JEE Main 2025 (Online) 3rd April Evening ShiftQuestion 1
Let the Mean and Variance of five observations $x_1=1$, $x_2=3$, $x_3=a$, $x_4=7$ and $x_5=b$, $a>b$, be $5$ and $10$ respectively. Then the Variance of the observations $n+x_n$, $n=1,2,\ldots,5$ is
Statistics
Solution
$1+3+a+7+b=25\Rightarrow a+b=14$. $\sum x^2=5(10+25)=175\Rightarrow a^2+b^2=175-59=116$. $ab=\frac{196-116}{2}=40$: $a=10,b=4$. New data $n+x_n$: $2,5,13,11,9$. Mean $8$, $\sum sq=4+25+169+121+81=400$, variance $=80-64=16$. Option A.