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,$\sigma + \mu = 3.5 + 1.5 = 5$.

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

$X=x$	$1$	$2$	$3$	$4$
$P(X=x)$	$0.1$	$0.2$	$0.3$	$0.4$

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

If $f(x) = \frac{x}{8}$ for $0 < x < 4$ and $f(x) = 0$ otherwise,is the probability density function (p.d.f.) of a continuous random variable $X$,and $F(x)$ is the cumulative distribution function (c.d.f.) associated with $f(x)$,then find $F(0.5)$.

If a random variable $X$ denotes the number that appears on the upper face of a die when it is rolled,then $\frac{\text{Variance of } X}{\text{Mean of } X}$ is equal to

Two cards are drawn at random from a box which contains $5$ cards numbered $1, 1, 2, 2, 3$. If $X$ denotes the sum of the numbers on the two cards, then the expected value of $X$ is...

Find the probability distribution of the number of heads in four tosses of a coin.

$A$ random variable $X$ has the following probability distribution:
$X=x$ $1$ $2$ $3$ $4$
$P(X=x)$ $0.1$ $0.2$ $0.3$ $0.4$

The mean and standard deviation of $X$ are respectively:

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