Question

three randomly selected children are surveyed. the ages of the children are 2, 7, and 12. assume that samples of size ( n = 2 ) are randomly selected with replacement from the population of 2, 7, and 12. listed below are the nine different samples. complete parts (a) through (d).
2,2 2,7 2,12 7,2 7,7 7,12 12,2 12,7 12,12
the proportion is 0.667
(round to three decimal places as needed.)
b. find the proportion of even numbers of each of the nine samples, then summarize the sampling distribution of the sample proportion of even numbers in the format of a table representing the probability distribution of the distinct proportion values.

c. find the mean of the sampling distribution of the sample proportion of even numbers.
the mean is (square)
(round to three decimal places as needed.)

Explanation:

Step1: Recall the formula for the mean of a probability distribution

The formula for the mean \(\mu\) of a probability distribution is \(\mu=\sum(x\times P(x))\), where \(x\) is the sample proportion and \(P(x)\) is the corresponding probability.

Step2: Calculate the product for each row

• For \(x = 0\) and \(P(x)=\frac{1}{9}\), the product is \(0\times\frac{1}{9}=0\).
• For \(x = 0.5\) and \(P(x)=\frac{4}{9}\), the product is \(0.5\times\frac{4}{9}=\frac{2}{9}\approx0.222\).
• For \(x = 1\) and \(P(x)=\frac{4}{9}\), the product is \(1\times\frac{4}{9}=\frac{4}{9}\approx0.444\).

Step3: Sum up the products

\(\mu=0+\frac{2}{9}+\frac{4}{9}=\frac{2 + 4}{9}=\frac{6}{9}=\frac{2}{3}\approx0.667\)

Answer:

\(0.667\)