Chapter 7.5, Problem 51E

CBS News reported that 4% of adult Americans have a food allergy (June 1, 2017, cbsnews.com/news/food-allergies-in-america-new-report-shellfish-peanut-dairy, retrieved March 25, 2018). Consider selecting 10 adult Americans at random. Define the random variable x as

$\begin{array}{l}x=\text{number}\text{of}\text{people}\text{in}\text{the}\text{sample}\text{of}10\text{that}\hfill \\ \text{have}\text{a}\text{food}\text{allergy}.\hfill \end{array}$

Find the following probabilities. (Hint: See Examples 7.19 and 7.20.)

1. a. p(x < 3)
2. b. p(x ≤ 3)
3. c. p(x ≥ 4)
4. d. p(1 ≤ x ≤ 3)

To determine

Find $p\left(x<3\right)$.

Answer to Problem 51E

The value of $p\left(x<3\right)$ is 0.9937.

Explanation of Solution

Calculation:

It is given that 4% of Country A adults have a food allergy and 10 Country A adults are randomly selected. The random variable x is defined as number of people in the sample of 10 that have a food allergy.

Here, there are fixed number of trials. That is, n = 10. There are two possible outcomes (adult with food allergy and adult without food allergy), outcomes are independent, and probability of success is fixed for each trial. Thus, the random variable x follows binomial distribution with n = 10 and p = 0.04.

The binomial probability is, $p\left(x\right)=\left(\begin{array}{l}n\\ x\end{array}\right)p^x{\left(1-p\right)}^{n-x},x=1,2,\mathrm{...},n$.

The required probability is obtained as given below:

$\begin{array}{c}p\left(x<3\right)=p\left(x=0\right)+p\left(x=1\right)+p\left(x=2\right)\\ =\left[\begin{array}{l}\left(\begin{array}{l}10\\ 0\end{array}\right){\left(0.04\right)}^0{\left(1-0.04\right)}^{10-0}+\left(\begin{array}{l}10\\ 1\end{array}\right){\left(0.04\right)}^1{\left(1-0.04\right)}^{10-1}\\ +\left(\begin{array}{l}10\\ 2\end{array}\right){\left(0.04\right)}^2{\left(1-0.04\right)}^{10-2}\end{array}\right]\\ =\left(1\right)\left(1\right){\left(0.96\right)}^{10}+\left(10\right)\left(0.04\right){\left(0.96\right)}^9+\left(45\right){\left(0.04\right)}^2{\left(0.96\right)}^8\\ =0.6648+0.2770+0.0519\\ =0.9937\end{array}$

Thus, the value of $p\left(x<3\right)$ is 0.9937.

To determine

Find $p\left(x\le 3\right)$.

Answer to Problem 51E

The value of $p\left(x\le 3\right)$ is 0.9995.

Explanation of Solution

Calculation:

The required probability is obtained as given below:

$\begin{array}{c}p\left(x\le 3\right)=p\left(x=0\right)+p\left(x=1\right)+p\left(x=2\right)+p\left(x=3\right)\\ =\left[\begin{array}{l}\left(\begin{array}{l}10\\ 0\end{array}\right){\left(0.04\right)}^0{\left(1-0.04\right)}^{10-0}+\left(\begin{array}{l}10\\ 1\end{array}\right){\left(0.04\right)}^1{\left(1-0.04\right)}^{10-1}\\ +\left(\begin{array}{l}10\\ 2\end{array}\right){\left(0.04\right)}^2{\left(1-0.04\right)}^{10-2}+\left(\begin{array}{l}10\\ 3\end{array}\right){\left(0.04\right)}^3{\left(1-0.04\right)}^{10-3}\end{array}\right]\\ =\left[\begin{array}{l}\left(1\right)\left(1\right){\left(0.96\right)}^{10}+\left(10\right)\left(0.04\right){\left(0.96\right)}^9+\\ \left(45\right){\left(0.04\right)}^2{\left(0.96\right)}^8+\left(120\right){\left(0.04\right)}^3{\left(0.96\right)}^7\end{array}\right]\\ =0.6648+0.2770+0.0519+\text{0}\text{.0058}\\ =\text{0}\text{.9995}\end{array}$

Thus, the value of $p\left(x\le 3\right)$ is 0.9995.

To determine

Find $p\left(x\ge 4\right)$.

Answer to Problem 51E

The value of $p\left(x\ge 4\right)$ is 0.0005.

Explanation of Solution

Calculation:

The required probability is obtained as given below:

$\begin{array}{c}p\left(x\ge 4\right)=1-p\left(x\le 3\right)\\ =1-0.9995\\ =0.0005\end{array}$

Thus, the value of $p\left(x\ge 4\right)$ is 0.0005.

To determine

Find $p\left(1\le x\le 3\right)$.

Answer to Problem 51E

The value of $p\left(1\le x\le 3\right)$ is 0.3347.

Explanation of Solution

Calculation:

The required probability is obtained as given below:

$\begin{array}{c}p\left(1\le x\le 3\right)=p\left(x=1\right)+p\left(x=2\right)+p\left(x=3\right)\\ =\left[\begin{array}{l}\left(\begin{array}{l}10\\ 1\end{array}\right){\left(0.04\right)}^1{\left(1-0.04\right)}^{10-1}+\left(\begin{array}{l}10\\ 2\end{array}\right){\left(0.04\right)}^2{\left(1-0.04\right)}^{10-2}\\ +\left(\begin{array}{l}10\\ 3\end{array}\right){\left(0.04\right)}^3{\left(1-0.04\right)}^{10-3}\end{array}\right]\\ =\left(10\right)\left(0.04\right){\left(0.96\right)}^9+\left(45\right){\left(0.04\right)}^2{\left(0.96\right)}^8+\left(120\right){\left(0.04\right)}^3{\left(0.96\right)}^7\\ =0.2770+0.0519+\text{0}\text{.0058}\\ =\text{0}\text{.3347}\end{array}$

Thus, the value of $p\left(1\le x\le 3\right)$ is 0.3347.