Uniform Data: Probabilities of Sample Means

Step 1 of 3

Let \(X\) be the discrete uniform random variable with the probability distribution function \(f\).

First, we need to determine the mean:

\(\mu_{X}=2 \cdot f(2)+4 \cdot f(4)+6 \cdot f(6)=2 \cdot \frac{1}{3}+4 \cdot \frac{1}{3}+6 \cdot \frac{1}{3}=\frac{2+4+6}{3}=4\)

and the variance of \(X\):

\(\begin{aligned}
\sigma_{X}^{2} & =\frac{1}{3} \cdot\left(\left(2-\mu_{X}\right)^{2}+\left(4-\mu_{X}\right)^{2}+\left(6-\mu_{X}\right)^{2}\right) \\
& =\frac{1}{3}\left((2-4)^{2}+(4-4)^{2}+(6-4)^{2}\right)=\frac{2^{2}+0^{2}+2^{2}}{3}=\frac{8}{3}
\end{aligned}\)

Now, by using the obtained values and the length \(n=54\) of the sample, we're able to compute the mean and the standard deviation of the sample mean:

\(\begin{array}{c}
\mu_{\hat{X}}=\mu_{X}=4 \\
\sigma_{\hat{X}}=\sqrt{\frac{\sigma_{X}^{2}}{n}}=\sqrt{\frac{\frac{8}{3}}{54}}=\frac{\mathbf{2}}{\mathbf{9}}
\end{array}\)