Question

the number of seconds x after the minute that class ends is uniformly distributed between 0 and 60. round all answers to 4 decimal places where possible.
a. what is the distribution of x? x - u(0, 60)
then the sampling distribution is
b. suppose that 40 classes are clocked. what is the distribution of \\(\bar{x}\\) for this group of classes? \\(\bar{x}\\) - n(15, 17.3205 x)
c. what is the probability that the average of 40 classes will end with the second hand between 29 and 33 seconds? 0.250 x
hint:

Explanation:

Step1: Recall uniform - distribution properties

For a uniform distribution $X\sim U(a,b)$, the mean $\mu=\frac{a + b}{2}$ and the standard deviation $\sigma=\frac{b - a}{\sqrt{12}}$. Here, $a = 0$ and $b = 60$, so $\mu=\frac{0+60}{2}=30$ and $\sigma=\frac{60 - 0}{\sqrt{12}}=\frac{60}{2\sqrt{3}} = 10\sqrt{3}\approx17.3205$.

Step2: Central - Limit Theorem for sample mean

If $X\sim U(a,b)$ and we take a sample of size $n$, the sampling distribution of the sample mean $\bar{X}$ is approximately normal for $n$ large (by the Central - Limit Theorem). For $n = 40$, $\bar{X}\sim N(\mu_{\bar{X}},\sigma_{\bar{X}})$, where $\mu_{\bar{X}}=\mu = 30$ and $\sigma_{\bar{X}}=\frac{\sigma}{\sqrt{n}}=\frac{10\sqrt{3}}{\sqrt{40}}\approx2.7386$.

Step3: Standardize the bounds

We want to find $P(29<\bar{X}<33)$. First, standardize the values: $z_1=\frac{29 - 30}{2.7386}\approx - 0.3651$ and $z_2=\frac{33 - 30}{2.7386}\approx1.0953$.

Step4: Use the standard normal table

$P(29<\bar{X}<33)=P(-0.3651

Answer:

a. $X\sim U(0,60)$
b. $\bar{X}\sim N(30,2.7386)$
c. $0.5065$