Let $\mu$ be the mean and $\sigma$ be the standard deviation of the distribution ${X}_{i}$ $0$ $1$ $2$ $3$ $4$ $5$ ${f}_{i}$ $k+2$ $2k$ ${k}^{2}-1$ ${k}^{2}-1$ ${k}^{2}+1$ $k-3$ where $\Sigma {f}_{i}=62$. If $[x]$ denotes the greatest integer $\leq x$, then$[{\mu }^{2}+{\sigma }^{2}]$ is equal to

JEE Main 2023 — Mathematics Probability & Statistics

2023mcqeasy

Let $\mu$ be the mean and $\sigma$ be the standard deviation of the distribution

${X}_{i}$	$0$	$1$	$2$	$3$	$4$	$5$
${f}_{i}$	$k+2$	$2k$	${k}^{2}-1$	${k}^{2}-1$	${k}^{2}+1$	$k-3$

where

\Sigma {f}_{i}=62

. If

[x]

denotes the greatest integer

\leq x

, then

[{\mu }^{2}+{\sigma }^{2}]

is equal to

Official previous-year question

Held on 10 Apr 2023 · Verified 6 Jul 2026.

Options

A
$9$
B
$8$
C
$7$
D
$6$

Did you get this right?

Solution

Given,

${X}_{i}$	$0$	$1$	$2$	$3$	$4$	$5$
${f}_{i}$	$k+2$	$2k$	${k}^{2}-1$	${k}^{2}-1$	${k}^{2}+1$	$k-3$

Now we know that,

$\text{Mean}=\frac{\sum {X}_{i}{f}_{i}}{\sum {f}_{i}}$

$\Rightarrow \mu =\frac{2k+2{k}^{2}–2+3{k}^{2}–3+4{k}^{2}+4+5k–15}{62}$

$\Rightarrow \mu =\frac{9{k}^{2}+7k-16}{62}.......(1)$

And $\displaystyle \sum {f}_{i}=62$

$\Rightarrow 3{k}^{2}+4k-2=62$

$\Rightarrow 3{k}^{2}+4k-64=0$

$\Rightarrow (3k+16)(k-4)=0$

$\Rightarrow k=4$

Now putting the value of $k$ in above equation $(1)$ we get,

$\mu =\frac{156}{62}$

Now finding variance we get,

${\sigma }^{2}=\frac{\sum {({X}_{i})}^{2}{f}_{i}}{\sum {f}_{i}}-{(\text{Mean})}^{2}$

$\Rightarrow {\sigma }^{2}=\frac{1\cdot (2k)+4({k}^{2}-1)+9({k}^{2}-1)+16({k}^{2}+1)+25(k-3)}{62}-{(\frac{156}{62})}^{2}$

$\Rightarrow {\sigma }^{2}=\frac{29{k}^{2}+27k-72}{62}-{\mu }^{2}$

Now putting the value of $k$ in above equation we get,

$\Rightarrow {\sigma }^{2}=\frac{500}{62}-{\mu }^{2}$

$\Rightarrow {\sigma }^{2}+{\mu }^{2}=\frac{500}{62}$

$\Rightarrow [{\sigma }^{2}+{\mu }^{2}]=8$

Your note

JEE Main Probability & Statistics in other years

Probability & Statistics 2026 Probability & Statistics 2025 Probability & Statistics 2024 Probability & Statistics 2022 Probability & Statistics 2021 Probability & Statistics 2020 Probability & Statistics 2019 Probability & Statistics 2018 Probability & Statistics 2017 Probability & Statistics 2016 Probability & Statistics 2015 Probability & Statistics 2014

More JEE Main Probability & Statistics questions

A bag contains 10 balls out of which $k$ are red and ($10-k$) are black, where $0 \leq k \leq 10$. If three balls are drawn at random without replacement and all of them are found to be black, then the probability that the bag contains 1 red and 9 black balls is:

