Statistics

Written and reviewed by Devendra Dubey, Mathematics faculty at UniDrill (M.Sc. Mathematics, Osmania University · 10 years teaching Mathematics). Based on NCERT Mathematics Class XI, Chapter 13.

Two batsmen, one average

Two batsmen, A and B, both average 53 runs. One of them scored 0, 100, 10, 120 and 35 in five innings, while the other scored 52, 54, 53, 53 and 53. Same average, but you would want the second one in a final when you need a steady score. The mean and the median tell you where the data is centred. They say nothing about how scattered it is. This chapter gives you numbers that measure the scatter.

Measures of dispersion

A measure of dispersion is a single number that says how far the data is spread around a centre. Four are named: range, quartile deviation, mean deviation and standard deviation. The chapter works with range, mean deviation and standard deviation.

The simplest is the range, the largest value minus the smallest.

$$\text{Range} = \text{Maximum} - \text{Minimum}$$

It is easy but fragile, because it depends only on the two extreme values and ignores everything in between. Take 12, 15, 18, 20. The range is $20 - 12 = 8$. Add one wild value, 100, and the range jumps to $88$ even though four of the five observations did not move.

Deviations, and why we take their modulus

A natural idea is to measure how far each observation $x_i$ is from the mean $\bar{x}$ and average those deviations $x_i - \bar{x}$. It fails for a simple reason: the deviations cancel.

$$\sum (x_i - \bar{x}) = \sum x_i - n\bar{x} = 0$$

The mean is exactly the balance point of the data, so the pulls above and below cancel. To stop the cancelling we must make every deviation positive. Taking the absolute value gives the mean deviation. Squaring them gives the variance.

Pause and predict: for the data 2, 4, 9 the mean is 5. What is $\sum (x_i - 5)$?

Check your answer: $(-3) + (-1) + 4 = 0$. Whatever the data, this sum is $0$ when taken about the mean. That is why a plain average of deviations is useless and why the modulus is compulsory.

Mean deviation

For $n$ observations and a central value $a$ (the mean or the median $M$), the mean deviation is the average distance of the observations from $a$.

$$\text{M.D.}(a) = \dfrac{1}{n}\sum \lvert x_i - a \rvert$$

When the data comes with frequencies $f_i$ and $N = \sum f_i$ is the total frequency,

$$\text{M.D.}(a) = \dfrac{1}{N}\sum f_i\,\lvert x_i - a \rvert$$

For a grouped (continuous) table, each class is replaced by its mid-point, the average of its two limits, and the same formula is used.

Worked example: mean deviation about the mean. Find the mean deviation about the mean of 6, 7, 10, 12, 13, 4, 8, 12.

Common wrong step. Subtracting the mean without taking the modulus. The deviations add to $0$ and the "answer" is $0$, which NTA likes to put in the options.

Worked example: mean deviation about the median. Find the mean deviation about the median of 3, 9, 5, 3, 12, 10, 18, 4, 7, 19, 21.

Common wrong step. Taking the middle term of the data as it was written, which is $10$, without sorting. The median needs ordered data. For an even number of terms it is the average of the two middle terms.

For a grouped table, the median is found from the cumulative frequency. The median class is the class whose cumulative frequency is the first to reach or pass $\dfrac{N}{2}$, and

$$M = l + \dfrac{\dfrac{N}{2} - C}{f}\times h$$

where $l$, $f$ and $h$ are the lower limit, frequency and width of the median class and $C$ is the cumulative frequency of the class just before it.

Beyond the textbook. For the exam: mean deviation is calculated about either the mean or the median, and the value about the median is never larger. In reality: the two centres each win a different contest. The sum $\sum \lvert x_i - a \rvert$ is smallest when $a$ is the median, and the sum $\sum (x_i - a)^2$ is smallest when $a$ is the mean. Why it matters: it explains why the mean is tied to variance and the median to mean deviation, and why the median resists outliers while the mean does not.

Variance and standard deviation

The modulus is hard to work with algebraically because it is not smooth at zero. Squaring is smoother and gives more weight to the far observations. The mean of the squared deviations is the variance $\sigma^2$, and its positive square root is the standard deviation $\sigma$.

$$\sigma^2 = \dfrac{1}{n}\sum (x_i - \bar{x})^2, \qquad \sigma = \sqrt{\dfrac{1}{n}\sum (x_i - \bar{x})^2}$$

We divide by $n$ so that the measure does not grow just because there are more data points. We take the square root so that the answer is in the same unit as the data. Variance is in squared units, such as cm squared.

