Let the mean and the variance of seven observ | vedclass

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(B) Given observations: $2, 4, \alpha, 8, \beta, 12, 14$. Number of observations $n = 7$.Mean $\bar{x} = \frac{2+4+\alpha+8+\beta+12+14}{7} = 8$.$40 +

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$x^2 - 41x + 420 = 0$

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(B) Given observations: $2, 4, \alpha, 8, \beta, 12, 14$. Number of observations $n = 7$.Mean $\bar{x} = \frac{2+4+\alpha+8+\beta+12+14}{7} = 8$.$40 + \alpha + \beta = 56 \Rightarrow \alpha + \beta = 16$ (Equation $1$).Variance $\sigma^2 = \frac{\sum x_i^2}{n} - (\bar{x})^2 = 16$.$\frac{4 + 16 + \alpha^2 + 64 + \beta^2 + 144 + 196}{7} - 8^2 = 16$.$\frac{424 + \alpha^2 + \beta^2}{7} = 16 + 64 = 80$.$424 + \alpha^2 + \beta^2 = 560 \Rightarrow \alpha^2 + \beta^2 = 136$ (Equation $2$).Using $(\alpha + \beta)^2 = \alpha^2 + \beta^2 + 2\alpha\beta$, we get $16^2 = 136 + 2\alpha\beta$.$256 - 136 = 2\alpha\beta \Rightarrow 2\alpha\beta = 120 \Rightarrow \alpha\beta = 60$.Since $\alpha + \beta = 16$ and $\alpha\beta = 60$, the values are $\alpha = 6$ and $\beta = 10$ (as $\alpha The roots of the required quadratic equation are $3\alpha + 2 = 3(6) + 2 = 20$ and $2\beta + 1 = 2(10) + 1 = 21$.Sum of roots $= 20 + 21 = 41$.Product of roots $= 20 	imes 21 = 420$.The quadratic equation is $x^2 - (	ext{sum of roots})x + (	ext{product of roots}) = 0$, which is $x^2 - 41x + 420 = 0$.

Let the mean and the variance of seven observations $2, 4, \alpha, 8, \beta, 12, 14$ (where $\alpha < \beta$) be $8$ and $16$ respectively. Then the quadratic equation whose roots are $3\alpha + 2$ and $2\beta + 1$ is:

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