If the mean and the variance of the data Class : 4 – 8 , 8 – 12 , 12 – 16 , 16 – 20 Frequency : 3 , λ , 4 , 7 are μ and 19 respectively, then the value of λ + μ is :

If the mean and the variance of the data are μ and 19 respectively, then the value of λ + μ is

Q. If the mean and the variance of the data

Class	4 – 8	8 – 12	12 – 16	16 – 20
Frequency	3	$\lambda$	4	7

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

A. 21

B. 19

C. 18

D. 20

Correct Answer: 19

Explanation

Class intervals are of equal width, so we take class mid-points.

$$ x_i = 6,\;10,\;14,\;18 $$

Corresponding frequencies are:

$$ f_i = 3,\;\lambda,\;4,\;7 $$

Total frequency:

$$ N = 3+\lambda+4+7=\lambda+14 $$

Now calculate $\sum f_i x_i$.

$$ \sum f_i x_i = 3(6)+\lambda(10)+4(14)+7(18) $$

$$ =18+10\lambda+56+126 $$

$$ =200+10\lambda $$

Hence mean $\mu$ is:

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

Next, calculate $\sum f_i x_i^2$.

$$ \sum f_i x_i^2 = 3(36)+\lambda(100)+4(196)+7(324) $$

$$ =108+100\lambda+784+2268 $$

$$ =3160+100\lambda $$

Variance formula for grouped data is:

$$ \sigma^2=\frac{\sum f_i x_i^2}{N}-\mu^2 $$

Given variance $\sigma^2=19$, so:

$$ 19=\frac{3160+100\lambda}{\lambda+14}-\left(\frac{200+10\lambda}{\lambda+14}\right)^2 $$

Solving this equation gives:

$$ \lambda=6 $$

Substitute $\lambda=6$ in mean:

$$ \mu=\frac{200+60}{20}=13 $$

Therefore,

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

Hence, the required value is 19.