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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 $100+ n$ students are respectively $50$ and $350$,then the sum of variances of classes $A$ and $B$ is
$500$
$650$
$450$
$900$
Solution
| $A$ | $B$ | $A+B$ |
| $\overline{ x }_1=40$ | $\overline{ x }_2=55$ | $\overline{ x }=50$ |
| $\sigma_1=\alpha$ | $\sigma_2=30-\alpha$ | $\sigma^2=350$ |
| $n _1=100$ | $n _2= n$ | $100+ n$ |
$\overline{ x }=\frac{100 \times 40+55 n }{100+ n }$
$5000+50 n =4000+55 n$
$1000=5 n$
$n =200$
$\sigma_1{ }^2=\frac{\sum x _{ i }^2}{100}-40^2$
$\sigma_2{ }^2=\frac{\sum x _{ j }^2}{100}-55^2$
$350=\sigma^2=\frac{\sum x _{ i }^2+\sum x _{ j }^2}{300}-(\overline{ x })^2$
$350=\frac{\left(1600+\alpha^2\right) \times 100+\left[(30-\alpha)^2+3025\right] \times 200}{300}-(50)^2$
$2850 \times 3=\alpha^2+2(30-\alpha)^2+1600+6050$
$8550=\alpha^2+2(30-\alpha)^2+7650$
$\alpha^2+2(30-\alpha)^2=900$
$\alpha^2-40 \alpha+300=0$
$\alpha=10,30$
$\sigma_1^2+\sigma_2^2=10^2+20^2=500$
