[shahkb4]

If the mean and the variance of the data

[t]{|l|l|l|l|l|} 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
[JEE MAIN 23 Jan 2026 S II]
(1) 18
(2) 21
(3) 20
(4) 19

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[shahkb4]

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 $\lambda=6, \mu=10+3=13$

\[ \therefore \lambda+\mu=19 \text { Answer (4) } \]

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