Match the List I with List II

List I (Population Mean ($$\mu$$) and  $$\frac{1}{N} \sum x_i^2$$)​		List II (Population Standard Deviation Mean ($$\sigma$$​))
A.	​$$\mu = 3, \quad \frac{1}{N} \sum x_i^2 = 90$$​​	I.	12
B.	​$$\mu = 4, \quad \frac{1}{N} \sum x_i^2 = 116$$​​	II.	11
C.	​$$\mu = 5, \quad \frac{1}{N} \sum x_i^2 = 146$$​​	III.	9
D.	​$$\mu = 6, \quad \frac{1}{N} \sum x_i^2 = 180$$​​	IV.	10

Correct option is C

The population standard deviation (σ) can be calculated using the following formula:

​$$\sigma = \sqrt{\frac{1}{N} \sum x_i^2 - \mu^2}$$

Where:

• $\mu$ is the population mean.
• ∑xi2\sum x_i^2$$\sum x_i^2$$​ is the sum of squares of the data points.
• $N$ is the number of data points (implicitly assumed in the formula).

A. $$\mu = 3, \quad \frac{1}{N} \sum x_i^2 = 90$$

$$\sigma = \sqrt{90 - 3^2} = \sqrt{90 - 9} = \sqrt{81} = 9$$​​​

B. $$\mu = 4, \quad \frac{1}{N} \sum x_i^2 = 116$$

$$\sigma = \sqrt{116 - 4^2} = \sqrt{116 - 16} = \sqrt{100} = 10$$​​

C. $$\mu = 5, \quad \frac{1}{N} \sum x_i^2 = 146$$

$$\sigma = \sqrt{146 - 5^2} = \sqrt{146 - 25} = \sqrt{121} = 11$$​​

D. $$\mu = 6, \quad \frac{1}{N} \sum x_i^2 = 180$$

$$\sigma = \sqrt{180 - 6^2} = \sqrt{180 - 36} = \sqrt{144} = 12$$​​