Statistics Question 9
Question 9 - 2024 (31 Jan Shift 2)
Let the mean and the variance of 6 observation $\mathrm{a}, \mathrm{b}, 68,44,48,60$ be 55 and 194 , respectively if $a>b$, then $a+3 b$ is
(1) 200
(2) 190
(3) 180
(4) 210
Show Answer
Answer (3)
Solution
$a, b, 68,44,48,60$
Mean $=55$
Variance $=194 \quad a>b$
$\frac{a+b+68+44+48+60}{6}=55$
$\Rightarrow 220+a+b=330$
$\therefore a+b=110$.
Also,
$\sum \frac{\left(x_{i}-\bar{x}\right)^{2}}{n}=194$
$\Rightarrow(a-55)^{2}+(b-55)^{2}+(68-55)^{2}+(44-55)^{2}$
$+(48-55)^{2}+(60-55)^{2}=194 \times 6$
$\Rightarrow(a-55)^{2}+(b-55)^{2}+169+121+49+25=1164$
$\Rightarrow(a-55)^{2}+(b-55)^{2}=1164-364=800$
$a^{2}+3025-110 a+b^{2}+3025-110 b=800$
$\Rightarrow a^{2}+b^{2}=800-6050+12100$
$a^{2}+b^{2}=6850$
Solve (1) $\backslash &$ (2);
$a=75, b=35$
$\therefore a+3 b=75+3(35)=75+105=180$
