continuous random variable x, the population mean and the population standard deviation are 54 and 12 respectively. you extract a sample of 36 elements from this population. the mean of the sampling distribution of the sample mean is:

Answer 1

Solution

Back-up Theory

CENTRAL LIMIT THEOREM ……………………………………………………….............……. (1)

Let {X₁, X₂, …, X_n} be a sequence of n independent and identically distributed (i.i.d) random variables drawn from a distribution [i.e., {x₁, x_2, …, x_n} is a random sample of size n] of expected value given by µ and finite variance given by σ². Then, as n gets larger, the distribution of

Z = {√n(Xbar − µ)/σ}, approximates the normal distribution with mean 0 and variance 1 …… (1a)

(i.e., Standard Normal Distribution)

i.e., sample average from any distribution follows Normal Distribution with mean µ and variance σ²/n ......... (1b)

Now to work out the solution,

Here, µ = 54 and σ = 12.

By (1a), {√n(Xbar − µ)/σ}, approximates the normal distribution with mean 0 and variance 1

This would imply that mean of Xbar = µ and standard deviation of Xbar = σ/√n. Thus,

mean of the sampling distribution of the sample mean is: 54 Answer

DONE