Statistics Ans 2
Q2 - 25 January - Shift 1
The mean and variance of the marks obtained by the students in a test are 10 and 4 respectively. Later, the marks of one of the students is increased from 8 to 12. If the new mean of the marks is 10.2. then their new variance is equal to :
(1) 4.04
(2) 4.08
(3) 3.96
(4) 3.92
Show Answer
Answer: (3)
Solution:
Formula: Arithmetic mean of individual series (ungrouped data), Variance of individual observations (ungrouped data)
$$\begin{aligned} & \sum_{\mathrm{i}=1}^{\mathrm{n}} \mathrm{x}{\mathrm{i}}=10 \mathrm{n} \ & \sum{\mathrm{i}=1}^{\mathrm{n}} \mathrm{x}{\mathrm{i}}-8+12=(10.2) \mathrm{n} \quad \therefore \mathrm{n}=20 \ & \text { Now } \frac{\sum{\mathrm{i}=1}^{20} \mathrm{x}{\mathrm{i}}{ }^2}{20}-(10)^2=4 \ & \Rightarrow \sum{\mathrm{i}=1}^{20} \mathrm{x}{\mathrm{i}}{ }^2=2080 \ & \frac{\sum{\mathrm{i}=1}^{20} \mathrm{x}_{\mathrm{i}}{ }^2-8^2+12^2}{20}-(10.2)^2 \ & =108-104.04=3.96 \end{aligned} $$
