Let the mean and the standard deviation of th | vedclass

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

Question

(A) For a probability distribution,the sum of probabilities is $1$:$\frac{1}{3} + K + \frac{1}{6} + \frac{1}{4} = 1$$K + \frac{4+2+3}{12} = 1 \Rightar

Vedclass · Accepted Answer

$5$

Vedclass · Answer

(A) For a probability distribution,the sum of probabilities is $1$:$\frac{1}{3} + K + \frac{1}{6} + \frac{1}{4} = 1$$K + \frac{4+2+3}{12} = 1 \Rightarrow K + \frac{9}{12} = 1 \Rightarrow K = 1 - \frac{3}{4} = \frac{1}{4}$.The mean $\mu = \sum X P(X)$:$\mu = \alpha(\frac{1}{3}) + 1(\frac{1}{4}) + 0(\frac{1}{6}) + (-3)(\frac{1}{4}) = \frac{\alpha}{3} + \frac{1}{4} - \frac{3}{4} = \frac{\alpha}{3} - \frac{1}{2}$.The variance $\sigma^2 = \sum X^2 P(X) - \mu^2$:$\sum X^2 P(X) = \alpha^2(\frac{1}{3}) + 1^2(\frac{1}{4}) + 0^2(\frac{1}{6}) + (-3)^2(\frac{1}{4}) = \frac{\alpha^2}{3} + \frac{1}{4} + \frac{9}{4} = \frac{\alpha^2}{3} + \frac{10}{4} = \frac{\alpha^2}{3} + \frac{5}{2}$.$\sigma^2 = (\frac{\alpha^2}{3} + \frac{5}{2}) - (\frac{\alpha}{3} - \frac{1}{2})^2 = \frac{\alpha^2}{3} + \frac{5}{2} - (\frac{\alpha^2}{9} - \frac{\alpha}{3} + \frac{1}{4}) = \frac{2\alpha^2}{9} + \frac{\alpha}{3} + \frac{9}{4}$.Given $\sigma - \mu = 2$,so $\sigma = \mu + 2$. Squaring both sides:$\sigma^2 = (\mu + 2)^2 = \mu^2 + 4\mu + 4$.Substituting $\mu = \frac{\alpha}{3} - \frac{1}{2}$:$\frac{2\alpha^2}{9} + \frac{\alpha}{3} + \frac{9}{4} = (\frac{\alpha}{3} - \frac{1}{2})^2 + 4(\frac{\alpha}{3} - \frac{1}{2}) + 4$$\frac{2\alpha^2}{9} + \frac{\alpha}{3} + \frac{9}{4} = \frac{\alpha^2}{9} - \frac{\alpha}{3} + \frac{1}{4} + \frac{4\alpha}{3} - 2 + 4$$\frac{\alpha^2}{9} - \frac{2\alpha}{3} = 0 \Rightarrow \alpha(\frac{\alpha}{9} - \frac{2}{3}) = 0$.Since $\alpha 
eq 0$ (as $X=0$ is already a distinct value),$\alpha = 6$.Then $\mu = \frac{6}{3} - \frac{1}{2} = 2 - 0.5 = 1.5$.Since $\sigma = \mu + 2 = 1.5 + 2 = 3.5$.Therefore,$\sigma + \mu = 3.5 + 1.5 = 5$.

$X$	$\alpha$	$1$	$0$	$-3$
$P(X)$	$\frac{1}{3}$	$K$	$\frac{1}{6}$	$\frac{1}{4}$

$X$	$0$	$1$	$2$	$3$	$4$	$5$	$6$	$7$
$P(X)$	$0$	$2p$	$2p$	$3p$	$p^2$	$2p^2$	$7p^2$	$2p$

$X=x$	$-3$	$6$	$9$
$P(X=x)$	$\frac{1}{6}$	$\frac{1}{2}$	$\frac{1}{3}$

Let the mean and the standard deviation of the probability distribution be $\mu$ and $\sigma$,respectively. If $\sigma - \mu = 2$,then $\sigma + \mu$ is equal to:

$X$ $\alpha$ $1$ $0$ $-3$

$P(X)$ $\frac{1}{3}$ $K$ $\frac{1}{6}$ $\frac{1}{4}$

Similar Questions

$A$ random variable $X$ has the following probability distribution:
$X$ $0$ $1$ $2$ $3$ $4$ $5$ $6$ $7$
$P(X)$ $0$ $2p$ $2p$ $3p$ $p^2$ $2p^2$ $7p^2$ $2p$

The value of $p$ 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

If $X$ is a random variable with the following probability distribution:
$X=x$ $-3$ $6$ $9$
$P(X=x)$ $\frac{1}{6}$ $\frac{1}{2}$ $\frac{1}{3}$

Then the variance of $X$ is:

Given the probability density function (p.d.f.) of the random variable $X$ as $f(x) = \frac{1}{2a}$ for $0 < x < 2a$ and $f(x) = 0$ otherwise, where $a > 0$, which of the following is correct?

Two numbers are selected at random from the first six positive integers. If $X$ denotes the larger of the two numbers,then $\operatorname{Var}(X) = $

Vedclass Test Series

Exam Paper Generator

Online Exam Module

Let the mean and the standard deviation of the probability distribution be $\mu$ and $\sigma$,respectively. If $\sigma - \mu = 2$,then $\sigma + \mu$ is equal to: $X$ $\alpha$ $1$ $0$ $-3$ $P(X)$ $\frac{1}{3}$ $K$ $\frac{1}{6}$ $\frac{1}{4}$