# Solved[Free]-Eighteen individuals are scheduled to take a driving test at a particular DMV office on a certain day, eight of whom wil

#### ByDr. Raju Chaudhari

Oct 6, 2020

Eighteen individuals are scheduled to take a driving test at a particular DMV office on a certain day, eight of whom will be taking the test for the first time. Suppose that six of these individuals are randomly assigned to a particular examiner, and let X be the number among the six who are taking the test for the first time.

(a) What kind of a distribution does X have (name and values of all parameters)?
(b) Compute $P(X=2)$, $P(X\leq 2)$, $P(X\geq 2)$.
(c) Calculate the mean value and standard deviation of X.

## Solution

Let $X$ be the number among the six who are taking the test for the first time.

$p=8/18 = 0.444$ be the probability of taking a driving test at a particular DMV office.

(a) Given that $p=0.444$ and $n =6$. Thus $X$ follows a Binomial distribution. That is, $X\sim B(6, 0.444)$.

The probability mass function of $X$ is
\begin{aligned} P(X=x) &= \binom{6}{x} (0.444)^x (1-0.444)^{6-x},\\ & \qquad x=0,1,\cdots, 6. \end{aligned}

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

\begin{aligned} P(X= 2) & =P(2)\\ & = 0.2826\\ \end{aligned}

\begin{aligned} P(X\leq 2) & =\sum_{x=0}^{2} P(x)\\ & =P(0) + P(1) + P(2)\\ & = 0.0295+0.1415+0.2826\\ &= 0.4537 \end{aligned}

\begin{aligned} P(X\geq 2) & =1-P(X\leq 1)\\ &= 1-(P(0)+P(1))\\ & = 1-\big(0.0295+0.1415 \big)\\ & = 0.8289 \\ \end{aligned}

(c) The mean of $X$ is $\mu = n*p = 6*0.444 = 2.664$ and the standard deviation of $X$ is $\sigma =\sqrt{n*p*(1-p)}=\sqrt{6*0.444* (1- 0.444)}= 1.217$.