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

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 + 3b is

Accepted Answer

Mean $=55$: $\dfrac{a+b+68+44+48+60}{6}=55 \implies a+b+220=330 \implies a+b=110$

Variance $=194$: $\dfrac{a^2+b^2+68^2+44^2+48^2+60^2}{6}-55^2=194$
$$a^2+b^2+4624+1936+2304+3600 = 6(194+3025)=19314$$
$a^2+b^2 = 19314-12464=6850$

$(a+b)^2=12100=6850+2ab \implies ab=2625$

$a,b$ are roots of $t^2-110t+2625=0$. Discriminant $=12100-10500=1600$, $\sqrt{}=40$.
$$t = \frac{110\pm40}{2} = 75 \text{ or } 35$$

Since $a>b$: $a=75, b=35$.

$a+3b = 75+105=180$

Solutions · 1