Find the variance of the number obtained on a | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(A) The sample space of the experiment is $S = \{1, 2, 3, 4, 5, 6\}$.Let $X$ denote the number obtained on the throw. Then $X$ is a random variable wh

Vedclass · Accepted Answer

$\frac{35}{12}$

Vedclass · Answer

(A) The sample space of the experiment is $S = \{1, 2, 3, 4, 5, 6\}$.Let $X$ denote the number obtained on the throw. Then $X$ is a random variable which can take values $1, 2, 3, 4, 5,$ or $6$.Since it is a fair die,$P(1) = P(2) = P(3) = P(4) = P(5) = P(6) = \frac{1}{6}$.						$X$			$1, 2, 3, 4, 5, 6$							$P(X)$			$\frac{1}{6}, \frac{1}{6}, \frac{1}{6}, \frac{1}{6}, \frac{1}{6}, \frac{1}{6}$			Now,$E(X) = \sum x_i p(x_i) = (1+2+3+4+5+6) 	imes \frac{1}{6} = \frac{21}{6} = \frac{7}{2}$.Also,$E(X^2) = \sum x_i^2 p(x_i) = (1^2+2^2+3^2+4^2+5^2+6^2) 	imes \frac{1}{6} = (1+4+9+16+25+36) 	imes \frac{1}{6} = \frac{91}{6}$.Thus,$Var(X) = E(X^2) - (E(X))^2 = \frac{91}{6} - (\frac{7}{2})^2 = \frac{91}{6} - \frac{49}{4} = \frac{182 - 147}{12} = \frac{35}{12}$.

$X$	$0$	$1$	$2$	$3$	$4$
$P(X)$	$0.1$	$0.5$	$0.2$	$-0.1$	$0.3$

Find the variance of the number obtained on a throw of a fair die.

Similar Questions

If the range of a random variable $X$ is $\{0, 1, 2, 3, 4, \ldots\}$ with $P(X=k) = \frac{(k+1)a}{3^k}$ for $k \geq 0$,then $a$ is equal to

In a book of $500$ pages, it is found that there are $250$ typing errors. Assume that Poisson law holds for the number of errors per page. Then, the probability that a random sample of $2$ pages will contain no error, is:

$A$ random variable $X$ takes the values $0, 1$ and $2$. If $P(X=1)=P(X=2)$ and $P(X=0)=0.4$, then the mean of the random variable $X$ is

State which of the following is not a probability distribution of a random variable. Give reasons for your answer.

$X$ $0$ $1$ $2$ $3$ $4$

$P(X)$ $0.1$ $0.5$ $0.2$ $-0.1$ $0.3$

The p.d.f. of a continuous random variable $X$ is $f(x)=\begin{cases} \frac{x^2}{18} & \text{if } -3 < x < 3 \\ 0 & \text{otherwise} \end{cases}$. Then $P[|X| < 2]=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module