Question

If the mean and the variance of the data

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

• A. 21
• B. 18
• C. 19
• D. 20

Hint:

Variance is independent of change of origin but depends on the scale. $\text{Var}(ax+b) = a^2 \text{Var}(x)$.

Answer 1

The Correct Option is C

Let the mid-points ($x_i$) of the classes be 6, 10, 14, 18.
The total frequency is $N = 3 + \lambda + 4 + 7 = 14 + \lambda$.
We use the step-deviation method with assumed mean $A=14$ and height $h=4$. Let $u_i = \frac{x_i - 14}{4}$.
The values for $u_i$ are $-2, -1, 0, 1$.
The mean of $u$ is $\bar{u} = \frac{\sum f_i u_i}{N} = \frac{3(-2) + \lambda(-1) + 4(0) + 7(1)}{14+\lambda} = \frac{1-\lambda}{14+\lambda}$.
The variance relation is $\sigma_x^2 = h^2 \sigma_u^2 = 16 \left[ \frac{\sum f_i u_i^2}{N} - (\bar{u})^2 \right] = 19$.
Calculate $\sum f_i u_i^2 = 3(4) + \lambda(1) + 4(0) + 7(1) = 19 + \lambda$.
Substitute into the variance equation:
$19 = 16 \left[ \frac{19+\lambda}{14+\lambda} - \left(\frac{1-\lambda}{14+\lambda}\right)^2 \right]$.
$\frac{19}{16} = \frac{(19+\lambda)(14+\lambda) - (1-\lambda)^2}{(14+\lambda)^2}$.
$19(14+\lambda)^2 = 16 [ (266 + 33\lambda + \lambda^2) - (1 - 2\lambda + \lambda^2) ]$.
$19(196 + 28\lambda + \lambda^2) = 16(265 + 35\lambda)$.
$3724 + 532\lambda + 19\lambda^2 = 4240 + 560\lambda$.
$19\lambda^2 - 28\lambda - 516 = 0$.
Solving for $\lambda$: $\lambda = \frac{28 \pm \sqrt{784 - 4(19)(-516)}}{38} = \frac{28 \pm 200}{38}$.
Since frequency is positive, $\lambda = \frac{228}{38} = 6$.
Now find $\mu$: $\bar{u} = \frac{1-6}{20} = -\frac{1}{4}$.
$\mu = 14 + 4(-\frac{1}{4}) = 13$.
Therefore, $\lambda + \mu = 6 + 13 = 19$.