Let the mean and the variance of 6 observatio | vedclass

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(C) Given observations: $a, b, 68, 44, 48, 60$. Mean $\overline{x} = 55$,Variance $\sigma^2 = 194$. Sum of observations: $a + b + 68 + 44 + 48 + 60 =

Vedclass · Accepted Answer

$180$

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(C) Given observations: $a, b, 68, 44, 48, 60$. Mean $\overline{x} = 55$,Variance $\sigma^2 = 194$. Sum of observations: $a + b + 68 + 44 + 48 + 60 = 6 	imes 55 = 330$. $a + b + 220 = 330 \Rightarrow a + b = 110$ (Equation $1$). Variance formula: $\frac{1}{n} \sum (x_i - \overline{x})^2 = 194$. $(a - 55)^2 + (b - 55)^2 + (68 - 55)^2 + (44 - 55)^2 + (48 - 55)^2 + (60 - 55)^2 = 194 	imes 6$. $(a - 55)^2 + (b - 55)^2 + 13^2 + (-11)^2 + (-7)^2 + 5^2 = 1164$. $(a - 55)^2 + (b - 55)^2 + 169 + 121 + 49 + 25 = 1164$. $(a - 55)^2 + (b - 55)^2 = 1164 - 364 = 800$. $a^2 - 110a + 3025 + b^2 - 110b + 3025 = 800$. $a^2 + b^2 - 110(a + b) + 6050 = 800$. Substitute $a + b = 110$: $a^2 + b^2 - 110(110) + 6050 = 800$. $a^2 + b^2 - 12100 + 6050 = 800 \Rightarrow a^2 + b^2 = 6850$ (Equation $2$). From $(a + b)^2 = a^2 + b^2 + 2ab$,we have $110^2 = 6850 + 2ab$. $12100 - 6850 = 2ab$ $\Rightarrow 2ab = 5250$ $\Rightarrow ab = 2625$. Since $a + b = 110$ and $ab = 2625$,$a$ and $b$ are roots of $t^2 - 110t + 2625 = 0$. $(t - 75)(t - 35) = 0$. Since $a > b$,$a = 75$ and $b = 35$. Therefore,$a + 3b = 75 + 3(35) = 75 + 105 = 180$.

No. of goals scored	$0, 1, 2, 3, 4$
No. of matches	$1, 9, 7, 5, 3$

Let the mean and the variance of $6$ observations $a, b, 68, 44, 48, 60$ be $55$ and $194$,respectively. If $a > b$,then the value of $a + 3b$ is:

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