If the mean and the variance of the data $ \begin{array}{|c|c|c|c|c|} \hline ext{Class} & 4 ext{-}8 & 8 ext{-}12 & 12 ext{-}16 & 16 ext{-}20 \ \hline ext{Frequency} & 3 & \lambda & 4 & 7 \ \hline \end{array}$ are $\mu$ and 19 respectively, then the value of $\lambda+\mu$ is

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If the mean and the variance of the data $ \begin{array}{|c|c|c|c|c|} \hline 	ext{Class} & 4	ext{-}8 & 8	ext{-}12 & 12	ext{-}16 & 16	ext{-}20 \ \hline 	ext{Frequency} & 3 & \lambda & 4 & 7 \ \hline \end{array}$ are $\mu$ and 19 respectively, then the value of $\lambda+\mu$ is

Aicharya · Accepted Answer

Class midpoints: $6, 10, 14, 18$.
Frequencies: $3, \lambda, 4, 7$. $N = 14 + \lambda$.
Mean $\mu = \frac{200 + 10\lambda}{14 + \lambda}$.
$\sum f_i x_i^2 = 3(36) + 100\lambda + 4(196) + 7(324) = 3160 + 100\lambda$.
Variance: $19 = \frac{3160+100\lambda}{N} - \mu^2$.
Let $u = 14+\lambda$:
$19u^2 = u(1760+100u) - (60+10u)^2 \Rightarrow 19u^2 - 560u + 3600 = 0$.
$u = \frac{560 \pm 200}{38}$, so $u = 20 \Rightarrow \lambda = 6$.
$\mu = \frac{260}{20} = 13$. Hence $\lambda + \mu = 6 + 13 = 19$. ^

If the mean and the variance of the data $ \begin{array}{|c|c|c|c|c|} \hline \text{Class} & 4\text{-}8 &…

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