Question

Let the mean and variance of the frequency distribution

xi	2	4	6	8	10	12	14	16
fi	4	4	α	15	8	β	4	5

are 9 and 15.08 respectively, then the value of α²+β²-αβ is ____.

Answer 1

Correct Answer: 25

We use the given values and apply the formula for mean and variance to find:

 \[ N = \sum f_i = 40 + \alpha + \beta \] \[ \sum f_i x_i = 360 + 6\alpha + 12\beta \] \[ \sum f_i x_i^2 = 3904 + 36\alpha + 144\beta \] Mean (\(\overline{x}\)) is: \[ \overline{x} = \frac{\sum f_i x_i}{\sum f_i} = 9 \] From this, we have the equation: \[ 360 + 6\alpha + 12\beta = 9(40 + \alpha + \beta) \] Simplifying: \[ 360 + 6\alpha + 12\beta = 9(40 + \alpha + \beta) \quad \Rightarrow \quad 3\alpha = 3\beta \quad \Rightarrow \quad \alpha = \beta \] Next, we calculate the variance (\(\sigma^2\)): \[ \sigma^2 = \frac{\sum f_i x_i^2}{\sum f_i} - \left( \frac{\sum f_i x_i}{\sum f_i} \right)^2 \] Substituting the values: \[ \sigma^2 = \frac{3904 + 36\alpha + 144\beta}{40 + \alpha + \beta} - (9)^2 = 15.08 \] From this, we get: \[ 3904 + 36\alpha + 144\beta = (40 + \alpha + \beta)(9)^2 = 15.08 \] Solving this: \[ 3904 + 36\alpha + 144\beta = 360 + 180\alpha \] \[ 3904 + 36\alpha + 144\beta = 360 + 180\alpha \] Now solving for \(\alpha = 5\) and \(\beta = 2\), we find: \[ \alpha^2 + \beta^2 - \alpha\beta = 25 \]

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