Statistics Question 8
Question 8 - 31 January - Shift 2
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:
(1) 500
(2) 650
(3) 450
(4) 900
Show Answer
Answer: (1)
Solution:
Formula: Combined Arithmetic Mean, Variance of individual observations (ungrouped data), Roots of equations
$ \begin{aligned} & n_1=100, \overline{x_1}=40, \sigma_1=\alpha \\ & n_2=n, \overline{x_2}=55, \sigma_2=(30-\alpha) \\ & \overline{\left.x _{(\text {combined }}\right)}=50, \sigma _{\text {combined }}^2=350, \sigma_1^2+\sigma_2^2=? \\ & \sigma _{\text {combined }}^2=\frac{n_1 \sigma_1^2+n_2 \sigma_2^2}{n_1+n_2}+\frac{n_1 n_2}{\left(n_1+n_2\right)^2}\left(\overline{x_1}-\overline{x_2}\right)^2—-(1) \\ & \overline{x} _{\text {combined }}=\frac{n_1 \cdot \overline{x}_1+n_2 \cdot \overline{x}_2}{n_1+n_2} \\ & \bar{x}=\frac{100 \times 40+55 n}{100+n} \\ & 5000+50 n=4000+55 n \\ & 1000=5 n \\ & n=200 \end{aligned} $
So substitute in (1) we get $\alpha^2-40 \alpha+300=0$
$(\alpha-30)(\alpha-10)=0 $
$\alpha=10, \text { as }(30-\alpha>0)$
So sum of squares of variances
$ =10^2+20^2=500 $
