Question

Using a Formula to Describe a Sampling Distribution Exercise 15 “Births” requires the construction of a table that describes the sampling distribution of the proportions of girls from two births. Consider the formula shown here, and evaluate that formula using sample proportions (represented by x) of 0, 0.5, and 1. Based on the results, does the formula describe the sampling distribution? Why or why not?

$P\left(x\right)=\frac{1}{2\left(2-2x\right)!\left(2x\right)!}\text{where}x=0,0.5,1$

Short answer

The probability values of occurrence of sample proportions using the given formula are listed below:

• P(0)=0.25
• P(0.5)=0.5
• P(1)=0.25

Because the probability values are the same as the values derived empirically by defining all the samples, the probability formula can be considered to represent the sampling distribution of sample proportions.

Step-by-step solution

Given information

The sample proportion of girls in two births hold the values: 0, 0.5, and 1.

The probability of occurrence of each of the proportions is to be computed using the given formula.

Refer to Exercise 15, all possible samples of size 2 given in the sample space are:

bb

bg

gb

gg

Probability of sample proportions

Let x denote the sample proportion of girls in two births.

The sample of proportion of girls in two births (x) can have the following values:

• 0
• 0.5
• 1

The total number of samples is equal to 4.

The probability formula given to compute the probability ofoccurrence of each of the proportions in all 4 samples is shown below:

$P\left(x\right)=\frac{1}{2\left(2-2x\right)!\left(2x\right)!}\text{where}x=0,0.5,1$

Substituting the values of x, the following probabilities are obtained:

$$\begin{gathered} P\left(0\right)=\frac{1}{2\left(2-2\left(0\right)\right)!\left(2\left(0\right)\right)!} \\ =\frac{1}{4} \\ =0.25 \end{gathered}$$

Thus, the probability of occurrence of sample proportion equal to 0 is equal to 0.25.

$$\begin{gathered} P\left(0.5\right)=\frac{1}{2\left(2-2\left(0.5\right)\right)!\left(2\left(0.5\right)\right)!} \\ =\frac{1}{2} \\ =0.5 \end{gathered}$$

Thus, the probability of occurrence of sample proportion equal to 0.5 is equal to 0.5.

$$\begin{gathered} P\left(1\right)=\frac{1}{2\left(2-2\left(1\right)\right)!\left(2\left(1\right)\right)!} \\ =\frac{1}{4} \\ =0.25 \end{gathered}$$

Thus, the probability of occurrence of sample proportion equal to 1 is equal to 0.25.

Since the probability values are the same as the values computed empirically by defining all the samples (obtained in Exercise 15), it can be concluded that the probability formula considered describes the sampling distribution of sample proportions.