Let x_1, x_2, ldots, x_{11} be the observatio | vedclass

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(B) Given $\sum_{i=1}^{11}(x_i-4)=22$. Dividing by $11$,we get $\bar{x}-4 = \frac{22}{11} = 2$,so $\bar{x} = \alpha = 6$. Now,the variance $\beta$ is

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$15 x^2-34 x+15=0$

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(B) Given $\sum_{i=1}^{11}(x_i-4)=22$. Dividing by $11$,we get $\bar{x}-4 = \frac{22}{11} = 2$,so $\bar{x} = \alpha = 6$. Now,the variance $\beta$ is given by $\beta = \frac{1}{n} \sum (x_i-\bar{x})^2$. Since $\bar{x}=6$,we have $x_i-\bar{x} = x_i-6 = (x_i-4)-2$. Thus,$\sum (x_i-6)^2 = \sum ((x_i-4)-2)^2 = \sum (x_i-4)^2 - 4\sum (x_i-4) + \sum 4$. Substituting the values: $154 - 4(22) + 11(4) = 154 - 88 + 44 = 110$. So,$\beta = \frac{110}{11} = 10$. The roots are $\frac{\alpha}{\beta} = \frac{6}{10} = \frac{3}{5}$ and $\frac{\beta}{\alpha} = \frac{10}{6} = \frac{5}{3}$. The quadratic equation is $x^2 - (	ext{sum of roots})x + (	ext{product of roots}) = 0$. Sum of roots $= \frac{3}{5} + \frac{5}{3} = \frac{9+25}{15} = \frac{34}{15}$. Product of roots $= \frac{3}{5} 	imes \frac{5}{3} = 1$. The equation is $x^2 - \frac{34}{15}x + 1 = 0$,which simplifies to $15x^2 - 34x + 15 = 0$.

Marks:	$2$	$3$	$5$	$7$
Frequency:	$(x+1)^2$	$2x-5$	$x^2-3x$	$x$

Let $x_1, x_2, \ldots, x_{11}$ be the observations satisfying $\sum_{i=1}^{11}(x_i-4)=22$ and $\sum_{i=1}^{11}(x_i-4)^2=154$. If the mean and variance of the observations are $\alpha$ and $\beta$,then the quadratic equation having the roots $\frac{\alpha}{\beta}$ and $\frac{\beta}{\alpha}$ is

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