JEE MainMathematics

Answer verified against the official NTA key

+4 marks−1 if incorrectNumericalPrevious-year question

Finding an Unknown Frequency from the Variance of a Distribution

Q: Why choose a = 5 as the assumed mean instead of any other value?

Because the unknown frequency α occurs at x_i = 5. Choosing a = 5 gives d_i = 0 at that point, which completely eliminates α from sum of f_i d_i and sum of f_i d_i^2, turning the problem into a simple linear equation.

If the variance of the frequency distribution

\begin{array}{cccccccc} x_{i} & 2 & 3 & 4 & 5 & 6 & 7 & 8 \\ Frequency f_{i} & 3 & 6 & 16 & α & 9 & 5 & 6 \end{array}

is 3, then α is equal to

Your answer stays private

What feels right?

No score. Commit to your first instinct. We’ll show what your mind noticed and what it missed.

Type the value - units or words beside it are fine.

✓ Source and academic review

Question type: Numerical
Exam relevance: JEE Main · Mathematics
Academic status: Answer verified against the official NTA key
Source: Previous-year question
Editorial review: 19 September 2026

Quick checks

Students also ask

Why choose

a = 5

as the assumed mean instead of any other value?

Because the unknown frequency $α$ occurs at $x_{i} = 5$ . Choosing $a = 5$ gives $d_{i} = 0$ at that point, which completely eliminates $α$ from $\sum f_{i} d_{i}$ and $\sum f_{i} d_{i}^{2}$ , turning the problem into a simple linear equation.

Answer

Accepted answer: 5

The unknown frequency $α$ is equal to $5$ .

Why each option works or fails

Step-by-step solution

Given: Given the discrete frequency distribution with values $x_{i} \in {2, 3, 4, 5, 6, 7, 8}$ and corresponding frequencies $f_{i} \in {3, 6, 16, α, 9, 5, 6}$ , with variance $σ^{2} = 3$ .
Goal: Find the value of the unknown frequency parameter $α$ .
Approach: Shift the origin by setting $d_{i} = x_{i} - 5$ . Choosing $a = 5$ eliminates $α$ from both $\sum f_{i} d_{i}$ and $\sum f_{i} d_{i}^{2}$ , simplifying the variance formula $σ^{2} = \frac{\sum f_{i} d_{i}^{2}}{N} - {(\frac{\sum f_{i} d_{i}}{N})}^{2}$ , where $N = \sum f_{i}$ .
Execute: Compute $N = \sum f_{i} = 3 + 6 + 16 + α + 9 + 5 + 6 = 45 + α$ . Compute deviations $d_{i} = x_{i} - 5$ : For $x_{i} = 2, 3, 4, 5, 6, 7, 8$ , the deviations $d_{i}$ are $- 3, - 2, - 1, 0, 1, 2, 3$ . Compute $\sum f_{i} d_{i}$ : $\sum f_{i} d_{i} = 3 (- 3) + 6 (- 2) + 16 (- 1) + α (0) + 9 (1) + 5 (2) + 6 (3) = - 9 - 12 - 16 + 0 + 9 + 10 + 18 = 0$ . Compute $\sum f_{i} d_{i}^{2}$ : $\sum f_{i} d_{i}^{2} = 3 (9) + 6 (4) + 16 (1) + α (0) + 9 (1) + 5 (4) + 6 (9) = 27 + 24 + 16 + 0 + 9 + 20 + 54 = 150$ .
Execute: Substitute into the variance formula: $σ^{2} = \frac{\sum f_{i} d_{i}^{2}}{N} - {(\frac{\sum f_{i} d_{i}}{N})}^{2} ⟹ 3 = \frac{150}{45 + α} - 0^{2}$ . Solve for $α$ : $3 (45 + α) = 150 ⟹ 45 + α = 50 ⟹ α = 5$ .
Verify: Verify: For $α = 5$ , total frequency $N = 50$ . $\sum f_{i} d_{i}^{2} / N = 150 / 50 = 3$ . Since $\sum f_{i} d_{i} = 0$ , mean $\overset{ˉ}{x} = 5$ , and variance is exactly $3 - 0 = 3$ .

Finding an Unknown Frequency from the Variance of a Distribution

What feels right?

The unknown frequency $α$ is equal to $5$ .

Why each option works or fails

Solution

01Given

02Goal

03Approach

04Execute

05Execute

✓Verify

We think you should solve this next.

Let $M$ be the maximum value of the product of two positive integers when their sum is 66…

If an unbiased die, marked with $- 2, - 1, 0, 1, 2, 3$ on its faces, is thrown five times, then…

If the sum and product of four positive consecutive terms of a G.P., are 126 and 1296 , r…

Students also ask

Finding an Unknown Frequency from the Variance of a Distribution

The unknown frequency αα is equal to 55.

01Given

02Goal

03Approach

04Execute

05Execute

✓Verify

Let MM be the maximum value of the product of two positive integers when their sum is 66…

If an unbiased die, marked with −2,−1,0,1,2,3−2,−1,0,1,2,3 on its faces, is thrown five times, then…

If the sum and product of four positive consecutive terms of a G.P., are 126 and 1296 , r…

Students also ask

The unknown frequency $α$ is equal to $5$ .

Let $M$ be the maximum value of the product of two positive integers when their sum is 66…

If an unbiased die, marked with $- 2, - 1, 0, 1, 2, 3$ on its faces, is thrown five times, then…