Statistics Question 2
Question 2 - 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)
$$\begin{aligned} & \sum _{i=1}^{n} x _i=10 n \\ & \sum _{i=1}^{n} x _i-8+12=(10.2) n \quad \therefore n=20 \\ & \text { Now } \frac{\sum _{i=1}^{20} x _i{ }^2}{20}-(10)^2=4 \\ & \Rightarrow \sum _{i=1}^{20} x _i{ }^2=2080 \\ & \frac{\sum _{i=1}^{20} x _i{ }^2-8^2+12^2}{20}-(10.2)^2 \\ & =108-104.04=3.96 \end{aligned} $$
