Question

a computer selects a number x from 5 to 12 randomly and uniformly. round all answers to 4 decimal places where possible.
a. what is the distribution of x? x ~ u(5, 12)
b. suppose that the computer randomly picks 37 such numbers. what is the distribution of \\(\bar{x}\\) for this selection of numbers. \\(\bar{x}\\) ~ n(8.5, 1.9728)
c. what is the probability that the average of 37 numbers will be more than 8.2? 0.5604
hint:
some helpful videos:

• finding the sampling distribution +
• finding a probability using the central limit theorem +
• finding value given a probability using the central limit theorem +
• the central limit theorem for sums +

hint

Explanation:

Step1: Recall mean and standard - deviation of uniform distribution

For a uniform distribution $X\sim U(a,b)$ where $a = 5$ and $b = 12$, the mean $\mu=\frac{a + b}{2}=\frac{5+12}{2}=8.5$ and the standard - deviation $\sigma=\sqrt{\frac{(b - a)^2}{12}}=\sqrt{\frac{(12 - 5)^2}{12}}=\sqrt{\frac{49}{12}}\approx2.0207$.

Step2: Apply the Central Limit Theorem for the sample mean

When $n = 37$ samples are taken, the sampling distribution of the sample mean $\bar{X}$ is approximately normal with mean $\mu_{\bar{X}}=\mu = 8.5$ and standard - deviation $\sigma_{\bar{X}}=\frac{\sigma}{\sqrt{n}}=\frac{\sqrt{\frac{(12 - 5)^2}{12}}}{\sqrt{37}}=\frac{\sqrt{\frac{49}{12}}}{\sqrt{37}}\approx\frac{2.0207}{\sqrt{37}}\approx0.3317$.

Step3: Standardize the value

We want to find $P(\bar{X}>8.2)$. First, we standardize $8.2$ using the formula $z=\frac{\bar{x}-\mu_{\bar{X}}}{\sigma_{\bar{X}}}=\frac{8.2 - 8.5}{0.3317}=\frac{- 0.3}{0.3317}\approx - 0.9044$.

Step4: Find the probability

$P(\bar{X}>8.2)=P(Z>-0.9044)$. Since $P(Z > z)=1 - P(Z\leq z)$, and from the standard normal table $P(Z\leq - 0.9044)\approx0.1834$, then $P(Z>-0.9044)=1 - 0.1834 = 0.8166$.

Answer:

$0.8166$