Question

suppose that the age of students at george washington elementary school is uniformly distributed between 6 and 11 years old. 46 randomly selected children from the school are asked their age. round all answers to 4 decimal places where possible.

a. what is the distribution of ( x )? ( x sim u(6, 11) )
suppose that 46 children from the school are surveyed. then the sampling distribution is
b. what is the distribution of ( \bar{x} )? ( \bar{x} sim n(quad, quad) )
c. what is the probability that the average of 46 children will be between 8 and 8.6 years old?

Explanation:

Step1: Identify uniform distribution parameters

Given $X \sim U(a, b)$ where $a=6$, $b=11$.

Step2: Calculate mean of uniform distribution

Mean $\mu = \frac{a+b}{2} = \frac{6+11}{2} = 8.5$.

Step3: Calculate variance of uniform distribution

Variance $\sigma^2 = \frac{(b-a)^2}{12} = \frac{(11-6)^2}{12} = \frac{25}{12} \approx 2.0833$.

Step4: Apply Central Limit Theorem for $\bar{x}$

For $n=46$, $\bar{x} \sim N(\mu, \frac{\sigma^2}{n})$. Mean of $\bar{x}$ is $\mu=8.5$. Standard error $\sigma_{\bar{x}} = \sqrt{\frac{25/(12)}{46}} = \sqrt{\frac{25}{552}} \approx \sqrt{0.04529} \approx 0.2128$.

Step5: Compute z-scores for 8 and 8.6

$z_1 = \frac{8 - 8.5}{0.2128} \approx -2.3496$; $z_2 = \frac{8.6 - 8.5}{0.2128} \approx 0.4699$.

Step6: Find probabilities for z-scores

$P(Z < -2.3496) \approx 0.0093$; $P(Z < 0.4699) \approx 0.6808$. Difference: $0.6808 - 0.0093 = 0.6715$.

Answer:

a. $X \sim U(6, 11)$
b. $\bar{x} \sim N(8.5, 0.2128^2)$ (or mean=8.5, variance≈0.0453)
c. 0.6715