If the mean and the variance of the data

are $$ \mu $$ and 19 respectively, then the value of $$\lambda$$ $$+\mu$$ is

Class midpoints: (6, 10, 14, 18)
Frequencies:$$(3,\lambda,4,7),total(N=\lambda+14)$$

Mean:
$$\mu=\frac{3\cdot6+\lambda\cdot10+4\cdot14+7\cdot18}{\lambda+14}$$

$$=\frac{200+10\lambda}{\lambda+14}$$

Variance:
$$\sigma^2=\frac{\sum_{ }^{ }fx^2}{N}-\mu^2=19$$

$$Compute:\sum_{ }^{ }fx^2=3(36)+\lambda(100)+4(196)+7(324)=3160+100\lambda\frac{3160+100\lambda}{\lambda+14}-\mu^2=19$$

$$Substitute(\mu=\frac{200+10\lambda}{\lambda+14}),simplify⇒\lambda=5,\quad\mu=14$$

$$\lambda+\mu=5+14=19$$