Let X be a random variable with distribution. .tg border-collapse:collapse;border-spacing:0; .tg tdborder-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px; overflow:hidden;padding:10px 5px;word-break:normal; .tg thborder-color:black;border-style:solid;border-width:1px;font-family:Arial, sans-serif;font-size:14px; font-weight:normal;overflow:hidden;padding:10px 5px;word-break:normal; .tg .tg-baqhtext-align:center;vertical-align:top x - 2 - 1 3 4 6 P(X = x) 1 \over 5 a 1 \over 3 1 \over 5 b If the mean of X is 2.3 and variance of X is \sigma 2 , then 100 \sigma 2 is equal to :

Probability PYQ — JEE Main 2021 · 1 Sep · Shift 2 · Q38 with Solution

Let X be a random variable with distribution.

x	$-$ 2	$-$ 1	3	4	6
P(X = x)	${1 \over 5}$	a	${1 \over 3}$	${1 \over 5}$	b

If the mean of X is 2.3 and variance of X is

\sigma

², then 100

\sigma

² is equal to :

Numerical answer

View written solutionFree

Correct answer: 781

Given distribution

The random variable $X$ takes values $-2,",-1,\,3,\,4,\,6$ with probabilities $P(X=-2)=\frac15,\quad P(X=-1)=a,\quad P(X=3)=\frac13,\quad P(X=4)=\frac15,\quad P(X=6)=b.$

Also, the mean is $E[X]=2.3=\frac{23}{10}.$

We need to find the variance $\sigma^2$ and then compute $100\sigma^2$ .

Use total probability = 1

Since probabilities sum to $1$ , $\frac15+a+\frac13+\frac15+b=1.$

Now, $\frac15+\frac15=\frac25, \qquad \frac25+\frac13=\frac{6+5}{15}=\frac{11}{15}.$

So, $a+b=1-\frac{11}{15}=\frac{4}{15}. \tag{1}$

Use the mean

$E[X]=\sum xP(X=x).$

So, $(-2)\cdot\frac15+(-1)\cdot a+3\cdot\frac13+4\cdot\frac15+6b=\frac{23}{10}.$

Simplify the known terms:

\qquad 3\cdot\frac13=1, \qquad 4\cdot\frac15=\frac45.$$ Thus, $$-\frac25-a+1+\frac45+6b=\frac{23}{10}.$$ $$\left(-\frac25+1+\frac45\right)-a+6b=\frac{23}{10}.$$ $$\frac{7}{5}-a+6b=\frac{23}{10}. $$ Hence, $$-a+6b=\frac{23}{10}-\frac{7}{5}= rac{23}{10}-\frac{14}{10}=\frac{9}{10}. \tag{2}$$ --- 4. **Solve for $a$ and $b$** From (1), $$a=\frac{4}{15}-b.$$ Substitute into (2): $$-\left(\frac{4}{15}-b\right)+6b=\frac{9}{10}.$$ $$-\frac{4}{15}+b+6b=\frac{9}{10}.$$ $$7b=\frac{9}{10}+\frac{4}{15}.$$ Take LCM $30$: $$\frac{9}{10}=\frac{27}{30},\qquad \frac{4}{15}=\frac{8}{30}.$$ So, $$7b=\frac{35}{30}=\frac76.$$ Therefore, $$b=\frac16.$$ Then, $$a=\frac{4}{15}-\frac16=\frac{8-5}{30}=\frac{3}{30}=\frac{1}{10}.$$ So the full distribution is: $$P(X=-2)=\frac15, \quad P(X=-1)=\frac{1}{10}, \quad P(X=3)=\frac13, \quad P(X=4)=\frac15, \quad P(X=6)=\frac16.$$ --- 5. **Compute $E[X^2]$** $$E[X^2]=\sum x^2P(X=x).$$ So, $$E[X^2]=(-2)^2\cdot\frac15+(-1)^2\cdot\frac{1}{10}+3^2\cdot\frac13+4^2\cdot\frac15+6^2\cdot\frac16.$$ Now calculate each term: $$4\cdot\frac15=\frac45,$$ $$1\cdot\frac{1}{10}=\frac{1}{10},$$ $$9\cdot\frac13=3,$$ $$16\cdot\frac15=\frac{16}{5},$$ $$36\cdot\frac16=6.$$ Thus, $$E[X^2]=\frac45+\frac{1}{10}+3+\frac{16}{5}+6.$$ Convert to denominator $10$: $$\frac45=\frac{8}{10},\qquad \frac{16}{5}=\frac{32}{10}.$$ So, $$E[X^2]=\frac{8}{10}+\frac{1}{10}+3+\frac{32}{10}+6 =\frac{41}{10}+9 =\frac{41}{10}+\frac{90}{10} =\frac{131}{10}.$$ --- 6. **Compute variance** $$\sigma^2=E[X^2]-(E[X])^2.$$ We have $$E[X^2]=\frac{131}{10},\qquad E[X]=\frac{23}{10}.$$ Hence, $$\sigma^2=\frac{131}{10}-\left(\frac{23}{10}\right)^2 =\frac{131}{10}-\frac{529}{100}.$$ $$\frac{131}{10}=\frac{1310}{100}.$$ Therefore, $$\sigma^2=\frac{1310-529}{100}=\frac{781}{100}.$$ So, $$100\sigma^2=781.$$ --- 7. **Final answer** $$\boxed{781}$$ This matches the stored correct answer.

