# Solved (Free): 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

#### ByDr. Raju Chaudhari

Mar 10, 2021

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

Let $X$ denote the number who say they go online to get news out of 5. In a typical day, 31% of people in the United States with Internet access go online to get news. That is $p = 0.31$.

Here $n = 5$ and $p=0.31$. The probability distribution of $X$ is Binomial distribution. That is $X\sim B(5,0.60)$.

The probability mass function of $X$ is


\begin{aligned} P(X=x) &= \binom{5}{x} (0.31)^x (1-0.31)^{5-x},\ &\quad x=0,1,\cdots, 5 \end{aligned}

(a) The probability that the number who say they go online to get news is exactly two $2$

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

(b) The probability that the number who say they go online to get news is at least two

 \begin{aligned} P(X\geq 2) & =1-P(X< 2)\\ & = 1- \bigg(\binom{5}{0} (0.31)^{0} (1-0.31)^{5-0}+\binom{5}{1} (0.31)^{1} (1-0.31)^{5-1}\bigg)\\ & = 1- \bigg(0.1564+ 0.3513\bigg)\\ &= 1- 0.5077\\ &=0.4923 \end{aligned} 

(c) The probability that the number who say they go online to get news is more than two

 \begin{aligned} P(X> 2) & =1-P(X\leq 2)\\ & = 1- \bigg(\binom{5}{0} (0.31)^{0} (1-0.31)^{5-0}+\binom{5}{1} (0.31)^{1} (1-0.31)^{5-1}+\binom{5}{2} (0.31)^{2} (1-0.31)^{5-2}\bigg)\\ & = 1- \bigg(0.1564+ 0.3513+0.3157\bigg)\\ &= 1- 0.8234\\ &=0.1766 \end{aligned} 

(a) The binomial distribution is

x px
0 0.1564031
1 0.3513404
2 0.3156971
3 0.1418350
4 0.0318615
5 0.0028629

(b) Histogram of Binomial Distribution

The histogram is positively skewed.

(c) The mean of $X$ is $E(X) = np = 5 \times 0.31 = 1.55$.

The variance of $X$ is $V(X) = np(1-p) = 5 \times 0.31\times (1-0.31) = 1.0695$.

The standard deviation of the probability distribution is $sd(X) = \sqrt{np(1-p)} = \sqrt{5 \times 0.31 \times (1- 0.31)} = 1.0342$`.

(d) The values of the random variable x that would be consider as unusual is $X=5$ because the $P(X=5)$ is very small.