JEE MainMathematics

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+4 marks−1 if incorrectSingle correctPrevious-year question

Mean and Variance of a Discrete Frequency Distribution

Q: Why didn't we compute μ explicitly?

Because σ^2 = (sum of f_i x_i^2)/N - μ^2 implies μ^2 + σ^2 = (sum of f_i x_i^2)/N, so μ^2 cancels out directly, saving computation time.

Let μ be the mean and σ be the standard deviation of the distribution

\begin{array}{lllllll} x_{i} & 0 & 1 & 2 & 3 & 4 & 5 \\ f_{i} & k + 2 & 2 k & k^{2} - 1 & k^{2} - 1 & k^{2} + 1 & k - 3 \end{array}

where sum of f_i=62. if [x] denotes the greatest integer ≤ x, then [μ^2+σ^2] is equal

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Step-by-step solutionView

Correct answer

By finding $k = 4$ from $\sum f_{i} = 62$ , we evaluate $μ^{2} + σ^{2} = \frac{\sum f_{i} x_{i}^{2}}{N} = \frac{506}{62} \approx 8.161$ , giving $[μ^{2} + σ^{2}] = 8$ .

Option analysis

Why each option works or fails

A · 8

The student correctly finds $k = 4$ , simplifies $μ^{2} + σ^{2}$ 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 $σ^{2}$ alone instead of $μ^{2} + σ^{2}$ , or miscalculated $\sum f_{i} x_{i}^{2}$ as a smaller value, yielding $\approx 7.8$ . Recognize that $σ^{2} = \frac{\sum f_{i} x_{i}^{2}}{N} - μ^{2}$ , which implies $μ^{2} + σ^{2} = \frac{\sum f_{i} x_{i}^{2}}{N}$ directly, avoiding unnecessary subtractions.

C · 6

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

D · 9

The student took the ceiling function $⌈ μ^{2} + σ^{2} ⌉$ 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 $f_{i} = [k + 2, 2 k, k^{2} - 1, k^{2} - 1, k^{2} + 1, k - 3]$ such that $\sum f_{i} = 62$ .

02Goal

Find the value of $[μ^{2} + σ^{2}]$ , where $μ$ is the mean, $σ$ 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 $μ^{2} + σ^{2} = \frac{\sum f_{i} x_{i}^{2}}{\sum f_{i}}$ to directly find the sum without separately computing and subtracting $μ^{2}$ .

04Execute

Sum the frequencies: $\sum f_{i} = (k + 2) + 2 k + (k^{2} - 1) + (k^{2} - 1) + (k^{2} + 1) + (k - 3) = 3 k^{2} + 4 k - 1 = 62$ . Thus $3 k^{2} + 4 k - 63 = 0 ⟹ (k - 4) (3 k + 16) = 0$ . Since $k > 3$ for $f_{5} = k - 3 \geq 0$ , 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 $σ^{2} = \frac{\sum f_{i} x_{i}^{2}}{N} - μ^{2}$ , we have $μ^{2} + σ^{2} = \frac{\sum f_{i} x_{i}^{2}}{N} = \frac{500}{62} \approx 8.0645$ . Taking the floor, $[μ^{2} + σ^{2}] = [8.0645] = 8$ .

✓Verify

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

Hints that build this answer step by step

What is the value of $k$ given that the sum of the frequencies $\sum f_{i} = 62$ ?
$k = 4$
How is $μ^{2} + σ^{2}$ directly related to the distribution values and frequencies?
$μ^{2} + σ^{2} = \frac{\sum f_{i} x_{i}^{2}}{N}$
With frequencies $f = [6, 8, 15, 15, 17, 1]$ for $x = [0, 1, 2, 3, 4, 5]$ , what is $[μ^{2} + σ^{2}]$ ?
8

✓ Source and academic review

Question type: Single correct
Exam relevance: JEE Main · Mathematics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 3 September 2026

Quick checks

Students also ask

Why didn't we compute

μ

explicitly?

