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( , )
b. suppose that the computer randomly picks 35 such numbers. what is the distribution of \\(\bar{x}\\) for this selection of numbers. \\(\bar{x}\\) ~ n( , )
c. what is the probability that the average of 35 numbers will be less than 9?

Explanation:

Step1: Identify distribution of $X$

For a uniform distribution on the interval $[a,b]$, if $X$ is uniformly distributed on $[5,12]$, the distribution of $X$ is $X\sim U(5,12)$.

Step2: Find mean and standard - deviation of $X$

The mean of a uniform distribution $U(a,b)$ is $\mu=\frac{a + b}{2}=\frac{5+12}{2}=\frac{17}{2}=8.5$. The standard deviation is $\sigma=\sqrt{\frac{(b - a)^2}{12}}=\sqrt{\frac{(12 - 5)^2}{12}}=\sqrt{\frac{49}{12}}\approx2.0207$.

Step3: Identify distribution of $\bar{X}$

By the Central Limit Theorem, if we have a sample of size $n = 35$ from a population with mean $\mu$ and standard - deviation $\sigma$, the sampling distribution of the sample mean $\bar{X}$ is approximately normal with mean $\mu_{\bar{X}}=\mu$ and standard - deviation $\sigma_{\bar{X}}=\frac{\sigma}{\sqrt{n}}$. So $\bar{X}\sim N(8.5,\frac{2.0207}{\sqrt{35}}\approx0.3422)$.

Step4: Calculate the $z$ - score

The $z$ - score is calculated as $z=\frac{\bar{x}-\mu_{\bar{X}}}{\sigma_{\bar{X}}}$, where $\bar{x} = 9$, $\mu_{\bar{X}}=8.5$, and $\sigma_{\bar{X}}\approx0.3422$. So $z=\frac{9 - 8.5}{0.3422}=\frac{0.5}{0.3422}\approx1.4612$.

Step5: Find the probability

We want to find $P(\bar{X}<9)$, which is equivalent to $P(Z<1.4612)$ using the standard normal distribution. Looking up the value in the standard - normal table, $P(Z < 1.4612)\approx0.9279$.

Answer:

a. 5, 12
b. 8.5, 0.3422
c. 0.9279