Mean & SD of Sampling Distribution

Question

A population consists of $4$ numbers $\{0, 2, 4, 6\}$. Consider drawing a random sample of size $n = 2$ with replacement.

(a) What is the sampling distribution of $\bar x$?

Is this a normal distribution ? Since $\bar x $~ $N\left(\mu, \dfrac{\sigma^2}{n}\right)$?

(b) Calculate the mean & standard deviation of the sampling distribution of $\bar x$.

I got the answer of mean $\mu$ by $\frac{0+2+4+6}{4} = 3$

Thereafter, I proceed to calculate $\sigma$

$$\sigma = \frac{(0 - 3)^2 + (2 - 3)^2 + (4 - 3)^2 + (6 - 3)^2}{4} = 5$$

Substituting it back into the sample distribution gives:

$$\bar x \sim N\left(3, \dfrac{5^2}{4}\right)$$

Thus, I derive the standard deviation to be:

$$\sqrt{\frac{\sigma^2}{n}} = \dfrac{5}{2}.$$

However, the answer given was $\dfrac{\sqrt{5}}{\sqrt{2}}$.

Can someone explain why is this so? I'm really quite confused with the whole concept of sampling distribution..

Thanks a lot!

Answer 1

You may have confused variance and standard deviation

With $n=2$, the possible values of $\bar{x}$ are $\{0,1,2,3,4,5,6\}$ with respective probabilities $\frac1{16},\frac2{16},\frac3{16},\frac4{16},\frac3{16},\frac2{16},\frac1{16}$

This is not Gaussian. If anything, you might describe it as a discrete triangular distribution

Your $3$ and $5$ are the expectation and variance of a sample with $n=1$

For a sample with $n=2$ the sum would have expectation $6$ and variance $10$, while the mean of the sample would have expectation $3$ and variance $\frac{5}{2}$ and so standard deviation $\frac{\sqrt 5}{\sqrt 2}$