JEE Main 2026 Apr 6 Shift 1, Mathematics Q9: Measure of Dispersion

JEE Main2026Apr 6, Shift 1Mathematics

A data consists of $20$ observations $x_{1}, x_{2}, \dots, x_{20}$ . If $i = 1 \sum 20 (x_{i} + 5)^{2} = 2500$ and $i = 1 \sum 20 (x_{i} - 5)^{2} = 100$ , then the ratio of mean to standard deviation of this data is:

A
$2 : 1$
B
$3 : 1$
C
$3 : 2$
D
$4 : 1$

Solution

Expand both sums. Subtracting:

$\sum (x_{i} + 5)^{2} - \sum (x_{i} - 5)^{2} = 20 \sum x_{i} = 2500 - 100 = 2400 \Rightarrow \sum x_{i} = 120$ , so $\overset{x}{ˉ} = 6$ .

Adding: $\sum (x_{i} + 5)^{2} + \sum (x_{i} - 5)^{2} = 2 \sum x_{i}^{2} + 2 \cdot 20 \cdot 25 = 2600 \Rightarrow \sum x_{i}^{2} = 800$ .

Variance $σ^{2} = \frac{800}{20} - 6^{2} = 40 - 36 = 4$ , so $σ = 2$ .

$∴ \overset{x}{ˉ} : σ = 6 : 2 = 3 : 1$ .

Concept behind this question

Measure of DispersionNotes, formulas and examples →

Practice more Mathematics

Concept-wise practice with instant solutions on Fundamenthol.

Start practicing →

JEE Main 2026 Apr 6 Shift 1, Mathematics Q9: Measure of Dispersion

Concept behind this question

More previous year questions on Measure of Dispersion

Practice more Mathematics