Q44 / 75 · JEE Main 2025 (3 Apr Shift 2) · Mathematics

Let the Mean and Variance of five observations $x_1=1, x_2=3, x_3=a, x_4=7$ and $x_5=b, a \gt b$, be 5 and 10 respectively. Then the Variance of the observations $n+x_n, n=1,2, \ldots \ldots . .5$ is

  • A
    $17$
  • B
    $16.4$
  • C
    $17.4$
  • Correct answer
    D
    $16$

Solution

$\begin{aligned} & \bar{x}=\frac{\sum x_i}{n}=\frac{1+3+a+7+b}{5}=5 \\ & a+b=14 \\ & \sigma^2=\frac{\sum x_i^2}{n}-(\bar{x})^2 \\ & \Rightarrow \frac{1^2+3^2+a^2+7^2+b^2}{5}-25=10 \\ & a^2+b^2=116 \\ & a \gt b \quad a=10 \quad b=4 \\ & n+x_n: 2,5,13,11,9\end{aligned}$
$\begin{aligned} & \sigma^2=\frac{2^2+5^2+13^2+11^2+9^2}{5}-\left(\frac{2+5+13+11+9}{5}\right)^2 \\ & =80-64=16\end{aligned}$
option 4

Hence, the correct answer is option: 'D'.

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