If the mean and variance of the frequency distribution are 9 and 15.08 respectively, then …

Question

If the mean and variance of the frequency distribution are 9 and 15.08 respectively, then the value of $\alpha^2+\beta^2-\alpha\beta$ is \_\_\_\_\_\_\_\_.

Answer: 25

Accepted Answer

Let $N=\sum f_i = 40+\alpha+\beta$, $\sum f_ix_i = 360+6\alpha+12\beta$, $\sum f_ix_i^2=3904+36\alpha+144\beta$.

Mean $=9$: $360+6\alpha+12\beta = 9(40+\alpha+\beta) \implies 3\alpha=3\beta \implies \alpha=\beta$

Variance $=15.08$: $\dfrac{3904+36\alpha+144\beta}{40+\alpha+\beta}-81=15.08$

With $\alpha=\beta$: $\dfrac{3904+180\alpha}{40+2\alpha}-81=15.08 \implies \alpha=5$

$$\alpha^2+\beta^2-\alpha\beta = \alpha^2 = 25$$

Solutions · 1