If the mean and the variance of the data are | vedclass

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(D) The class marks $(x_i)$ are $6, 10, 14, 18$. The frequencies $(f_i)$ are $3, \lambda, 4, 7$. The sum of frequencies is $N = 14+\lambda$.Mean $\mu

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$19$

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(D) The class marks $(x_i)$ are $6, 10, 14, 18$. The frequencies $(f_i)$ are $3, \lambda, 4, 7$. The sum of frequencies is $N = 14+\lambda$.Mean $\mu = \frac{\sum f_i x_i}{N} = \frac{3(6) + 10\lambda + 4(14) + 7(18)}{14+\lambda} = \frac{18 + 10\lambda + 56 + 126}{14+\lambda} = \frac{200+10\lambda}{14+\lambda}$.Variance $\sigma^2 = \frac{\sum f_i x_i^2}{N} - \mu^2 = 19$.Calculating $\sum f_i x_i^2 = 3(36) + 100\lambda + 4(196) + 7(324) = 108 + 100\lambda + 784 + 2268 = 3160 + 100\lambda$.So, $\frac{3160+100\lambda}{14+\lambda} - \left(\frac{200+10\lambda}{14+\lambda}\right)^2 = 19$.Solving for $\lambda$, we find $\lambda = 6$.Substituting $\lambda = 6$ into the mean formula: $\mu = \frac{200+60}{14+6} = \frac{260}{20} = 13$.Therefore, $\lambda + \mu = 6 + 13 = 19$.

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

Daily wage (Rs.)	$30$-$40$	$40$-$50$	$50$-$60$	$60$-$70$	$70$-$80$	$80$-$90$
No. of workers	$17$	$28$	$21$	$15$	$13$	$6$

If the mean and the variance of the data are $\mu$ and $19$ respectively, then the value of $\lambda+\mu$ is:
Class $4-8$ $8-12$ $12-16$ $16-20$
Frequency $3$ $\lambda$ $4$ $7$

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If the mean and the variance of the data are $\mu$ and $19$ respectively, then the value of $\lambda+\mu$ is:Class$4-8$$8-12$$12-16$$16-20$Frequency$3$$\lambda$$4$$7$