Let the mean and the variance of 6 observations a, b, 68,44,48,60 be 55 and 194, respectiv…

Question

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

Accepted Answer

$\begin{gathered} a,b,68,44,48,60 \ =55\, a>b \ =194\, a+3b \ \frac{a+b+68+44+48+60}{6}=55 \ \Rightarrow 220+a+b=330 \ 	herefore a+b=110\ldots ..(1) \end{gathered}$

Also,

$\begin{gathered} \sum \frac{{({x}_{i}-\vec{x})}^{2}}{n}=194 \ \Rightarrow (a-55{)}^{2}+(b-55{)}^{2}+(68-55{)}^{2}+(44-55{)}^{2} \ +(48-55{)}^{2}+(60-55{)}^{2}=194	imes 6 \ \Rightarrow (a-55{)}^{2}+(b-55{)}^{2}+169+121+49+25=1164 \ \Rightarrow (a-55{)}^{2}+(b-55{)}^{2}=1164-364=800 \ {a}^{2}+3025-110a+{b}^{2}+3025-110b=800 \ \Rightarrow {a}^{2}+{b}^{2}=800-6050+12100 \ {a}^{2}+{b}^{2}=6850\ldots \ldots ..(2) \end{gathered}$

Solve (1) \& (2);

$\begin{gathered} a=75,b=35 \ 	herefore a+3b=75+3(35)=75+105=180 \end{gathered}$

Solutions · 1