Question

Carbon Tax. A poll commissioned by Friends of the Earth and conducted by the Mellman Group found that 72% of American voters are in favor of a carbon tax. Suppose that six voters in the United States are randomly sampled and asked whether they favor a carbon tax. Determine the probability that the number answering in the affirmative is

(a) exactly two. (b) exactly four. (c) at least two.

(d) Determine the probability distribution of the number of American voters in a sample of six who favor a carbon tax.

(e) Strictly speaking, why is the probability distribution that you obtained in part (d) only approximately correct? What is the exact distribution called?

Short answer

Part (a) 0.0478

Part (b) 0.3160

Part (c) 0.9921

Part (d)

X	0	1	2	3	4	5	6
P(X)	0.0005	0.0074	0.0478	0.1639	0.3160	0.3251	0.1393

Part (e) Hypergeometric distribution.

Step-by-step solution

Part (a) Step 1. Given information.

The given statement is:

A carbon tax would be supported by 72 percent of Americans.

Six Americans are randomly selected and asked whether they support a carbon price.

$p=0.70,n=6$

Part (a) Step 2. Find the probability that the chosen number answering in the affirmative is exactly two.

$$\begin{gathered} P\left(X=x\right)=\left(\begin{array}{c}n\\ x\end{array}\right)p^x{\left(1-p\right)}^{n-x} \\ P\left(X=2\right)=\left(\begin{array}{c}6\\ 2\end{array}\right){\left(0.72\right)}^2{\left(1-0.72\right)}^{6-2} \\ =\left(15\right){\left(0.72\right)}^2{\left(0.28\right)}^4 \\ =15\left(0.5184\right)\left(0.00614656\right) \\ =0.0478 \end{gathered}$$

Part (b) Step 1. Find the probability that the chosen number answering in the affirmative is exactly four.

$$\begin{gathered} P\left(X=x\right)=\left(\begin{array}{c}n\\ x\end{array}\right)p^x{\left(1-p\right)}^{n-x} \\ P\left(X=4\right)=\left(\begin{array}{c}6\\ 4\end{array}\right){\left(0.72\right)}^4{\left(1-0.72\right)}^{6-4} \\ =\left(15\right){\left(0.72\right)}^4{\left(0.28\right)}^2 \\ =15\left(0.26873856\right)\left(0.0784\right) \\ =0.3160 \end{gathered}$$

Part (c) Step 1. Find the probability that the chosen number answering in the affirmative is at least two.

$$\begin{gathered} P\left(X\ge 2\right)=1-P\left(X<2\right) \\ =1-\left[P\left(X=0\right)+P\left(X=1\right)\right] \\ =1-\left[\left(\begin{array}{c}6\\ 0\end{array}\right){\left(0.72\right)}^0{\left(1-0.72\right)}^{6-0}+\left(\begin{array}{c}6\\ 1\end{array}\right){\left(0.72\right)}^1{\left(1-0.72\right)}^{6-1}\right] \\ =1-\left[\left(1\right){\left(0.72\right)}^0{\left(0.28\right)}^6+\left(6\right){\left(0.72\right)}^0{\left(0.28\right)}^5\right] \\ =1-\left[0.00791676928\right] \\ =0.9921 \end{gathered}$$

Part (d) Step 1. Determine the probability distribution of six American voters who support a carbon tax.

The probability distribution is given below:

X	Probability
0	$$\begin{gathered} \left(\begin{array}{c}6\\ 0\end{array}\right){\left(0.72\right)}^0{\left(0.28\right)}^6=\left(1\right)\left(1\right){\left(0.28\right)}^6 \\ =0.0005 \end{gathered}$$
1	$$\begin{gathered} \left(\begin{array}{c}6\\ 1\end{array}\right){\left(0.72\right)}^1{\left(0.28\right)}^5=\left(6\right)\left(0.72\right){\left(0.28\right)}^5 \\ =0.0074 \end{gathered}$$
2	$$\begin{gathered} \left(\begin{array}{c}6\\ 2\end{array}\right){\left(0.72\right)}^2{\left(0.28\right)}^4=\left(15\right){\left(0.72\right)}^2{\left(0.28\right)}^4 \\ =0.0478 \end{gathered}$$
3	$$\begin{gathered} \left(\begin{array}{c}6\\ 3\end{array}\right){\left(0.72\right)}^3{\left(0.28\right)}^3=\left(20\right){\left(0.72\right)}^3{\left(0.28\right)}^3 \\ =0.1639 \end{gathered}$$
4	$$\begin{gathered} \left(\begin{array}{c}6\\ 4\end{array}\right){\left(0.72\right)}^4{\left(0.28\right)}^2=\left(15\right){\left(0.72\right)}^4{\left(0.28\right)}^2 \\ =0.3160 \end{gathered}$$
5	$$\begin{gathered} \left(\begin{array}{c}6\\ 5\end{array}\right){\left(0.72\right)}^5{\left(0.28\right)}^1=\left(6\right){\left(0.72\right)}^5\left(0.28\right) \\ =0.3251 \end{gathered}$$
6	$$\begin{gathered} \left(\begin{array}{c}6\\ 6\end{array}\right){\left(0.72\right)}^6{\left(0.28\right)}^0=\left(1\right){\left(0.72\right)}^6\left(1\right) \\ =0.1393 \end{gathered}$$

Part (e) Step 1. Explanation.

Because the sampling from a finite population is done without replacement and the number of successes follows a hypergeometric distribution.

Therefore, the that we have obtained in part (d) is only approximately correct.