Let the mean and standard deviation of marks of class A of 100 students be respectively 40 and \alpha( 0 ), and the mean and standard deviation of marks of class B of n students be respectively 55 and 30 -\alpha. If the mean and variance of the marks of the combined class of 100+\mathrm{n} studants

Question

Let the mean and standard deviation of marks of class A of 100 students be respectively 40 and \alpha(> 0 ), and the mean and standard deviation of marks of class B of n students be respectively 55 and 30 -\alpha. If the mean and variance of the marks of the combined class of 100+\mathrm{n} studants are respectively 50 and 350 , then the sum of variances of classes A and B is :

Accepted Answer

Answer: D. 500 \begin{array}{cll} \quad \mathbf{A} & \mathbf{B} & \mathbf{A}+\mathbf{B} \ \overline{\mathrm{x}}_{1}=40 & \overline{\mathrm{x}}_{2}=55 & \overline{\mathrm{x}}=50 \ \sigma_{1}=\alpha & \sigma_{2}=30-\alpha & \sigma^{2}=350 \ \mathrm{n}_{1}=100 & \mathrm{n}_{2}=\mathrm{n} & 100+\mathrm{n} \\ \overline{\mathrm{x}}=\frac{100 	imes 40+55 \mathrm{n}}{100+\mathrm{n}} & \end{array} \begin{aligned} & 5000+50 \mathrm{n}=4000+55 \mathrm{n} \\ & 1000=5 \mathrm{n} \\ & \mathrm{n}=200 \\ & \sigma_{1}^{2}=\frac{\sum \mathrm{x}_{\mathrm{i}}^{2}}{100}-40^{2} \\ & \sigma_{2}^{2}=\frac{\sum \mathrm{x}_{\mathrm{j}}^{2}}{100}-55^{2} \\ & 350=\sigma^{2}=\frac{\sum \mathrm{x}_{\mathrm{i}}{ }^{2}+\sum \mathrm{x}_{\mathrm{j}}^{2}}{300}-(\overline{\mathrm{x}})^{2} \\ & 350=\frac{\left(1600+\alpha^{2}ight) 	imes 100+\left[(30-\alpha)^{2}+3025ight] 	imes 200}{300}-(50)^{2} \\ & 2850 	imes 3=\alpha^{2}+2(30-\alpha)^{2}+1600+6050 \\ & 8550=\alpha^{2}+2(30-\alpha)^{2}+7650 \\ & \alpha^{2}+2(30-\alpha)^{2}=900 \\ & \alpha^{2}-40 \alpha+300=0 \\ & \alpha=10,30 \\ & \sigma_{1}^{2}+\sigma_{2}^{2}=10^{2}+20^{2}=500 \end{aligned}

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