Statistics Ans 8
Q8 - 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} & \mathrm{n}_1=100, \overline{\mathrm{x}1}=40, \sigma_1=\alpha \ & \mathrm{n}2=\mathrm{n}, \overline{\mathrm{x}2}=55, \sigma_2=(30-\alpha) \ & \overline{\left.\mathrm{x}{(\text {combined }}\right)}=50, \sigma{\text {combined }}^2=350, \sigma_1^2+\sigma_2^2=? \ & \sigma{\text {combined }}^2=\frac{\mathrm{n}_1 \sigma_1^2+\mathrm{n}_2 \sigma_2^2}{\mathrm{n}_1+\mathrm{n}_2}+\frac{\mathrm{n}_1 \mathrm{n}_2}{\left(\mathrm{n}_1+\mathrm{n}_2\right)^2}\left(\overline{\mathrm{x}_1}-\overline{\mathrm{x}2}\right)^2—-(1) \ & \overline{\mathrm{x}}{\text {combined }}=\frac{\mathrm{n}_1 \cdot \overline{\mathrm{x}}_1+\mathrm{n}_2 \cdot \overline{\mathrm{x}}_2}{\mathrm{n}_1+\mathrm{n}_2} \ & \bar{x}=\frac{100 \times 40+55 \mathrm{n}}{100+\mathrm{n}} \ & 5000+50 \mathrm{n}=4000+55 \mathrm{n} \ & 1000=5 \mathrm{n} \ & \mathrm{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 $
