# Solved (Free): A federal report find that a lie detector test given to truthful persons have a probability of 0.2 of suggesting that the person is deceptive

#### ByDr. Raju Chaudhari

Mar 30, 2021

A federal report find that a lie detector test given to truthful persons have a probability of 0.2 of suggesting that the person is deceptive. A company asks 12 job applicants to take a lie detect test. Suppose that all 12 applicants answer truthfully.

a. What is the probability that exactly one is being deceptive?
b. What is the probability that at most one is being deceptive?
c. What is the mean and standard deviation of this distribution?

#### Solution

Let $p$ be the probability that a lie detector test suggests that the person is deceptive. Given that $p=0.2$. And $n=12$ job applicants are asked to take a lie detect test. Thus $X$ denote the number of deceptive person.

Here $X\sim B(12, 0.2)$.

The probability mass function of $X$ is

 \begin{aligned} P(X=x) &= \binom{12}{x} (0.2)^x (1-0.2)^{12-x},\\ &\quad x=0,1,\cdots, 12. \end{aligned}

a. The probability that exactly one is begin deceptive is

 \begin{aligned} P(X= 1) &= \binom{12}{1}(0.2)^1(1-0.2)^{11}\\ & = 0.2062 \end{aligned}

b. The probability that at the most 1 is being deceptive is

 \begin{aligned} P(X\leq 1) &= \sum_{x=0}^{1}\binom{12}{x}(0.2)^x(1-0.2)^{12-x}\\ &=\binom{12}{0}(0.2)^0(1-0.2)^{12}+\binom{12}{1}(0.2)^1(1-0.2)^{11}\\ & = 0.0687+0.2062\\ & = 0.2749 \end{aligned}

c. The mean of the distribution is

 \begin{aligned} E(X) &= np \\ &= 12 * 0.2\\ &= 2.4 \end{aligned}

The standard deviation of the distribution is

 \begin{aligned} sd &= \sqrt{np(1-p)}\\ &= \sqrt{ 12 * 0.2 *0.8 }\\ &= 1.3856406 \end{aligned}