Why didn't we compute explicitly?

Because , so cancels out directly, saving computation time.

JEE MainMathematics

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Statistics and Probability: Mathematics | JEE Main

Let

\mu

be the mean and

\sigma

be the standard deviation of the distribution

\begin{array}{|l|l|l|l|l|l|l|} \hline x_i & 0 & 1 & 2 & 3 & 4 & 5 \\ \hline f_i & k+2 & 2k & k^2-1 & k^2-1 & k^2+1 & k-3 \\ \hline \end{array}

where

\sum \mathrm{f}_{i}=62

. if

[\mathrm{x}]

denotes the greatest integer

\leq \mathrm{x}

, then

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

is equal

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Step-by-step solutionView

Correct answer

By finding $k = 4$ from $\sum f_i = 62$ , we evaluate $\mu^2 + \sigma^2 = \frac{\sum f_i x_i^2}{N} = \frac{506}{62} \approx 8.161$ , giving $\left[\mu^2 + \sigma^2\right] = 8$ .

Option analysis

Why each option works or fails

A · 8

None. The student correctly finds $k=4$ , simplifies $\mu^2+\sigma^2$ to $\frac{\sum f_i x_i^2}{N} = \frac{506}{62} \approx 8.161$ , and applies the floor function to get $8$ . Correct.

B · 7

The student computed $\sigma^2$ alone instead of $\mu^2 + \sigma^2$ , or miscalculated $\sum f_i x_i^2$ as a smaller value, yielding $\approx 7.8$ . Recognize that $\sigma^2 = \frac{\sum f_i x_i^2}{N} - \mu^2$ , which implies $\mu^2 + \sigma^2 = \frac{\sum f_i x_i^2}{N}$ directly, avoiding unnecessary subtractions.

C · 6

The student calculated only $\mu^2 = (2.29)^2 \approx 5.24$ or made an arithmetic error when evaluating $\sum f_i x_i^2$ . Ensure you add both $\mu^2$ and $\sigma^2$ , which together equal the raw second moment $\frac{1}{N}\sum f_i x_i^2$ .

D · 9

The student took the ceiling function $\lceil \mu^2 + \sigma^2 \rceil$ or rounded to the nearest integer instead of taking the greatest integer function $[\cdot]$ . The greatest integer function $[x]$ always rounds down to the largest integer less than or equal to $x$ .

Reviewed route

Solution

StepWorking

01given

A frequency distribution with values $x_i \in \{0, 1, 2, 3, 4, 5\}$ and frequencies $f_i = [k+2, 2k, k^2-1, k^2-1, k^2+1, k-3]$ such that $\sum f_i = 62$ .

02goal

Find the value of $[\mu^2 + \sigma^2]$ , where $\mu$ is the mean, $\sigma$ is the standard deviation, and $[\cdot]$ is the greatest integer function.

03approach

First solve for $k$ using $\sum f_i = 62$ (noting frequencies must be non-negative integers). Then use the identity $\mu^2 + \sigma^2 = \frac{\sum f_i x_i^2}{\sum f_i}$ to directly find the sum without separately computing and subtracting $\mu^2$ .

04execute

Sum the frequencies: $\sum f_i = (k+2) + 2k + (k^2-1) + (k^2-1) + (k^2+1) + (k-3) = 3k^2 + 4k - 1 = 62$ . Thus $3k^2 + 4k - 63 = 0 \implies (k-4)(3k+16) = 0$ . Since $k > 3$ for $f_5 = k-3 \ge 0$ , we get $k = 4$ .

05execute

Substitute $k = 4$ into each frequency: $f_0 = 6, f_1 = 8, f_2 = 15, f_3 = 15, f_4 = 17, f_5 = 1$ . Check: $\sum f_i = 6 + 8 + 15 + 15 + 17 + 1 = 62$ .

06execute

Compute $\sum f_i x_i^2 = 6(0)^2 + 8(1)^2 + 15(4) + 15(9) + 17(16) + 1(25) = 0 + 8 + 60 + 135 + 272 + 25 = 500$ .

07execute

Since $\sigma^2 = \frac{\sum f_i x_i^2}{N} - \mu^2$ , we have $\mu^2 + \sigma^2 = \frac{\sum f_i x_i^2}{N} = \frac{500}{62} \approx 8.0645$ . Taking the floor, $[\mu^2 + \sigma^2] = [8.0645] = 8$ .

✓verify

Notice $62 \times 8 = 496 < 500 < 558 = 62 \times 9$ , so $500/62 \in [8, 9)$ , giving greatest integer equal to 8.

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Concepts assessed: Mathematics
Academic status: Reviewed by official_key
Source: pyq
Editorial review: 3 September 2026

Quick checks

Students also ask

Why didn't we compute

\mu

explicitly?

