​Match the LIST-I with LIST-II

LIST-I Population mean$$(\mu)$$ and $$\left({1}{N} \right) \sum X_i^2$$​​	LIST-II Population standard deviation$$(\sigma)$$​​
A. $$\mu = 5,\ \left({1}/{N} \right) \sum X_i^2 = 61$$​​	I. 7
B. $$\mu = 6,\ \left({1}/{N} \right) \sum X_i^2 = 85$$​​	II. 9
C. $$\mu = 7,\ \left({1}/{N} \right) \sum X_i^2 = 113$$​​	III. 8
D. $$\mu = 8,\ \left( \frac{1}{N} \right)\sum X_i^2 = 145$$​​	IV. 6

​Choose the correct answer from the options given below:​

Correct option is D

Solution: 

Population standard deviation: $$\sigma = \sqrt{ \left( \frac{1}{N} \sum X_i^2 \right) - \mu^2 }$$

A. $$\mu = 5,\ \frac{1}{N} \sum X_i^2 = 61\\[1em]\sigma = \sqrt{61 - 5^2} = \sqrt{61 - 25} = \sqrt{36} = 6 \Rightarrow \text{IV}$$​​

B. $$\mu = 6,\ \frac{1}{N} \sum X_i^2 = 85 \\\sigma = \sqrt{85 - 6^2} = \sqrt{85 - 36} = \sqrt{49} = 7 \Rightarrow \text{I}$$​​

C. $$\mu = 7,\ \frac{1}{N} \sum X_i^2 = 113 \\\sigma = \sqrt{113 - 7^2} = \sqrt{113 - 49} = \sqrt{64} = 8 \Rightarrow \text{III}$$​​

D. $$\mu = 8,\ \frac{1}{N} \sum X_i^2 = 145 \\\sigma = \sqrt{145 - 8^2} = \sqrt{145 - 64} = \sqrt{81} = 9 \Rightarrow \text{II}$$

$$\boxed{\text{Final Match: } A \rightarrow \text{IV},\ B \rightarrow \text{I},\ C \rightarrow \text{III},\ D \rightarrow \text{II}}$$​​​

​Final Answer:
Option D: A–IV, B–I, C–III, D–II