Let x₁, x₂, , x₁1} be the observations satisfying _{i=1}¹1} (x_i-4)=22 and _{i=1}¹1} (x_i-4)²=154. If the mean and variance of the observations are α and β, then the quadratic equation having the roots \;α/β and \;β/α is [AP EAPCET 23 May 2025 S2]

Correct Answer is : 15x2−34x+15=0\;15x² - 34x + 15 = 015x2−34x+15=0

Correct Answer is : 15x2−34x+15=0\;15x² - 34x + 15 = 015x2−34x+15=0

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Statistics

Section: Mathematics

Question : 2 of 16

Marks: +1, -0

Let

x_1, x_2, \dots, x_{11}

\sum\limits_{i=1}^{11} (x_i-4)=22

\sum\limits_{i=1}^{11} (x_i-4)^2=154

\alpha

\beta

\;\frac{\alpha}{\beta}

\;\frac{\beta}{\alpha}

Solution:

\;\sum\limits_{i=1}^{11} (x_i-4)=22

\;\Rightarrow \;\; \sum\limits_{i=1}^{11} x_i - 4 \times 11 = 22

\;\Rightarrow \;\; \sum\limits_{i=1}^{11} x_i = 66

So, mean

(\overline{x}) = \frac{66}{11} = 6

and

\sum\limits_{i=1}^{11} (x_i-4)^2 = 154

\because \;\; x_i - \overline{x} = x_i - 6

= (x_i-4) - 2

\therefore \;\; [x_i - \overline{x}]^2 = [ (x_i-4) - 2 ]^2

= (x_i-4)^2 - 4(x_i-4) + 4

So,

\sum\limits_{i=1}^{11} (x_i-\overline{x})^2 = \sum\limits_{i=1}^{11} (x-4)^2

-4 \sum\limits_{i=1}^{11} (x_i-4) + 11 \times 4

= 154 - 4 \times 22 + 44 = 110

\;\text{Variance}\; (\sigma) = \frac{1}{n} \sum (x_i-\overline{x})^2 \\ = \frac{1}{11} \times 110 = 10

Thus,

\alpha = 6, \beta = 10

Therefore,

\;\frac{\alpha}{\beta} + \;\frac{\beta}{\alpha} = \;\frac{6}{10} + \;\frac{10}{6} = \;\frac{3}{5} + \;\frac{5}{3}

= \frac{9+25}{15} = \frac{34}{15} = \;\text{Sum of roots}\;

and

\;\frac{\alpha}{\beta} \cdot \;\frac{\beta}{\alpha} = 1 =

product of roots

Hence, the quadratic is

\;x^2 - \left(\;\frac{34}{15}\right) x + 1 = 0

\Rightarrow 15x^2 - 34x + 15 = 0

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