Because $\sigma^2 = \frac{\sum f_i x_i^2}{N} - \mu^2 \implies \mu^2 + \sigma^2 = \frac{\sum f_i x_i^2}{N}$ , so $\mu^2$ cancels out directly, saving computation time.

Answer

By finding $k = 4$ from $\sum f_i = 62$ , we evaluate $\mu^2 + \sigma^2 = \frac{\sum f_i x_i^2}{N} = \frac{506}{62} \approx 8.161$ , giving $\left[\mu^2 + \sigma^2\right] = 8$ .

Why each option works or fails

A · correct: 8 - None. The student correctly finds $k=4$ , simplifies $\mu^2+\sigma^2$ to $\frac{\sum f_i x_i^2}{N} = \frac{506}{62} \approx 8.161$ , and applies the floor function to get $8$ . Correct.
B: 7 - The student computed $\sigma^2$ alone instead of $\mu^2 + \sigma^2$ , or miscalculated $\sum f_i x_i^2$ as a smaller value, yielding $\approx 7.8$ . Recognize that $\sigma^2 = \frac{\sum f_i x_i^2}{N} - \mu^2$ , which implies $\mu^2 + \sigma^2 = \frac{\sum f_i x_i^2}{N}$ directly, avoiding unnecessary subtractions.
C: 6 - The student calculated only $\mu^2 = (2.29)^2 \approx 5.24$ or made an arithmetic error when evaluating $\sum f_i x_i^2$ . Ensure you add both $\mu^2$ and $\sigma^2$ , which together equal the raw second moment $\frac{1}{N}\sum f_i x_i^2$ .
D: 9 - The student took the ceiling function $\lceil \mu^2 + \sigma^2 \rceil$ or rounded to the nearest integer instead of taking the greatest integer function $[\cdot]$ . The greatest integer function $[x]$ always rounds down to the largest integer less than or equal to $x$ .

Step-by-step solution

given: A frequency distribution with values $x_i \in \{0, 1, 2, 3, 4, 5\}$ and frequencies $f_i = [k+2, 2k, k^2-1, k^2-1, k^2+1, k-3]$ such that $\sum f_i = 62$ .
goal: Find the value of $[\mu^2 + \sigma^2]$ , where $\mu$ is the mean, $\sigma$ is the standard deviation, and $[\cdot]$ is the greatest integer function.
approach: First solve for $k$ using $\sum f_i = 62$ (noting frequencies must be non-negative integers). Then use the identity $\mu^2 + \sigma^2 = \frac{\sum f_i x_i^2}{\sum f_i}$ to directly find the sum without separately computing and subtracting $\mu^2$ .
execute: Sum the frequencies: $\sum f_i = (k+2) + 2k + (k^2-1) + (k^2-1) + (k^2+1) + (k-3) = 3k^2 + 4k - 1 = 62$ . Thus $3k^2 + 4k - 63 = 0 \implies (k-4)(3k+16) = 0$ . Since $k > 3$ for $f_5 = k-3 \ge 0$ , we get $k = 4$ .
execute: Substitute $k = 4$ into each frequency: $f_0 = 6, f_1 = 8, f_2 = 15, f_3 = 15, f_4 = 17, f_5 = 1$ . Check: $\sum f_i = 6 + 8 + 15 + 15 + 17 + 1 = 62$ .
execute: Compute $\sum f_i x_i^2 = 6(0)^2 + 8(1)^2 + 15(4) + 15(9) + 17(16) + 1(25) = 0 + 8 + 60 + 135 + 272 + 25 = 500$ .
execute: Since $\sigma^2 = \frac{\sum f_i x_i^2}{N} - \mu^2$ , we have $\mu^2 + \sigma^2 = \frac{\sum f_i x_i^2}{N} = \frac{500}{62} \approx 8.0645$ . Taking the floor, $[\mu^2 + \sigma^2] = [8.0645] = 8$ .
verify: Notice $62 \times 8 = 496 < 500 < 558 = 62 \times 9$ , so $500/62 \in [8, 9)$ , giving greatest integer equal to 8.

Statistics and Probability: Mathematics | JEE Main

By finding k=4k = 4k=4 from ∑fi=62\sum f_i = 62∑fi​=62, we evaluate μ2+σ2=∑fixi2N=50662≈8.161\mu^2 + \sigma^2 = \frac{\sum f_i x_i^2}{N} = \frac{506}{62} \approx 8.161μ2+σ2=N∑fi​xi2​​=62506​≈8.161, giving [μ2+σ2]=8\left[\mu^2 + \sigma^2\right] = 8[μ2+σ2]=8.

01given

02goal

03approach

04execute

05execute

06execute

07execute

✓verify

Students also ask

By finding $k = 4$ from $\sum f_i = 62$ , we evaluate $\mu^2 + \sigma^2 = \frac{\sum f_i x_i^2}{N} = \frac{506}{62} \approx 8.161$ , giving $\left[\mu^2 + \sigma^2\right] = 8$ .