JEE MAIN - Mathematics (2024 - 31st January Evening Shift - No. 20)
Let the mean and the variance of 6 observations $$a, b, 68,44,48,60$$ be $$55$$ and $$194$$, respectively. If $$a>b$$, then $$a+3 b$$ is
180
210
190
200
Explanation
$$\begin{aligned} & a, b, 68,44,48,60 \\ & \text { Mean }=55 \quad a>b \\ & \text { Variance }=194 \quad a+3 b \\ & \frac{a+b+68+44+48+60}{6}=55 \\ & \Rightarrow 220+a+b=330 \\ & \therefore a+b=110 \ldots . .(1) \end{aligned}$$
Also,
$$\begin{aligned} & \sum \frac{\left(x_i-\bar{x}\right)^2}{n}=194 \\ & \Rightarrow(a-55)^2+(b-55)^2+(68-55)^2+(44-55)^2 \\ & +(48-55)^2+(60-55)^2=194 \times 6 \\ & \Rightarrow(a-55)^2+(b-55)^2+169+121+49+25=1164 \\ & \Rightarrow(a-55)^2+(b-55)^2=1164-364=800 \\ & a^2+3025-110 a+b^2+3025-110 b=800 \\ & \Rightarrow a^2+b^2=800-6050+12100 \\ & a^2+b^2=6850 \ldots \ldots . .(2) \end{aligned}$$
Solve (1) & (2);
$$\begin{aligned} & a=75, b=35 \\ & \therefore a+3 b=75+3(35)=75+105=180 \end{aligned}$$
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