If the mean and variance of the frequency distribution, are 9 and 15.08 respectively, then the value of is ________ . [2023]

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Solution

Q.
If the mean and variance of the frequency distribution

$x_{i}$ 2 4 6 8 10 12 14 16

$f_{i}$ 4 4 $α$ 15 8 $β$ 4 5

are 9 and 15.08 respectively, then the value of $α^{2} + β^{2} - α β$ is ________ . [2023]

Ans.
(25)

$x_{i}$ $f_{i}$ $x_{i} f_{i}$ $f_{i} x_{i}^{2}$

2 4 8 16

4 4 16 64

6 $α$ $6 α$ $36 α$

8 15 120 960

10 8 80 800

12 $β$ $12 β$ $144 β$

14 $4$ 56 784

16 5 80 1280

Total $40 + α + β$ $360 + 6 α + 12 β$ $3904 + 36 α + 144 β$

$Mean (\bar{x}) = \frac{\sum f_{i} x_{i}}{\sum f_{i}} \Rightarrow \frac{360 + 6 α + 12 β}{40 + α + β} = 9$

$\Rightarrow 360 + 6 α + 12 β = 360 + 9 α + 9 β$

$\Rightarrow 3 α = 3 β \Rightarrow α = β$ ... $(i)$

$Also, variance σ^{2} = \frac{\sum f_{i} x_{i}^{2}}{\sum f_{i}} - {(\frac{\sum f_{i} x_{i}}{\sum f_{i}})}^{2}$

$\Rightarrow \frac{3904 + 36 α + 144 β}{40 + α + β} - {(\bar{x})}^{2} = 15.08$

$\Rightarrow \frac{3904 + 180 α}{40 + 2 α} - {(9)}^{2} = 15.08$ $\Rightarrow α = 5$

From (i), $β = 5$

$∴$ $α^{2} + β^{2} - α β = 25 + 25 - 25 = 25$

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