Because $σ^{2} = \frac{\sum f_{i} x_{i}^{2}}{N} - μ^{2} ⟹ μ^{2} + σ^{2} = \frac{\sum f_{i} x_{i}^{2}}{N}$ , so $μ^{2}$ cancels out directly, saving computation time.

Answer

By finding $k = 4$ from $\sum f_{i} = 62$ , we evaluate $μ^{2} + σ^{2} = \frac{\sum f_{i} x_{i}^{2}}{N} = \frac{506}{62} \approx 8.161$ , giving $[μ^{2} + σ^{2}] = 8$ .

Why each option works or fails

A · correct: 8 - The student correctly finds $k = 4$ , simplifies $μ^{2} + σ^{2}$ 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 $σ^{2}$ alone instead of $μ^{2} + σ^{2}$ , or miscalculated $\sum f_{i} x_{i}^{2}$ as a smaller value, yielding $\approx 7.8$ . Recognize that $σ^{2} = \frac{\sum f_{i} x_{i}^{2}}{N} - μ^{2}$ , which implies $μ^{2} + σ^{2} = \frac{\sum f_{i} x_{i}^{2}}{N}$ directly, avoiding unnecessary subtractions.
C: 6 - The student calculated only $μ^{2} = (2.29)^{2} \approx 5.24$ or made an arithmetic error when evaluating $\sum f_{i} x_{i}^{2}$ . Ensure you add both $μ^{2}$ and $σ^{2}$ , which together equal the raw second moment $\frac{1}{N} \sum f_{i} x_{i}^{2}$ .
D: 9 - The student took the ceiling function $⌈ μ^{2} + σ^{2} ⌉$ 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 $f_{i} = [k + 2, 2 k, k^{2} - 1, k^{2} - 1, k^{2} + 1, k - 3]$ such that $\sum f_{i} = 62$ .
Goal: Find the value of $[μ^{2} + σ^{2}]$ , where $μ$ is the mean, $σ$ 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 $μ^{2} + σ^{2} = \frac{\sum f_{i} x_{i}^{2}}{\sum f_{i}}$ to directly find the sum without separately computing and subtracting $μ^{2}$ .
Execute: Sum the frequencies: $\sum f_{i} = (k + 2) + 2 k + (k^{2} - 1) + (k^{2} - 1) + (k^{2} + 1) + (k - 3) = 3 k^{2} + 4 k - 1 = 62$ . Thus $3 k^{2} + 4 k - 63 = 0 ⟹ (k - 4) (3 k + 16) = 0$ . Since $k > 3$ for $f_{5} = k - 3 \geq 0$ , 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 $σ^{2} = \frac{\sum f_{i} x_{i}^{2}}{N} - μ^{2}$ , we have $μ^{2} + σ^{2} = \frac{\sum f_{i} x_{i}^{2}}{N} = \frac{500}{62} \approx 8.0645$ . Taking the floor, $[μ^{2} + σ^{2}] = [8.0645] = 8$ .
Verify: Notice $62 \times 8 = 496 < 500 < 558 = 62 \times 9$ , so $500 / 62 \in [8, 9)$ , giving greatest integer equal to 8.

Mean and Variance of a Discrete Frequency Distribution

What feels right?

By finding $k = 4$ from $\sum f_{i} = 62$ , we evaluate $μ^{2} + σ^{2} = \frac{\sum f_{i} x_{i}^{2}}{N} = \frac{506}{62} \approx 8.161$ , giving $[μ^{2} + σ^{2}] = 8$ .

Why each option works or fails

Solution

01Given

02Goal

03Approach

04Execute

05Execute

06Execute

07Execute

✓Verify

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Mean and Variance of a Discrete Frequency Distribution

By finding k=4k=4 from ∑fi=62∑fi​=62, we evaluate μ2+σ2=∑fixi2N=50662≈8.161μ2+σ2=N∑fi​xi2​​=62506​≈8.161, giving [μ2+σ2]=8[μ2+σ2]=8.

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