Working on summer vacation. Recall (Exercise 3.13, p. 169) that a Harris Interactive (July 2013) poll found that 22% of U.S. adults do not work at all while on summer vacation. In a random sample of 10 U.S. adults, let x represent the number who do not work during summer vacation.

a. For this experiment, define the event that represents a “success.”

b. Explain why x is (approximately) a binomial random variable.

c. Give the value of p for this binomial experiment.

d. Find P(x=3)

e. Find the probability that 2 or fewer of the 10 U.S. adults do not work during summer vacation.

The poll found that 61% of the respondents work during the summer vacation, 22% does not work at all while on vacation, and 17% were unemployed.

In a random sample of 10 U.S. adults, let x represent the number who do not work during summer vacation.

Here let X is the random variable

According to the question, the event of success is defined as the adults who do not work in summer vacation.

b.

There are only two possible outcomes

1. Adults who do not work in summer vacation (success).
2. Adults who do work in summer vacation (failure).

The binomial random variable x is the number of successes in n trials

Here,

p= probability of success

p= probability that adults do not work in summer vacation

Therefore

p = 22%

=0.22

Thus, x=0.22.

When

$$\begin{gathered} n=10 \\ x=3 \\ P\left(x=3\right)=\left(\begin{array}{c}n\\ x\end{array}\right)p^x{\left(1-p\right)}^{n-x} \\ =\left(\begin{array}{c}10\\ 3\end{array}\right){\left(0.22\right)}^3{\left(1-0.22\right)}^7 \\ =120x0.010648x0.1756556885 \\ =0.22444 \\ \mathbf{Thus}\mathbf{,}\mathbf{}\mathbf{}\mathrm{P}\left(\mathrm{x}=3\right)=0.22444 \end{gathered}$$

e.

When $$\begin{gathered} WhenP\left(x\le 2\right) \\ P\left(x\le 2\right)=P\left(x=1\right)+P\left(x=2\right) \\ =\left(\begin{array}{c}10\\ 1\end{array}\right)p^1{\left(1-p\right)}^{10-1}x\left(\begin{array}{c}10\\ 2\end{array}\right)p^2{\left(1-p\right)}^{10-2} \\ =10x0.22x{\left(0.78\right)}^{9}x45x{\left(0.22\right)}^{2}x{\left(0.78\right)}^8 \\ =0.6169 \end{gathered}$$,

Thus, the required probability is 0.6169.