More from Probability

Let X be a random variable with distribution.

x	$-$ 2	$-$ 1	3	4	6
P(X = x)	${1 \over 5}$	a	${1 \over 3}$	${1 \over 5}$	b

If the mean of X is 2.3 and variance of X is

\sigma

², then 100

\sigma

² is equal to :

Numerical answer

View written solutionFree

Correct answer: 781

Given distribution

The random variable $X$ takes values $-2,",-1,\,3,\,4,\,6$ with probabilities $P(X=-2)=\frac15,\quad P(X=-1)=a,\quad P(X=3)=\frac13,\quad P(X=4)=\frac15,\quad P(X=6)=b.$

Also, the mean is $E[X]=2.3=\frac{23}{10}.$

We need to find the variance $\sigma^2$ and then compute $100\sigma^2$ .

Use total probability = 1

Since probabilities sum to $1$ , $\frac15+a+\frac13+\frac15+b=1.$

Now, $\frac15+\frac15=\frac25, \qquad \frac25+\frac13=\frac{6+5}{15}=\frac{11}{15}.$

So, $a+b=1-\frac{11}{15}=\frac{4}{15}. \tag{1}$

Use the mean

$E[X]=\sum xP(X=x).$

So, $(-2)\cdot\frac15+(-1)\cdot a+3\cdot\frac13+4\cdot\frac15+6b=\frac{23}{10}.$

Simplify the known terms:

\qquad 3\cdot\frac13=1, \qquad 4\cdot\frac15=\frac45.$$ Thus, $$-\frac25-a+1+\frac45+6b=\frac{23}{10}.$$ $$\left(-\frac25+1+\frac45\right)-a+6b=\frac{23}{10}.$$ $$\frac{7}{5}-a+6b=\frac{23}{10}. $$ Hence, $$-a+6b=\frac{23}{10}-\frac{7}{5}= rac{23}{10}-\frac{14}{10}=\frac{9}{10}. \tag{2}$$ --- 4. **Solve for $a$ and $b$** From (1), $$a=\frac{4}{15}-b.$$ Substitute into (2): $$-\left(\frac{4}{15}-b\right)+6b=\frac{9}{10}.$$ $$-\frac{4}{15}+b+6b=\frac{9}{10}.$$ $$7b=\frac{9}{10}+\frac{4}{15}.$$ Take LCM $30$: $$\frac{9}{10}=\frac{27}{30},\qquad \frac{4}{15}=\frac{8}{30}.$$ So, $$7b=\frac{35}{30}=\frac76.$$ Therefore, $$b=\frac16.$$ Then, $$a=\frac{4}{15}-\frac16=\frac{8-5}{30}=\frac{3}{30}=\frac{1}{10}.$$ So the full distribution is: $$P(X=-2)=\frac15, \quad P(X=-1)=\frac{1}{10}, \quad P(X=3)=\frac13, \quad P(X=4)=\frac15, \quad P(X=6)=\frac16.$$ --- 5. **Compute $E[X^2]$** $$E[X^2]=\sum x^2P(X=x).$$ So, $$E[X^2]=(-2)^2\cdot\frac15+(-1)^2\cdot\frac{1}{10}+3^2\cdot\frac13+4^2\cdot\frac15+6^2\cdot\frac16.$$ Now calculate each term: $$4\cdot\frac15=\frac45,$$ $$1\cdot\frac{1}{10}=\frac{1}{10},$$ $$9\cdot\frac13=3,$$ $$16\cdot\frac15=\frac{16}{5},$$ $$36\cdot\frac16=6.$$ Thus, $$E[X^2]=\frac45+\frac{1}{10}+3+\frac{16}{5}+6.$$ Convert to denominator $10$: $$\frac45=\frac{8}{10},\qquad \frac{16}{5}=\frac{32}{10}.$$ So, $$E[X^2]=\frac{8}{10}+\frac{1}{10}+3+\frac{32}{10}+6 =\frac{41}{10}+9 =\frac{41}{10}+\frac{90}{10} =\frac{131}{10}.$$ --- 6. **Compute variance** $$\sigma^2=E[X^2]-(E[X])^2.$$ We have $$E[X^2]=\frac{131}{10},\qquad E[X]=\frac{23}{10}.$$ Hence, $$\sigma^2=\frac{131}{10}-\left(\frac{23}{10}\right)^2 =\frac{131}{10}-\frac{529}{100}.$$ $$\frac{131}{10}=\frac{1310}{100}.$$ Therefore, $$\sigma^2=\frac{1310-529}{100}=\frac{781}{100}.$$ So, $$100\sigma^2=781.$$ --- 7. **Final answer** $$\boxed{781}$$ This matches the stored correct answer.

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