# Solved: 10. In a typical day, 31% of people in the United States with Internet access go online to get news. You randomly select five people

#### ByDr. Raju Chaudhari

Oct 17, 2020
1. In a typical day, 31% of people in the United States with Internet access go online to get news. You randomly select five people in the United States with Internet access and ask them if they go online to get news. Find the probability that the number who say they go online to get news is (a) exactly two, (b) at least two, and (c) more than two. (Source: Pew Research Center)

(a) construct a binomial distribution,
(b) graph the binomial distribution using a histogram and describe its shape,
(c) find the mean, variance, and standard deviation of the binomial distribution and interpret the results in the context of the real-life situation, and
(d) determine the values of the random variable x that you would consider unusual.

## Solution

Here $X$ denote the number of people who say they go online to get news.

$p$ be the probability that people in US with internet access go online to get news.

Given that $p=0.31$ and $n =5$. Thus$X\sim B(5, 0.31)$.

The probability mass function of $X$ is
 \begin{aligned} P(X=x) &= \binom{5}{x} (0.31)^x (1-0.31)^{5-x},\\ &\qquad \; x=0,1,\cdots, 5. \end{aligned}

(a) The probability that $X$ is exactly $2$ is

 \begin{aligned} P(X= 2) & =\binom{5}{2} (0.31)^{2} (1-0.31)^{5-2}\\ & = 0.3157\\ \end{aligned}

(b) The probability that $X$ is at least $2$ is

 \begin{aligned} P(X\geq 2) & =1-P(X\leq 1)\\ &= 1-\sum_{x=0}^{1} P(x)\\ & = 1-0.5077 \\ & = 0.4923 \\ \end{aligned}

(c) The probability that $X$ is more than $2$ is

 \begin{aligned} P(X > 2) & =1-P(X\leq 2)\\ & =1- \big( P(X=0) + P(X=1)+P(X=2)\big)\\ & = 1- \big(0.1564 + 0.3513+ 0.3157\big)\\ & = 1- 0.8234\\ &= 0.1766\\ \end{aligned}

a) Construct a binomial distribution

The probability that $X$ is exactly $0$ is

 \begin{aligned} P(X= 0) & =\binom{5}{0} (0.31)^{0} (1-0.31)^{5-0}\\ & = 0.1564\\ \end{aligned}

The probability that $X$ is exactly $1$ is

 \begin{aligned} P(X= 1) & =\binom{5}{1} (0.31)^{1} (1-0.31)^{5-1}\\ & = 0.3513\\ \end{aligned}

The probability that $X$ is exactly $2$ is

 \begin{aligned} P(X= 2) & =\binom{5}{2} (0.31)^{2} (1-0.31)^{5-2}\\ & = 0.3157\\ \end{aligned}

The probability that $X$ is exactly $3$ is

 \begin{aligned} P(X= 3) & =\binom{5}{3} (0.31)^{3} (1-0.31)^{5-3}\\ & = 0.1418\\ \end{aligned}

The probability that $X$ is exactly $4$ is

 \begin{aligned} P(X= 4) & =\binom{5}{4} (0.31)^{4} (1-0.31)^{5-4}\\ & = 0.0319\\ \end{aligned}

The probability that $X$ is exactly $5$ is

 \begin{aligned} P(X= 5) & =\binom{5}{5} (0.31)^{5} (1-0.31)^{5-5}\\ & = 0.0029\\ \end{aligned}

x Px
0 0.1564
1 0.3513
2 0.3157
3 0.1418
4 0.0319
5 0.0029

(b) Graph of Binomial Distribution

(c) The mean of Binomial Distribution is

$E(X) = n*p = 5\times 0.31 = 1.55$

The variance of Binomial distribution is

$V(X) = np(1-p) = 5\times 0.31\times (1- 0.31) = 1.0695$

The standard deviation of Binomial distribution is

$sd(X) = \sqrt{V(X)}= \sqrt{1.0695} = 1.0341663$

(d) The probability of $X= 5$ is 0.0029 which is very small. So the value of $X=5$ is unusual.