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Pets. According to JAVMA News, a publication of the American Veterinary Medical Association, roughly 60% of U.S. households own one or more pets. Four U.S. households are selected at random. Use Table VII in Appendix A to solve the following problems.

(a) Find the probability that, of the four households sampled, the number that own one or more pets is exactly three; at least three; at most three.

(b) Find the probability distribution of the random variable X. the number of U.S. households in a random sample of four that own one or more pets.

(c) Without referring to the probability distribution obtained in part (b) or constructing a probability histogram, decide whether the probability distribution is right-skewed, symmetric, or left-skewed. Explain your answer. *

Step by step solution

01

Part (a) Step 1. Given information.

The given statement is:

According to JAVMA News, a publication of the American Veterinary Medical Association, roughly 60% of U.S. households own one or more pets. Four U.S. households are selected at random.

02

Part (a) Step 2. Find the probability that three households are likely to have one or more pets.

$$\begin{gathered} n=4,p=0.60,\mathrm{and}x=3 \\ P\left(X=x\right)=\left(\begin{array}{c}n\\ x\end{array}\right)p^x{\left(1-p\right)}^{n-x} \\ P\left(X=3\right)=\left(\begin{array}{c}4\\ 3\end{array}\right){\left(0.6\right)}^3{\left(1-0.6\right)}^{4-3} \\ =\left(4\right)\left(0.216\right)\left(0.4\right) \\ =0.3456 \end{gathered}$$

03

Part (a) Step 3. Find the probability that at least three households are likely to have one or more pets.

$$\begin{gathered} x鈮�3 \\ P\left(X鈮�3\right)=\left(\begin{array}{c}4\\ 3\end{array}\right){\left(0.6\right)}^3{\left(1-0.6\right)}^{4-1}+\left(\begin{array}{c}4\\ 4\end{array}\right){\left(0.6\right)}^4{\left(1-0.6\right)}^{4-4} \\ =0.3456+\left(1\right)\left(0.1296\right)\left(1\right) \\ =0.3456+0.1296 \\ =0.4752 \end{gathered}$$

04

Part (a) Step 4. Find the probability that at most three households are likely to have one or more pets.

$$\begin{gathered} P\left(X鈮�3\right)=1-P\left(X>3\right) \\ =1-\left(\begin{array}{c}4\\ 4\end{array}\right){\left(0.6\right)}^4{\left(1-0.6\right)}^{4-4} \\ =1-0.1296 \\ =0.8704 \end{gathered}$$

05

Part (b) Step 1. Find the probability distribution of X.

X	$P(X=x)$
0	$$\begin{gathered} P\left(X=0\right)=\left(\begin{array}{c}4\\ 0\end{array}\right){\left(0.6\right)}^0{\left(0.4\right)}^4 \\ =\left(1\right)\left(1\right)\left(0.0256\right) \\ =0.0256 \end{gathered}$$
1	$$\begin{gathered} P\left(X=1\right)=\left(\begin{array}{c}4\\ 1\end{array}\right){\left(0.6\right)}^1{\left(0.4\right)}^3 \\ =\left(4\right)\left(0.6\right)\left(0.064\right) \\ =0.1536 \end{gathered}$$
2	$$\begin{gathered} P\left(X=2\right)=\left(\begin{array}{c}4\\ 2\end{array}\right){\left(0.6\right)}^2{\left(0.4\right)}^2 \\ =\left(6\right)\left(0.36\right)\left(0.16\right) \\ =0.3456 \end{gathered}$$
3	$$\begin{gathered} P\left(X=3\right)=\left(\begin{array}{c}4\\ 3\end{array}\right){\left(0.6\right)}^3{\left(0.4\right)}^1 \\ =\left(4\right)\left(0.256\right)\left(0.4\right) \\ =0.3456 \end{gathered}$$
4	$$\begin{gathered} P\left(X=4\right)=\left(\begin{array}{c}4\\ 4\end{array}\right){\left(0.6\right)}^4{\left(0.4\right)}^0 \\ =\left(1\right)\left(0.1296\right)\left(1\right) \\ =0.1296 \end{gathered}$$

06

Part (c) Step 1. Identify whether the probability distribution is right-skewed, symmetric, or left-skewed.

$p=0.6$in this scenario which is greater than 0.5. As a result, the distribution is left-skewed.

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