The probability distribution of a random vari | vedclass

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

Question

(A) The expected value $E(X)$ is calculated as:$E(X) = \sum x_i \cdot P(x_i)$$E(X) = 0(0.4) + 1(0.3) + 2(0.1) + 3(0.1) + 4(0.1)$$E(X) = 0 + 0.3 + 0.2

Vedclass · Accepted Answer

$1.76$

Vedclass · Answer

(A) The expected value $E(X)$ is calculated as:$E(X) = \sum x_i \cdot P(x_i)$$E(X) = 0(0.4) + 1(0.3) + 2(0.1) + 3(0.1) + 4(0.1)$$E(X) = 0 + 0.3 + 0.2 + 0.3 + 0.4 = 1.2$The expected value of the square of the random variable $E(X^2)$ is:$E(X^2) = \sum x_i^2 \cdot P(x_i)$$E(X^2) = 0^2(0.4) + 1^2(0.3) + 2^2(0.1) + 3^2(0.1) + 4^2(0.1)$$E(X^2) = 0(0.4) + 1(0.3) + 4(0.1) + 9(0.1) + 16(0.1)$$E(X^2) = 0 + 0.3 + 0.4 + 0.9 + 1.6 = 3.2$The variance of $X$ is given by:$	ext{Var}(X) = E(X^2) - [E(X)]^2$$	ext{Var}(X) = 3.2 - (1.2)^2$$	ext{Var}(X) = 3.2 - 1.44 = 1.76$

$X = x_i$	$0$	$1$	$2$	$3$	$4$
$P(X = x_i)$	$0.4$	$0.3$	$0.1$	$0.1$	$0.1$

$X = x$	$0$	$1$	$2$	$3$	$4$	$5$	$6$	$7$
$P(x)$	$0.01$	$0.10$	$0.26$	$0.33$	$0.18$	$0.06$	$K$	$0.04$

The probability distribution of a random variable $X$ is given by:
$X = x_i$ $0$ $1$ $2$ $3$ $4$
$P(X = x_i)$ $0.4$ $0.3$ $0.1$ $0.1$ $0.1$

Then the variance of $X$ is:

Similar Questions

If $X$ is a Poisson variate such that $P(X=1) = 2P(X=2)$,then $P(X=3)$ is equal to:

Three defective oranges are accidentally mixed with seven good ones and on looking at them,it is not possible to differentiate between them. Two oranges are drawn at random from the lot. If $x$ denotes the number of defective oranges,then the variance of $x$ is:

The probability distribution of a random variable $X$ is given below.
$X = x$ $0$ $1$ $2$ $3$ $4$ $5$ $6$ $7$
$P(x)$ $0.01$ $0.10$ $0.26$ $0.33$ $0.18$ $0.06$ $K$ $0.04$

Then $P(X \geq 3) - P(X < 6) =$

If $X$ is a Poisson variate satisfying the condition $3 P(X=2)=P(X=4)$,then find $P(X=6)$.

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 probability distribution of a random variable $X$ is given by:$X = x_i$$0$$1$$2$$3$$4$$P(X = x_i)$$0.4$$0.3$$0.1$$0.1$$0.1$Then the variance of $X$ is: