### Solution Found!

# Uniform Data: Probabilities of Sample Means

**Chapter 9, Problem 20**

(choose chapter or problem)

**Get Unlimited Answers! Check out our subscriptions**

**QUESTION:**

Given the discrete uniform population

\(f(x)=\left\{\begin{array}{ll} \frac{1}{3}, & x=2,4,6, \\ 0, & \text { elsewhere } \end{array}\right.\)

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.

#### Watch The Answer!

##### Uniform Data: Probabilities of Sample Means

Want To Learn More? To watch the entire video and ALL of the videos in the series:

This video offers a comprehensive look into the discrete uniform population, analyzing a sample size of 54 selected with replacement. Through calculated steps, viewers learn how to determine probabilities for a sample mean within specified limits. Utilizing Z-scores and standard normal tables, the session illuminates statistical principles in real-world scenarios.

###
Not The Solution You Need? Search for *Your* Answer Here:

### Questions & Answers

**QUESTION:**

Given the discrete uniform population

\(f(x)=\left\{\begin{array}{ll} \frac{1}{3}, & x=2,4,6, \\ 0, & \text { elsewhere } \end{array}\right.\)

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:**

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}\)