Question

The standard deviation of 1, 2, 3, ..., 100 is:

• A. $\frac{1}{2}\sqrt{333}$
• B. $\frac{1}{4}\sqrt{333}$
• C. $\frac{1}{6}\sqrt{333}$
• D. $\frac{1}{8}\sqrt{333}$
• E. $\frac{1}{4}\sqrt{1111}$

Hint:

The variance of first $n$ natural numbers is $\frac{n^2-1}{12}$; the standard deviation is its square root.

Answer 1

The Correct Option is A

Step 1: Understanding the Concept:
We need to find the standard deviation of the first 100 natural numbers. Standard deviation is a measure of the amount of variation or dispersion of a set of values.
Step 2: Key Formula or Approach:
The standard deviation (\( \sigma \)) for the first n natural numbers is given by the formula for the variance (\( \sigma^2 \)) first:
\[ \sigma^2 = \frac{n^2 - 1}{12} \] The standard deviation is the square root of the variance:
\[ \sigma = \sqrt{\frac{n^2 - 1}{12}} \] This formula is derived from the standard variance formula \( \sigma^2 = \frac{\sum x_i^2}{n} - \left(\frac{\sum x_i}{n}\right)^2 \) and the formulas for the sum of the first n natural numbers and the sum of their squares.
Step 3: Detailed Explanation:
In this problem, we have the first 100 natural numbers, so \( n = 100 \).
We can directly use the shortcut formula for the variance:
\[ \sigma^2 = \frac{100^2 - 1}{12} \] \[ \sigma^2 = \frac{10000 - 1}{12} = \frac{9999}{12} \] Now, we simplify this fraction. Both numerator and denominator are divisible by 3.
\[ \sigma^2 = \frac{9999 \div 3}{12 \div 3} = \frac{3333}{4} \] The standard deviation is the square root of the variance:
\[ \sigma = \sqrt{\frac{3333}{4}} = \frac{\sqrt{3333}}{\sqrt{4}} = \frac{\sqrt{3333}}{2} \] This can be written as:
\[ \sigma = \frac{1}{2}\sqrt{3333} \] Step 4: Final Answer:
The standard deviation of the first 100 natural numbers is \( \frac{1}{2}\sqrt{3333} \).