For a frequency distribution, replace $n$ by $N = \sum f_i$ and weight each term by $f_i$.

$$\sigma^2 = \dfrac{1}{N}\sum f_i (x_i - \bar{x})^2$$

A shortcut that avoids the mean. Expand $\sum f_i(x_i - \bar{x})^2 = \sum f_i x_i^2 - 2\bar{x}\sum f_i x_i + N\bar{x}^2 = \sum f_i x_i^2 - N\bar{x}^2$. Divide by $N$ and use $\bar{x} = \dfrac{\sum f_i x_i}{N}$ to get

$$\sigma^2 = \dfrac{\sum f_i x_i^2}{N} - \bar{x}^2, \qquad \sigma = \dfrac{1}{N}\sqrt{N\sum f_i x_i^2 - \left(\sum f_i x_i\right)^2}$$

Worked example: variance of raw data. Find the variance and SD of 6, 8, 10, 12, 14, 16, 18, 20, 22, 24.

As a check with the shortcut, $\sum x_i^2 = 2580$, so $\dfrac{2580}{10} - 15^2 = 258 - 225 = 33$.

Common wrong step. Stopping at $33$ and calling it the standard deviation. The variance is in squared units, and the SD is its square root.

Grouped data and the step deviation method

When the class mid-points are large, subtract an assumed mean $A$ and divide by the class width $h$ to get small numbers $y_i = \dfrac{x_i - A}{h}$. Then

$$\bar{x} = A + h\,\bar{y}, \qquad \sigma_x = h\,\sigma_y, \qquad \sigma_x = \dfrac{h}{N}\sqrt{N\sum f_i y_i^2 - \left(\sum f_i y_i\right)^2}$$

Worked example: a grouped table. Find the median, mean and SD of the data below. Take $A = 25$ and $h = 10$ with mid-points $5, 15, 25, 35, 45$.

Totals: $N = 50$, $\sum fy = 10$, $\sum fy^2 = 68$.

Common wrong step. Reporting $\sqrt{1.32} \approx 1.15$ as the SD, which leaves out the factor $h$. The step deviation scale must be multiplied back for the SD. For the mean, you also add $A$ back.

Pause and predict: every observation in a data set is increased by $10$. What happens to the mean and to the SD?

Check your answer: the mean increases by $10$, but the SD does not change. Every deviation $x_i - \bar{x}$ stays exactly the same because both $x_i$ and $\bar{x}$ moved together. Shifting a data set slides it along the number line without changing its spread.

What shifting and scaling do

Replace every $x_i$ by $y_i = kx_i + a$. The mean goes to $k\bar{x} + a$. Each deviation becomes $k(x_i - \bar{x})$ because the $a$ cancels.

Worked example: a linear change. The data $x$ has mean $20$ and SD $4$. Find the mean and SD of $2x + 5$.

Common wrong step. Computing the SD as $2 \times 4 + 5 = 13$. Adding a constant moves the data but does not spread it.

Correcting a wrong observation

If a value was copied wrongly, rebuild the two sums, $\sum x_i$ for the mean and $\sum x_i^2$ for the variance, and recompute. Variance cannot be fixed by an adjustment of the old SD, because it depends on the squares.

Worked example: a wrong entry. The mean and SD of $100$ observations were found to be $40$ and $5.1$. Later it was found that $50$ had been used instead of $40$. Find the correct mean and SD.

Common wrong step. Correcting $\sum x_i$ but leaving $\sum x_i^2$ alone. The squares of the wrong and the right value are different numbers, so they must be swapped too.

Comparing two data sets: coefficient of variation

If two sets have the same mean, the one with the smaller SD is more consistent. When the means or units differ, SD alone is unfair, so we compare the SD as a percentage of the mean.

$$\text{C.V.} = \dfrac{\sigma}{\bar{x}} \times 100$$

The C.V. has no unit. The set with the smaller C.V. is more consistent. For example, a series with mean $50$ and SD $5$ has C.V. $10\%$, and one with mean $20$ and SD $3$ has C.V. $15\%$. The second series has the smaller SD but the larger relative spread, so the first is the more consistent.

Beyond the textbook. For the exam: variance divides by $n$. In reality: when the data is a sample drawn from a larger group and you want to estimate the spread of the whole group, statisticians divide by $n - 1$ instead. Calculators therefore have two SD keys, one for each. Why it matters: if a question gives a full list and asks for its own variance, divide by $n$; the second version belongs to later study.

Beyond the textbook. For the exam: SD is the square root of the mean of squared deviations. In reality: SD is also a ruler. For data with a bell shaped distribution, about 68 percent of values lie within one SD of the mean and about 95 percent within two. So a value more than two SDs above the mean is rare: only about 2.5 percent of values lie there. Why it matters: it shows that an SD is not just a number from a formula; it tells you how unusual a value is.

Exam traps

Formula sheet

60-second revision

Where this chapter connects

• Before: Limits and Derivatives. Both chapters ask what a quantity is heading towards, one for functions and one for data.
• Next in Class XI: Probability turns the counts behind a table of frequencies into chances.
• Class XII: Probability later uses mean and variance for the outcomes of random experiments, the same two ideas with probabilities in place of frequencies.