If the mean and variance of the frequency distribution x_i 2 4 6 8 10 12 14 16 f_i 4 4 15 …

Question

If the mean and variance of the frequency distribution

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

${f}_{i}$
 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 \_\_\_\_\_\_\_\_\_\_\_.

Answer: 25

Accepted Answer

$\begin{gathered} {x}_{i} \quad  {f}_{i} \quad  {x}_{2}^{i} \quad  {f}_{i}{x}_{i} \quad  {f}_{i}{x}_{2}^{i} \ 2 \quad  4 \quad  4 \quad  8 \quad  16 \ 4 \quad  4 \quad  16 \quad  16 \quad  64 \ 6 \quad  \alpha  \quad  36 \quad  6\alpha  \quad  36\alpha \ 8 \quad  15 \quad  64 \quad  120 \quad  960 \ 10 \quad  8 \quad  100 \quad  80 \quad  800 \ 12 \quad  \beta  \quad  144 \quad  12\beta  \quad  144\beta \ 14 \quad  4 \quad  196 \quad  56 \quad  784 \ 16 \quad  5 \quad  256 \quad  80 \quad  1280 \end{gathered}$

$	herefore $ $\begin{gathered} \Sigma {f}_{i}=40+\alpha +\beta \end{gathered}$

$\begin{gathered} \Sigma {f}_{i}{x}_{i}=360+6\alpha +12\beta \end{gathered}$

$\begin{gathered} \Sigma {f}_{i}{x}_{2}^{i}=3904+36\alpha +144\beta \end{gathered}$

Given, mean $=9$

$\begin{gathered} \Rightarrow \frac{\Sigma {f}_{i}{x}_{i}}{\Sigma {f}_{i}}=9 \ \Rightarrow \frac{360+6\alpha +12\beta }{40+\alpha +\beta }=9 \ \Rightarrow 360+6\alpha +12\beta =9(40+\alpha +\beta ) \ \Rightarrow 3\beta =3\alpha \ \Rightarrow \alpha =\beta .........(i) \end{gathered}$

$\begin{gathered} =15.08 \ \Rightarrow \frac{\Sigma {f}_{i}{x}_{2}^{i}}{\Sigma {f}_{i}}-{(\frac{\Sigma {f}_{i}{x}_{i}}{\Sigma {f}_{i}})}^{2}=15.08 \ \Rightarrow \frac{3904+36\alpha +144\beta }{40+\alpha +\beta }-(9{)}^{2}=15.08 \ \Rightarrow \frac{3904+180\alpha }{40+2\alpha }=81+15.08\, [\because \alpha =\beta ] \ \Rightarrow 3904+180\alpha =96.08(40+2\alpha ) \ \Rightarrow 3904+180\alpha =3843.2+192.16\alpha \ \Rightarrow 60.8=12.16\alpha \ \Rightarrow \alpha =5=\beta \ 	herefore {\alpha }^{2}+{\beta }^{2}-\alpha \beta =25+25-25=25 \end{gathered}$

Solutions · 1