2026hard

If X follows a Poisson distribution with P(X=1) = P(X=2) then the mean of the distribution is:

2026hard

If the mean and the variance of the data $\begin{array}{|c|c|c|c|c|} \hline \text{Class} & 4\text{-}8 & 8\text{-}12 & 12\text{-}16 & 16\text{-}20 \\ \hline \text{Frequency} & 3 & \lambda & 4 & 7 \\ \hline \end{array}$ are $\mu$ and 19 respectively, then the value of $\lambda+\mu$ is

2026hard

The mean and variance of $n$ observations are $8$ and $16$, respectively. If the sum of the first $(n-1)$ observations is $48$ and the sum of squares of the first $(n-1)$ observations is $496$, then the value of $n$ is:

2026medium

Let the mean and the variance of seven observations $2, 4, \alpha, 8, \beta, 12, 14$, $\alpha < \beta$, be $8$ and $16$ respectively. Then the quadratic equation whose roots are $3\alpha + 2$ and $2\beta + 1$ is :

2026medium

View all Probability & Statistics questions

Keep practicing Probability & Statistics

Work through every JEE Main Probability & Statistics PYQ, year by year.

Browse all Mathematics topics

Solution

Given,

${X}_{i}$	$0$	$1$	$2$	$3$	$4$	$5$
${f}_{i}$	$k+2$	$2k$	${k}^{2}-1$	${k}^{2}-1$	${k}^{2}+1$	$k-3$

Now we know that,

$\text{Mean}=\frac{\sum {X}_{i}{f}_{i}}{\sum {f}_{i}}$

$\Rightarrow \mu =\frac{2k+2{k}^{2}–2+3{k}^{2}–3+4{k}^{2}+4+5k–15}{62}$

$\Rightarrow \mu =\frac{9{k}^{2}+7k-16}{62}.......(1)$

And $\displaystyle \sum {f}_{i}=62$

$\Rightarrow 3{k}^{2}+4k-2=62$

$\Rightarrow 3{k}^{2}+4k-64=0$

$\Rightarrow (3k+16)(k-4)=0$

$\Rightarrow k=4$

Now putting the value of $k$ in above equation $(1)$ we get,

$\mu =\frac{156}{62}$

Now finding variance we get,

${\sigma }^{2}=\frac{\sum {({X}_{i})}^{2}{f}_{i}}{\sum {f}_{i}}-{(\text{Mean})}^{2}$

$\Rightarrow {\sigma }^{2}=\frac{1\cdot (2k)+4({k}^{2}-1)+9({k}^{2}-1)+16({k}^{2}+1)+25(k-3)}{62}-{(\frac{156}{62})}^{2}$

$\Rightarrow {\sigma }^{2}=\frac{29{k}^{2}+27k-72}{62}-{\mu }^{2}$

Now putting the value of $k$ in above equation we get,

$\Rightarrow {\sigma }^{2}=\frac{500}{62}-{\mu }^{2}$

$\Rightarrow {\sigma }^{2}+{\mu }^{2}=\frac{500}{62}$

$\Rightarrow [{\sigma }^{2}+{\mu }^{2}]=8$

JEE Main 2023 — Mathematics Probability & Statistics

Options

Did you get this right?

Solution

Your note

JEE Main Probability & Statistics in other years

More JEE Main Probability & Statistics questions

Other Mathematics topics for JEE Main

Algebra

Calculus

Coordinate Geometry

Vectors & 3D Geometry

Trigonometry

Keep practicing Probability & Statistics

JEE Main 2023 — Mathematics Probability & Statistics

Options

Did you get this right?

Solution

Your note

JEE Main Probability & Statistics in other years

More JEE Main Probability & Statistics questions

Other Mathematics topics for JEE Main

Algebra

Calculus

Coordinate Geometry

Vectors & 3D Geometry

Trigonometry

Keep practicing Probability & Statistics