Let the mean and standard deviation of marks | vedclass

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(A) Given for class $A$: $n_1 = 100, \overline{x}_1 = 40, \sigma_1 = \alpha$. Variance $\sigma_1^2 = \alpha^2$.Given for class $B$: $n_2 = n, \overlin

Vedclass · Accepted Answer

$500$

Vedclass · Answer

(A) Given for class $A$: $n_1 = 100, \overline{x}_1 = 40, \sigma_1 = \alpha$. Variance $\sigma_1^2 = \alpha^2$.Given for class $B$: $n_2 = n, \overline{x}_2 = 55, \sigma_2 = 30-\alpha$. Variance $\sigma_2^2 = (30-\alpha)^2$.Combined mean $\overline{x} = \frac{n_1\overline{x}_1 + n_2\overline{x}_2}{n_1+n_2} = 50$.$\frac{100(40) + n(55)}{100+n} = 50 \implies 4000 + 55n = 5000 + 50n \implies 5n = 1000 \implies n = 200$.Combined variance $\sigma^2 = \frac{n_1(\sigma_1^2 + d_1^2) + n_2(\sigma_2^2 + d_2^2)}{n_1+n_2}$,where $d_1 = \overline{x}_1 - \overline{x} = 40-50 = -10$ and $d_2 = \overline{x}_2 - \overline{x} = 55-50 = 5$.$350 = \frac{100(\alpha^2 + (-10)^2) + 200((30-\alpha)^2 + 5^2)}{100+200}$.$350 	imes 300 = 100(\alpha^2 + 100) + 200((30-\alpha)^2 + 25)$.$105000 = 100\alpha^2 + 10000 + 200(900 - 60\alpha + \alpha^2 + 25)$.$1050 = \alpha^2 + 100 + 2(925 - 60\alpha + \alpha^2) = \alpha^2 + 100 + 1850 - 120\alpha + 2\alpha^2$.$3\alpha^2 - 120\alpha + 900 = 0 \implies \alpha^2 - 40\alpha + 300 = 0$.$(\alpha - 10)(\alpha - 30) = 0 \implies \alpha = 10$ or $\alpha = 30$.If $\alpha = 30$,$\sigma_2 = 30-30 = 0$. If $\alpha = 10$,$\sigma_1^2 = 100, \sigma_2^2 = (30-10)^2 = 400$.Sum of variances = $\sigma_1^2 + \sigma_2^2 = 100 + 400 = 500$.

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