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

<p>$$\begin{aligned} & a, b, 68,44,48,60 \\ & \text { Mean }=55 \quad a>b \\ & \text { Variance }=194 \quad a+3 b \\ & \frac{a+b+68+44+48+60}{6}=55 \\ & \Rightarrow 220+a+b=330 \\ & \therefore a+b=110 \ldots . .(1) \end{aligned}$$</p> <p>Also,</p> <p>$$\begin{aligned} & \sum \frac{\left(x_i-\bar{x}\right)^2}{n}=194 \\ & \Rightarrow(a-55)^2+(b-55)^2+(68-55)^2+(44-55)^2 \\ & +(48-55)^2+(60-55)^2=194 \times 6 \\ & \Rightarrow(a-55)^2+(b-55)^2+169+121+49+25=1164 \\ & \Rightarrow(a-55)^2+(b-55)^2=1164-364=800 \\ & a^2+3025-110 a+b^2+3025-110 b=800 \\ & \Rightarrow a^2+b^2=800-6050+12100 \\ & a^2+b^2=6850 \ldots \ldots . .(2) \end{aligned}$$</p> <p>Solve (1) & (2);</p> <p>$$\begin{aligned} & a=75, b=35 \\ & \therefore a+3 b=75+3(35)=75+105=180 \end{aligned}$$</p>