If the mean and the variance of the data Class, 4--8, 8--12, 12--16, 16--20 | ---|---|---|…

Question

If the mean and the variance of the data
 Class, 4--8, 8--12, 12--16, 16--20 |
---|---|---|---|---|
 Frequency, 3, $\lambda$, 4, 7 |

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

Accepted Answer

Using class marks $x_i = 6, 10, 14, 18$:

$$\sum f_i = 14+\lambda, \qquad \sum f_ix_i = 18+10\lambda+56+126 = 200+10\lambda$$
$$\mu = \frac{200+10\lambda}{14+\lambda}$$

$$\sum f_ix_i^2 = 3(36)+\lambda(100)+4(196)+7(324) = 108+100\lambda+784+2268 = 3160+100\lambda$$

Variance: $\dfrac{3160+100\lambda}{14+\lambda} - \mu^2 = 19$

$$19(14+\lambda)^2 = (3160+100\lambda)(14+\lambda) - (200+10\lambda)^2$$
$$19(196+28\lambda+\lambda^2) = 44240+4560\lambda+100\lambda^2-40000-4000\lambda-100\lambda^2$$
$$3724+532\lambda+19\lambda^2 = 4240+560\lambda$$
$$19\lambda^2 - 28\lambda - 516 = 0$$

By quadratic formula: $\lambda = \dfrac{28\pm\sqrt{784+4(19)(516)}}{38} = \dfrac{28\pm200}{38}$. Taking the positive value, $\lambda = \dfrac{228}{38} = 6$.

Then $\mu = \dfrac{200+60}{20} = \dfrac{260}{20} = 13$.

$$\lambda + \mu = 6+13 = 19$$

Solutions · 1