Variance, Standard Deviation and Moments | NIMCET

Quick answer: Variance measures spread around the mean: $\sigma^2 = \frac{\sum f_i(x_i - \bar{x})^2}{N} = \frac{\sum f_ix_i^2}{N} - \bar{x}^2$. The standard deviation is $\sigma = \sqrt{\sigma^2}$. Adding a constant to every value leaves σ unchanged, while multiplying by k multiplies σ by $\lvert k\rvert$. The coefficient of variation is $\frac{\sigma}{\bar{x}} \times 100$. The rth central moment is $\mu_r = \frac{\sum f_i(x_i - \bar{x})^r}{N}$, so $\mu_2 = \sigma^2$.

This continues from mean, median and mode.

Measures of dispersion

Mean deviation is least when taken about the median.

Shortcut with step deviations

With $u_i = \frac{x_i - A}{h}$:

$\sigma^2 = h^2\left[\frac{\sum f_iu_i^2}{N} - \left(\frac{\sum f_iu_i}{N}\right)^2\right]$

Effect of changes in data

Standard results

Combined variance of two groups with sizes $n_1, n_2$, means $\bar{x}_1, \bar{x}_2$, variances $\sigma_1^2, \sigma_2^2$ and combined mean $\bar{x}$:

$\sigma^2 = \frac{n_1(\sigma_1^2 + d_1^2) + n_2(\sigma_2^2 + d_2^2)}{n_1 + n_2}, \quad d_i = \bar{x}_i - \bar{x}$

Moments, skewness and kurtosis

Common coefficients: $\beta_1 = \frac{\mu_3^2}{\mu_2^3}$ and $\beta_2 = \frac{\mu_4}{\mu_2^2}$. A normal curve has $\beta_2 = 3$.

Karl Pearson's coefficient of skewness: $\frac{\text{Mean} - \text{Mode}}{\sigma}$. It is positive for right-skewed data and negative for left-skewed data.

Moments about an arbitrary point A ($\mu_r'$) convert to central moments by $\mu_2 = \mu_2' - \mu_1'^2$.

Worked examples

Example 1: raw data

Find the variance and SD of 2, 4, 6, 8, 10.

Step 1: Mean $= 6$.

Step 2: Squared deviations: 16, 4, 0, 4, 16, sum 40.

Step 3: Variance $= 8$, SD $= 2\sqrt{2}$.

Check: these are $2 \times (1, 2, 3, 4, 5)$, and the variance of 1 to 5 is $\frac{25 - 1}{12} = 2$; multiplied by $2^2$ gives 8 ✓.

Example 2: effect of change

The SD of a data set is 4. Each value is multiplied by 3 and then 5 is added. Find the new SD and variance.

New SD $= 12$, new variance $= 144$.

Example 3: coefficient of variation

Two batsmen have means 50 and 40 with SDs 10 and 6. Who is more consistent?

CV: $\frac{10}{50} \times 100 = 20$ and $\frac{6}{40} \times 100 = 15$. The second batsman is more consistent.

Example 4: moments

The first two moments about 4 are $\mu_1' = 1$ and $\mu_2' = 5$. Find the mean and variance.

Mean $= 4 + 1 = 5$, variance $= 5 - 1^2 = 4$.

Summary table

Exam tips

• Use $\frac{\sum x^2}{n} - \bar{x}^2$; it is faster than summing squared deviations.
• Lower CV means more consistent data.
• Learn $\frac{n^2 - 1}{12}$ for consecutive integers; it saves time.

For more practice, see NIMCET probability and statistics: distributions and standard deviation and the NIMCET Probability and Statistics chapter.

Common mistakes

• Thinking that adding a constant changes the SD.
• Multiplying the variance by k instead of $k^2$.
• Writing variance as $\frac{\sum x^2}{n} - \bar{x}$ (the mean must be squared).
• Comparing consistency with SD alone when the means differ.

Frequently Asked Questions

What is the formula for variance?

σ² = Σfᵢxᵢ²/N − x̄², the mean of the squares minus the square of the mean.

What happens to the standard deviation if a constant is added to every value?

It does not change.

What is the coefficient of variation?

CV = (σ/x̄) × 100. The smaller the CV, the more consistent the data.

What is the variance of the first n natural numbers?

(n² − 1)/12.

What is the second central moment?

It equals the variance.