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  • If the mean and variance of the frequency distribution. <img alt="\mathrm{x_{i}}" src="https:

If the mean and variance of the frequency distribution.

\mathrm{x_{i}} 2 4 6 8 10 12 14 16
\mathrm{f_{i}} 4 4 \alpha 15 8 \beta 4 5


are 9 and 15.08 respectively, then the value of \alpha ^{2}+\beta ^{2}-\alpha \beta is _____:

Option: 1

25


Option: 2

-


Option: 3

-


Option: 4

-


Answers (1)

best_answer
\mathrm{x_{i}} \mathrm{f_{i}} \mathrm{f_{i}x_{i}} \mathrm{f_{i}x_{i}^{2}}
2 4 8 16
4 4 16 64
6 \alpha 6\alpha 36\alpha
8 15 120 960
10 8 80 800
12 \beta 12\beta 144\beta
14 4 56 784
16 5 80 1280


\mathrm{N}=\sum \mathrm{f}_{\mathrm{i}}=40+\alpha+\beta \\
\sum \mathrm{f}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}=360+6 \alpha+12 \beta \\
\sum \mathrm{f}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}^2=3904+36 \alpha+144 \beta \\
\text { Mean }(\overline{\mathrm{x}})=\frac{\sum \mathrm{f}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}}{\sum \mathrm{f}_{\mathrm{i}}}=9 \\
\Rightarrow 360+6 \alpha+12 \beta=9(40+\alpha+\beta) \\
3 \alpha=3 \beta \Rightarrow \alpha=\beta \\
\sigma^2=\frac{\sum \mathrm{f}_{\mathrm{i}} \mathrm{x}_1^2}{\sum \mathrm{f}_{\mathrm{i}}}-\left(\frac{\sum \mathrm{f}_{\mathrm{i}} \mathrm{x}_{\mathrm{i}}}{\sum \mathrm{f}_{\mathrm{i}}}\right)^2 \\
\Rightarrow \frac{3904+36 \alpha+144 \beta}{40+\alpha+\beta}-(\bar{x})^2=15.08 \\
\Rightarrow \frac{3904+180 \alpha}{40+2 \alpha}-(9)^2=15.08 \\
\Rightarrow \alpha=5 \\
\text { Now, } \alpha^2+\beta^2-\alpha \beta=\alpha^2=25

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