For a statistical data $\mathrm{x}_{1}, \mathrm{x}_{2}, \ldots, \mathrm{x}_{10}$ of 10 values, a student obtained the mean as 5.5 and $\sum_{\mathrm{i}=1}^{10} \mathrm{x}_{\mathrm{i}}^{2}=371$. 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

JEE Main · 2025 · 24 Jan 2025 · Shift 1 · Statistics

For a statistical data

x_{1}, x_{2}, \dots, x_{10}

of 10 values, a student obtained the mean as 5.5 and

\sum_{i = 1}^{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✓ Correct
(B)4
(C)9
(D)5

Answer: (A)

Solution

From the mean,

i = 1 \sum 10 x_{i} = 5.5 \times 10 = 55

.

Correcting the two wrong entries (removing

4, 5

and inserting

6, 8

(\sum x_{i})_{new} = 55 - (4 + 5) + (6 + 8) = 60.

(\sum x_{i}^{2})_{new} = 371 - (4^{2} + 5^{2}) + (6^{2} + 8^{2}) = 371 - 41 + 100 = 430.

σ^{2} = \frac{\sum x _{i}^{2}}{10} - (\frac{\sum x _{i}}{10})^{2} = \frac{430}{10} - (\frac{60}{10})^{2} = 43 - 36 = 7.

JEE Main · 2025 · 24 Jan 2025 · Shift 1 · Single correct option

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