Chapter 7.5, Problem 43E

a.

To determine

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

Answer to Problem 43E

$\underset{\_}{p\left(x<3\right)=0.9884}$.

Explanation of Solution

Calculation:

It is given that 1 out of 20 children in Country U has food allergy and 10 children are randomly selected. The random variable x is defined as the number of children in the sample of 10 who has food allergy.

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

The probability of success p is obtained as follows:

$\begin{array}{c}p=\frac{1}{20}\\ =0.05\end{array}$

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:

$\begin{array}{c}p\left(x<3\right)=p\left(x=0\right)+p\left(x=1\right)+p\left(x=2\right)\\ =\left[\begin{array}{l}\left(\begin{array}{l}10\\ 0\end{array}\right){\left(0.05\right)}^0{\left(1-0.05\right)}^{10-0}+\left(\begin{array}{l}10\\ 1\end{array}\right){\left(0.05\right)}^1{\left(1-0.05\right)}^{10-1}\\ +\left(\begin{array}{l}10\\ 2\end{array}\right){\left(0.05\right)}^2{\left(1-0.05\right)}^{10-2}\end{array}\right]\\ =\left(1\right)\left(1\right){\left(0.95\right)}^{10}+\left(10\right)\left(0.05\right){\left(0.95\right)}^9+\left(45\right){\left(0.05\right)}^2{\left(0.95\right)}^8\\ =0.5987+0.3151+0.0746\\ =0.9884\end{array}$

Thus, $\underset{\_}{p\left(x<3\right)=0.9884}$.

b.

To determine

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

Answer to Problem 43E

The value of $p\left(x\le 3\right)$ is 0.9989

Explanation of Solution

Calculation:

The required probability is obtained as given below:

$\begin{array}{c}p\left(x\le 3\right)=p\left(x=0\right)+p\left(x=1\right)+p\left(x=2\right)+p\left(x=3\right)\\ =\left[\begin{array}{l}\left(\begin{array}{l}10\\ 0\end{array}\right){\left(0.05\right)}^0{\left(1-0.05\right)}^{10-0}+\left(\begin{array}{l}10\\ 1\end{array}\right){\left(0.05\right)}^1{\left(1-0.05\right)}^{10-1}\\ +\left(\begin{array}{l}10\\ 2\end{array}\right){\left(0.05\right)}^2{\left(1-0.05\right)}^{10-2}+\left(\begin{array}{l}10\\ 3\end{array}\right){\left(0.05\right)}^3{\left(1-0.05\right)}^{10-3}\end{array}\right]\\ =\left[\begin{array}{l}\left(1\right)\left(1\right){\left(0.95\right)}^{10}+\left(10\right)\left(0.05\right){\left(0.95\right)}^9+\\ \left(45\right){\left(0.05\right)}^2{\left(0.95\right)}^8+\left(120\right){\left(0.05\right)}^3{\left(0.95\right)}^7\end{array}\right]\\ =0.5987+0.3151+0.0746+0.0105\\ =0.9989\end{array}$

Thus, the value of $p\left(x\le 3\right)$ is 0.9989.

c.

To determine

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

Answer to Problem 43E

$\underset{\_}{p\left(x\ge 4\right)=0.0011}$.

Explanation of Solution

Calculation:

The required probability is obtained as given below:

$\begin{array}{c}p\left(x\ge 4\right)=1-p\left(x\le 3\right)\\ =1-0.9989\\ =0.0011\end{array}$

Thus, $\underset{\_}{p\left(x\ge 4\right)=0.0011}$.

d.

To determine

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

Answer to Problem 43E

$\underset{\_}{p\left(1\le x\le 3\right)=0.4002}$

Explanation of Solution

Calculation:

The required probability is obtained as given below:

$\begin{array}{c}p\left(1\le x\le 3\right)=p\left(x=1\right)+p\left(x=2\right)+p\left(x=3\right)\\ =\left[\begin{array}{l}\left(\begin{array}{l}10\\ 1\end{array}\right){\left(0.05\right)}^1{\left(1-0.05\right)}^{10-1}+\left(\begin{array}{l}10\\ 2\end{array}\right){\left(0.05\right)}^2{\left(1-0.05\right)}^{10-2}\\ +\left(\begin{array}{l}10\\ 3\end{array}\right){\left(0.05\right)}^3{\left(1-0.05\right)}^{10-3}\end{array}\right]\\ =\left[\left(10\right)\left(0.05\right){\left(0.95\right)}^9+\left(45\right){\left(0.05\right)}^2{\left(0.95\right)}^8+\left(120\right){\left(0.05\right)}^3{\left(0.95\right)}^7\right]\\ =0.3151+0.0746+0.0105\\ =0.4002\end{array}$

Thus, $\underset{\_}{p\left(1\le x\le 3\right)=0.4002}$.