JEE Main 2025 Jan 24 Shift 1, Mathematics Q14: Measure of Dispersion

JEE Main2025Jan 24, Shift 1Mathematics

For a statistical data $x_{1}, x_{2}, \dots, x_{10}$ of 10 values, a student obtained the mean as 5.5 and $i = 1 \sum 10 x_{i}^{2} = 371$ . He later found that he had noted two values in the data incorrectly as 4 and 5, instead of the correct values 6 and 8, respectively. The variance of the corrected data is

A
$7$
B
$4$
C
$9$
D
$5$

Solution

From the (incorrect) mean: $\sum x_{i} = 10 \cdot 5.5 = 55$ .

Corrected sum (remove 4 and 5, add 6 and 8): $\sum x_{i}^{new} = 55 - 9 + 14 = 60$ . New mean $= 6$ .

Corrected $\sum x_{i}^{2}$ (subtract $4^{2} + 5^{2} = 41$ , add $6^{2} + 8^{2} = 100$ ):

$\sum x_{i}^{2, new} = 371 - 41 + 100 = 430$ .

Variance $= \frac{\sum x _{i}^{2}}{n} - \overset{x}{ˉ}^{2} = \frac{430}{10} - 6^{2} = 43 - 36 = 7$ .

$∴$ Option (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 2025 Jan 24 Shift 1, Mathematics Q14: Measure of Dispersion

Concept behind this question

More previous year questions on Measure of Dispersion

Practice more Mathematics