Solution
[t]{|l|l|l|l|l|}
class |
$x_{i}$ |
$f_{i}$ |
$x_{i} f_{i}$ |
$f_{i} x_{i}^{2}$ |
| 4-8 |
6 |
3 |
18 |
$18 \times 6$ |
| 8-12 |
10 |
$\lambda$ |
$10 \lambda$ |
$100 \lambda$ |
| 12-16 |
14 |
4 |
56 |
14 × 56 |
| 16-20 |
18 |
$\frac{7}{14+\lambda}$ |
126 |
18 × 126 |
\[
\begin{aligned}
& \mu=\frac{\sum f_{i} x_{i}}{\sum f_{i}}=\frac{18+10 \lambda+56+126}{14+\lambda} \\
& \Rightarrow \mu=\frac{200+10 \lambda}{14+\lambda}=\frac{10(\lambda+14)+60}{14+\lambda} \\
& \Rightarrow \mu=10+\frac{60}{14+\lambda} ; \lambda \in \mathrm{N} \\
& \lambda=1, \mu=10+4=14 ;
\end{aligned}
\]
\[
\begin{aligned}
& \lambda=6, \mu=10+3=13 \\
& \lambda=16, \mu=10+2=12 \\
& \lambda=46, \mu=10+1=11
\end{aligned}
\]
\[
\begin{aligned}
& \text { Variance }=\frac{\sum x_{i}^{2} f_{i}}{\sum f_{i}}(\text { Mean })^{2}=19 \\
& \Rightarrow \frac{108+100 \lambda+14 \times 56+18 \times 126}{14+\lambda}-\mu^{2}=19 \\
& \frac{108+100 \lambda+560+4 \times 56+1260+8 \times 126}{14+\lambda}-\left(10+\frac{60}{14+\lambda}\right)^{2}=19 \\
& \Rightarrow \frac{3160+100 \lambda}{14+\lambda}-\left(10+\frac{60}{14+\lambda}\right)^{2}=19
\end{aligned}
\]
we get $\lambda=6, \mu=10+3=13$
\[
\therefore \lambda+\mu=19 \text { Answer (4) }
\]
.
