The variance of the random variable X having | vedclass

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

Question

(D) The variance of a random variable $X$ is given by the formula:$\operatorname{Var}(X) = E(X^2) - [E(X)]^2$First,we calculate the expected value $E(

Vedclass · Accepted Answer

$\frac{5}{3}$

Vedclass · Answer

(D) The variance of a random variable $X$ is given by the formula:$\operatorname{Var}(X) = E(X^2) - [E(X)]^2$First,we calculate the expected value $E(X) = \sum P_i x_i$:$E(X) = (-2) 	imes \frac{1}{6} + (-1) 	imes \frac{1}{6} + 0 	imes \frac{1}{3} + 1 	imes \frac{1}{6} + 2 	imes \frac{1}{6}$$E(X) = -\frac{2}{6} - \frac{1}{6} + 0 + \frac{1}{6} + \frac{2}{6} = 0$Next,we calculate $E(X^2) = \sum P_i x_i^2$:$E(X^2) = (-2)^2 	imes \frac{1}{6} + (-1)^2 	imes \frac{1}{6} + 0^2 	imes \frac{1}{3} + 1^2 	imes \frac{1}{6} + 2^2 	imes \frac{1}{6}$$E(X^2) = 4 	imes \frac{1}{6} + 1 	imes \frac{1}{6} + 0 + 1 	imes \frac{1}{6} + 4 	imes \frac{1}{6}$$E(X^2) = \frac{4+1+0+1+4}{6} = \frac{10}{6} = \frac{5}{3}$Finally,the variance is:$\operatorname{Var}(X) = \frac{5}{3} - (0)^2 = \frac{5}{3}$

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

The variance of the random variable $X$ having the following distribution is:
$X = k$ $-2$ $-1$ $0$ $1$ $2$
$P(X = k)$ $\frac{1}{6}$ $\frac{1}{6}$ $\frac{1}{3}$ $\frac{1}{6}$ $\frac{1}{6}$

Similar Questions

The random variable $X$ has the following probability distribution:
| $X$ | $8$ | $12$ | $16$ | $20$ | $24$ |
|---|---|---|---|---|---|
| $P(X)$ | $K$ | $\frac{1}{6}$ | $\frac{3}{8}$ | $2K$ | $\frac{1}{12}$ |
Then the value of $K$ is:

If the mean of a Poisson distribution is $\frac{1}{3}$,then the ratio $P(X=1) : P(X=2)$ is

In a hospital, on an average if there are $35$ births in a week, then the probability that there will be less than $3$ births in a day is:

An unbiased coin is tossed $5$ times. Suppose that a variable $X$ is assigned the value $k$ when $k$ consecutive heads are obtained for $k=3, 4, 5$; otherwise,$X$ takes the value $-1$. Then the expected value of $X$ is:

From a lot of $20$ baskets,which includes $6$ defective baskets,a sample of $2$ baskets is drawn at random one by one without replacement. The expected value of the number of defective baskets is

Vedclass Test Series

Exam Paper Generator

Online Exam Module

The variance of the random variable $X$ having the following distribution is:$X = k$$-2$$-1$$0$$1$$2$$P(X = k)$$\frac{1}{6}$$\frac{1}{6}$$\frac{1}{3}$$\frac{1}{6}$$\frac{1}{6}$