For a statistical data x_1, x_2, , x_ 10 of 10 values, a student obtained the mean as 5.5 and ∑_ 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 answer: C. 7 aligned & Mean x =5.5 & =∑_ i=1 ^ 10 x_ i =5.5 × 10=55 & =∑_ i=1 ^ 10 x_ i^2=371 & (∑ x_ i )_ new =55-(4+5)+(6+8)=60 & & (∑ x_ i^2)_ new =371-(4^2+5^2)+(6^2+8^2)=430 aligned aligned & Variance σ^2= ∑ x_ i^2 10 -( ∑ x_ i 10 )^2 & σ^2=(430)/(10)-((60)/(10))^2 & σ^2=43-36 & σ^2=7 aligned

JEE Main 202524 JanuaryMorning Shift

Exam: JEE Main
Year: 2025
Date: 24 January
Shift: Morning Shift
Subject: Mathematics
Chapter: Statistics

JEE Main 2025 · 24 January · Morning Shift · Mathematics · Statistics

For a statistical data x_1, x_2, , x_ 10 of 10 values, a student...

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_{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

Options

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

Correct answer

C. $7$

Step-by-step solution

\begin{aligned} & \text{ Mean } \overline{\mathrm{x}}=5.5 \\ & =\sum_{\mathrm{i}=1}^{10} \mathrm{x}_{\mathrm{i}}=5.5 \times 10=55 \\ & =\sum_{\mathrm{i}=1}^{10} \mathrm{x}_{\mathrm{i}^2=371 \\ & \left(\sum \mathrm{x}_{\mathrm{i}}\right)_{\text{new }}=55-(4+5)+(6+8)=60 \\ & \\ & \left(\sum \mathrm{x}_{\mathrm{i}^2\right)_{\text{new }}=371-\left(4^2+5^2\right)+\left(6^2+8^2\right)=430\end{aligned}

\begin{aligned} & \text{ Variance } \sigma^2=\frac{\sum \mathrm{x}_{\mathrm{i}^2}{10}-\left(\frac{\sum \mathrm{x}_{\mathrm{i}}}{10}\right)^2 \\ & \sigma^2=\frac{430}{10}-\left(\frac{60}{10}\right)^2 \\ & \sigma^2=43-36 \\ & \sigma^2=7\end{aligned}

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