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JEE Main 20253 AprilEvening Shift
Exam
JEE Main
Year
2025
Date
3 April
Shift
Evening Shift
Subject
Mathematics
Chapter
Statistics

JEE Main 2025 · 3 April · Evening Shift · Mathematics · Statistics

Let the Mean and Variance of five observations x_1=1, x_2=3, x_3=a,...

Let the Mean and Variance of five observations x1=1,x2=3,x3=a,x4=7x_1=1, x_2=3, x_3=a, x_4=7 and x5=b,a>bx_5=b, a \gt b, be 5 and 10 respectively. Then the Variance of the observations n+xn,n=1,2,……..5n+x_n, n=1,2, \ldots \ldots . .5 is

Options

  1. A1717
  2. B16.416.4
  3. C17.417.4
  4. D1616

Correct answer

D. 1616

Step-by-step solution

xˉ=∑xin=1+3+a+7+b5=5a+b=14σ2=∑xi2n−(xˉ)2⇒12+32+a2+72+b25−25=10a2+b2=116a>ba=10b=4n+xn:2,5,13,11,9\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}
σ2=22+52+132+112+925−(2+5+13+11+95)2=80−64=16\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
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