Find variance and standard deviation of first 10 natural numbers

Question

Find the variance and standard deviation of the first 10 natural numbers: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10.

Solution — Step by Step

The first 10 natural numbers are: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10.

\bar{x} = \frac{\sum x_i}{n} = \frac{1+2+3+\cdots+10}{10} = \frac{n(n+1)/2}{n} = \frac{10 \times 11}{2 \times 10} = \frac{55}{10} = 5.5

We need $\sum x_i^2 = 1^2 + 2^2 + 3^2 + \cdots + 10^2$ .

Using the formula: $\sum_{i=1}^{n} i^2 = \dfrac{n(n+1)(2n+1)}{6}$

\sum x_i^2 = \frac{10 \times 11 \times 21}{6} = \frac{2310}{6} = 385

The computational formula for variance is:

\sigma^2 = \frac{\sum x_i^2}{n} - \bar{x}^2

\sigma^2 = \frac{385}{10} - (5.5)^2 = 38.5 - 30.25 = \mathbf{8.25}

\sigma = \sqrt{\sigma^2} = \sqrt{8.25} = \sqrt{\frac{33}{4}} = \frac{\sqrt{33}}{2}

\sigma = \frac{5.745\ldots}{2} \approx \mathbf{2.87}

Variance = 8.25, Standard Deviation = $\dfrac{\sqrt{33}}{2} \approx 2.87$

Why This Works

Variance measures how spread out the data is from its mean. The formula $\sigma^2 = \frac{\sum x_i^2}{n} - \bar{x}^2$ comes from expanding the definitional formula $\sigma^2 = \frac{\sum (x_i - \bar{x})^2}{n}$ .

The computational form avoids calculating $(x_i - \bar{x})$ for each value individually — which is tedious. Instead, we compute the mean of squares minus the square of the mean. This is algebraically identical but computationally faster.

Standard deviation is just the positive square root of variance. It has the same units as the data, making it interpretable: our standard deviation of ~2.87 tells us the typical deviation from the mean of 5.5 is about 2.87 units.

Alternative Method — Direct Deviation Method

Calculate $(x_i - \bar{x})^2$ for each value:

$x_i$	$x_i - \bar{x}$	$(x_i - \bar{x})^2$
1	-4.5	20.25
2	-3.5	12.25
3	-2.5	6.25
4	-1.5	2.25
5	-0.5	0.25
6	0.5	0.25
7	1.5	2.25
8	2.5	6.25
9	3.5	12.25
10	4.5	20.25
Sum		82.5

\sigma^2 = \frac{82.5}{10} = 8.25 \quad \checkmark

Both methods agree. The computational method (Step 3) is faster for exams.

Common Mistake

Using $n-1$ instead of $n$ in the denominator. In CBSE/NCERT, population variance uses $n$ in the denominator: $\sigma^2 = \frac{\sum(x_i - \bar{x})^2}{n}$ . The formula with $n-1$ is sample variance (used in statistics courses). For board problems on the first $n$ natural numbers, always use $n$ .

For the first $n$ natural numbers, there’s a direct formula: variance $= \dfrac{n^2 - 1}{12}$ . For $n = 10$ : $\sigma^2 = \dfrac{100-1}{12} = \dfrac{99}{12} = 8.25$ ✓. Memorise this shortcut — it works for any arithmetic series with common difference 1 starting at 1.

Question

Solution — Step by Step

Why This Works

Alternative Method — Direct Deviation Method

Common Mistake

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