JEE Main 2024 — Statistics Question with Solution

$$\begin{array}{rl}& a,b,68,44,48,60\\ & \text{ Mean }=55a>b\\ & \text{ Variance }=194a+3b\\ & \frac{a+b+68+44+48+60}{6}=55\\ & \Rightarrow 220+a+b=330\\ & \therefore a+b=110\dots ..(1)\end{array}$$

Also,

$$\begin{array}{rl}& \sum \frac{{\left(x_i-\bar{x}\right)}^2}{n}=194\\ & \Rightarrow (a-55{)}^2+(b-55{)}^2+(68-55{)}^2+(44-55{)}^2\\ & +(48-55{)}^2+(60-55{)}^2=194\times 6\\ & \Rightarrow (a-55{)}^2+(b-55{)}^2+169+121+49+25=1164\\ & \Rightarrow (a-55{)}^2+(b-55{)}^2=1164-364=800\\ & a^2+3025-110a+b^2+3025-110b=800\\ & \Rightarrow a^2+b^2=800-6050+12100\\ & a^2+b^2=6850\dots \dots ..(2)\end{array}$$

Solve (1) & (2);

$$\begin{array}{rl}& a=75,b=35\\ & \therefore a+3b=75+3(35)=75+105=180\end{array}$$