Question

suppose that the age of students at george washington elementary school is uniformly distributed between 5 and 11 years old. 50 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 - u(5,11) )
suppose that 50 children from the school are surveyed. then the sampling distribution is
b. what is the distribution of ( overline{x} )? ( overline{x}-n(quad,quad) )
c. what is the probability that the average of 50 children will be between 8 and 8.2 years old?

Explanation:

Step1: Find the mean and standard deviation of \(X\)

For a uniform distribution \(X - U(a,b)\) (here \(a = 5\), \(b=11\)), the mean \(\mu=\frac{a + b}{2}=\frac{5+11}{2}=8\) and the standard deviation \(\sigma=\sqrt{\frac{(b - a)^2}{12}}=\sqrt{\frac{(11 - 5)^2}{12}}=\sqrt{3}\approx1.7321\)

Step2: Find the mean and standard deviation of \(\bar{X}\)

For a sample of size \(n = 50\), the mean of the sampling distribution \(\mu_{\bar{X}}=\mu = 8\) and the standard deviation of the sampling distribution \(\sigma_{\bar{X}}=\frac{\sigma}{\sqrt{n}}=\frac{\sqrt{3}}{\sqrt{50}}\approx\frac{1.7321}{7.0711}\approx0.245\)

Step3: Standardize the values

For \(x_1 = 8\), \(z_1=\frac{x_1-\mu_{\bar{X}}}{\sigma_{\bar{X}}}=\frac{8 - 8}{0.245}=0\)
For \(x_2 = 8.2\), \(z_2=\frac{x_2-\mu_{\bar{X}}}{\sigma_{\bar{X}}}=\frac{8.2-8}{0.245}=\frac{0.2}{0.245}\approx0.8163\)

Step4: Find the probability

We want to find \(P(8<\bar{X}<8.2)\), which is equivalent to \(P(0<Z<0.8163)\)
Using the standard normal table \(P(0<Z<0.8163)=\Phi(0.8163)-\Phi(0)\)
where \(\Phi(z)\) is the cumulative distribution function of the standard normal distribution. \(\Phi(0) = 0.5\) and \(\Phi(0.8163)\approx0.7939\)
So \(P(0<Z<0.8163)=0.7939 - 0.5=0.2939\)

Answer:

a. \(X - U(5,11)\)
b. \(\bar{X}-N(8,0.245)\)
c. \(0.2939\)