The following is the probability distribution | vedclass

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

Question

(B) For a probability distribution,the sum of probabilities must be $1$:$\frac{1+p}{5} + \frac{2-2p}{5} + \frac{2-p}{5} + \frac{2p}{5} = 1$$\frac{1+p+

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(B) For a probability distribution,the sum of probabilities must be $1$:$\frac{1+p}{5} + \frac{2-2p}{5} + \frac{2-p}{5} + \frac{2p}{5} = 1$$\frac{1+p+2-2p+2-p+2p}{5} = 1$$\frac{5}{5} = 1$. This is always true for any $p$.Since $P(X=x) \ge 0$ for all $x$,we have:$1+p \ge 0 \implies p \ge -1$$2-2p \ge 0 \implies p \le 1$$2-p \ge 0 \implies p \le 2$$2p \ge 0 \implies p \ge 0$Combining these,$0 \le p \le 1$. The minimum value of $p$ is $0$.Now,calculate $E(X) = \sum x P(X=x)$:$E(X) = 0 \cdot \frac{1+p}{5} + 1 \cdot \frac{2-2p}{5} + 2 \cdot \frac{2-p}{5} + 3 \cdot \frac{2p}{5}$$E(X) = \frac{2-2p + 4-2p + 6p}{5} = \frac{6+2p}{5}$For $p=0$,$E(X) = \frac{6}{5}$.Therefore,$5 E(X) = 5 \cdot \frac{6}{5} = 6$.

$X = x$	$-1.5$	$-0.5$	$0.5$	$1.5$	$2.5$
$P(X = x)$	$0.05$	$0.2$	$0.15$	$0.25$	$0.35$

The following is the probability distribution of $X$:

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

$P(X=x)$ $\frac{1+p}{5}$ $\frac{2-2p}{5}$ $\frac{2-p}{5}$ $\frac{2p}{5}$

For a minimum value of $p$,the value of $5 E(X)$ is:

Similar Questions

If the probability mass function (p.m.f.) of a random variable $X$ is $P(X=x) = \frac{1}{10}$ for $x = 1, 2, 3, \ldots, 10$,and $0$ otherwise,then $\operatorname{Var}(X)$ is equal to:

An urn contains $3$ black and $5$ red balls. If $3$ balls are drawn at random from the urn,the mean of the probability distribution of the number of red balls drawn is

If three dice are thrown, then the mean of the sum of the numbers appearing on them is

If $X$ is a random variable with cumulative distribution function $F(x)$ and its probability distribution is given by the following table:
$X = x$ $-1.5$ $-0.5$ $0.5$ $1.5$ $2.5$
$P(X = x)$ $0.05$ $0.2$ $0.15$ $0.25$ $0.35$

Then,find the value of $F(1.5) - F(-0.5)$.

The probability mass function of a random variable $X$ is given by $P(X=x) = \frac{{}^{5}C_{x}}{2^{5}}$ for $x = 0, 1, 2, 3, 4, 5$ and $0$ otherwise. Then,$P(X \leq 2)$ is equal to:

Vedclass Test Series

Exam Paper Generator

Online Exam Module

The following is the probability distribution of $X$: $X$ $0$ $1$ $2$ $3$ $P(X=x)$ $\frac{1+p}{5}$ $\frac{2-2p}{5}$ $\frac{2-p}{5}$ $\frac{2p}{5}$ For a minimum value of $p$,the value of $5 E(X)$ is: