The probability distribution of a discrete ra | vedclass

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

Question

(C) First, we calculate $E(X)$ and $E(X^2)$ using the given distribution:$E(X) = \Sigma x P(X=x) = (-1)(\frac{1}{3}) + (0)(\frac{1}{6}) + (1)(\frac{1}

Vedclass · Accepted Answer

$\frac{113}{12}$

Vedclass · Answer

(C) First, we calculate $E(X)$ and $E(X^2)$ using the given distribution:$E(X) = \Sigma x P(X=x) = (-1)(\frac{1}{3}) + (0)(\frac{1}{6}) + (1)(\frac{1}{6}) + (2)(\frac{1}{3}) = -\frac{1}{3} + 0 + \frac{1}{6} + \frac{2}{3} = \frac{-2+0+1+4}{6} = \frac{3}{6} = \frac{1}{2}$$E(X^2) = \Sigma x^2 P(X=x) = (-1)^2(\frac{1}{3}) + (0)^2(\frac{1}{6}) + (1)^2(\frac{1}{6}) + (2)^2(\frac{1}{3}) = \frac{1}{3} + 0 + \frac{1}{6} + \frac{4}{3} = \frac{2+0+1+8}{6} = \frac{11}{6}$Now, we find $\operatorname{Var}(X) = E(X^2) - (E(X))^2 = \frac{11}{6} - (\frac{1}{2})^2 = \frac{11}{6} - \frac{1}{4} = \frac{22-3}{12} = \frac{19}{12}$Finally, we calculate $6 E(X^2) - \operatorname{Var}(X)$:$6 E(X^2) - \operatorname{Var}(X) = 6(\frac{11}{6}) - \frac{19}{12} = 11 - \frac{19}{12} = \frac{132-19}{12} = \frac{113}{12}$

$X=x$	$-1$	$0$	$1$	$2$
$P(X=x)$	$\frac{1}{3}$	$\frac{1}{6}$	$\frac{1}{6}$	$\frac{1}{3}$

$X$	$1$	$2$	$3$	$4$	$5$	$6$	$7$
$P(X)$	$k-1$	$3k$	$k$	$3k$	$3k^2$	$k^2$	$k^2+k$

The probability distribution of a discrete random variable $X$ is given below. If $E(X^2) = \Sigma x^2 P(X=x)$, then $6 E(X^2) - \operatorname{Var}(X) =$
$X=x$ $-1$ $0$ $1$ $2$
$P(X=x)$ $\frac{1}{3}$ $\frac{1}{6}$ $\frac{1}{6}$ $\frac{1}{3}$

Similar Questions

$A$ biased coin with probability $p$ $(0 < p < 1)$ of getting a head is tossed until a head appears for the first time. If the probability that the number of tosses required is even is $\frac{2}{5}$, then $p=$

If $X$ follows a Poisson distribution with variance $2$,then $P(X \geq 3) = $

$A$ random variable $X$ has the following probability distribution:
$X$ $1$ $2$ $3$ $4$ $5$ $6$ $7$
$P(X)$ $k-1$ $3k$ $k$ $3k$ $3k^2$ $k^2$ $k^2+k$

Then the value of $k$ is:

Let $X$ denote the sum of the numbers obtained when two fair dice are rolled. Find the variance and standard deviation of $X$.

The p.d.f. of a continuous random variable $X$ is given by $f(x) = \frac{x+2}{18}$ for $-2 < x < 4$ and $f(x) = 0$ otherwise. Then $P[|x| < 1] = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module

The probability distribution of a discrete random variable $X$ is given below. If $E(X^2) = \Sigma x^2 P(X=x)$, then $6 E(X^2) - \operatorname{Var}(X) =$$X=x$$-1$$0$$1$$2$$P(X=x)$$\frac{1}{3}$$\frac{1}{6}$$\frac{1}{6}$$\frac{1}{3}$