Elementary Statistics

The arithmetic mean (simply "the mean", written x̄) is the Average you already know: add all the observations and divide by how many there are.

x̄ = Σxn

Frequency table (discrete data). If a value x occurs f times, it contributes f × x to the sum. So multiply each value by its frequency, add the products, and divide by the total frequency.

x̄ = ΣfxΣf

In a quiz, 4 students scored 2 marks, 5 scored 3 marks and 1 scored 8 marks. Σf = 10 and Σfx = 8 + 15 + 8 = 31, so the mean is 3.1 marks.

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Grouped data. Take the class mark of each class as its x and use the same formula, x̄ = ΣfxΣf. This is the direct method.

Assumed mean method. When the class marks are large, the products f × x become heavy. Pick a class mark from the middle as the assumed mean A, find each deviation d = x − A, and use

x̄ = A + ΣfdΣf

Why does this work? Every x is A + d, so the mean is A plus the mean of the d's. The answer is the same; only the numbers are smaller.

Step-deviation method. If all classes have the same width h, divide the deviations by h as well: u = x − Ah, and x̄ = A + ΣfuΣf × h. Now u becomes −2, −1, 0, 1, 2 …, and the calculation is almost mental.

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Two properties you must know.

• The sum of the deviations of all observations from their mean is always zero: Σ(x − x̄) = 0. The values above the mean exactly balance the values below it.
• If a few values were read wrongly, correct the sum, not the mean: correct sum = wrong sum − wrong values + correct values. Then divide by n again.

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Weighted and combined mean. If n₁ items have mean x̄₁ and n₂ items have mean x̄₂, the combined mean is n₁x̄₁ + n₂x̄₂n₁ + n₂. A frequency table is really a weighted mean, with the frequencies acting as the weights.

When the data comes as a bar graph or a pie chart. Nothing new is needed, only a careful first step: turn the picture back into a frequency table. In a bar graph, the value is written under each bar and the height of the bar is its frequency. In a pie chart, each sector gives the share of a value. If the shares are percentages, the mean = Σ(x × percent)100; if they are central angles, the mean = Σ(x × angle)360. The total number of people is not needed at all, because the shares act as the weights. Say a pie chart shows that 25% of Radhika's classmates walk 1 km to school, 50% walk 2 km and 25% walk 4 km. The mean distance = 25 × 1 + 50 × 2 + 25 × 4100 = 225100 = 2.25 km. Read the picture into a table first; then it is the same Σfx ÷ Σf.

The median is the middle value of the data after it is arranged in ascending (or descending) order. Half the observations lie below it and half lie above it. The mean depends on the size of every value; the median depends only on position.

Raw data. Step 1: arrange in order. Step 2: count n.

• If n is odd, median = the n + 12th observation.
• If n is even, median = the average of the n2th and (n2 + 1)th observations.

Take Radhika's marks in seven tests: 12, 7, 15, 9, 18, 11, 9. In order they are 7, 9, 9, 11, 12, 15, 18. n = 7 is odd, so the median is the 4th value, 11. If an eighth test with 20 marks is added, n = 8 and the median = 11 + 122 = 11.5.

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Frequency table (discrete data). Write the values in ascending order, make the cumulative frequency (cf) column, and find the position as above using N = Σf. The median is the first value whose cf reaches that position.

Grouped data. Find N2. The class in which the cf first reaches or crosses N2 is the median class. Then

Median = l + N2 − cff × h

where l = lower limit of the median class, cf = cumulative frequency of the class before it, f = frequency of the median class and h = class width. The reasoning is simple: we need (N2 − cf) more observations from inside the median class, and we assume its f values are spread evenly over its width h.

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Unknowns in ordered data. When the data is already written in ascending order with x in it, write the middle term (or terms) in terms of x and equate it to the given median. Then check that your x keeps the data in order.

Median of an AP. For values in arithmetic progression (such as x + 2, x + 4, x + 6, …), the median equals the mean, because the values sit symmetrically around the centre.

The mode is the value that occurs the maximum number of times. For raw data no formula is needed; you only count.

Radhika noted the shoe sizes of her friends: 5, 6, 6, 7, 6, 5, 8. Size 6 appears three times, more than any other, so the mode is 6. A shopkeeper needs exactly this number, because it tells him which size to stock the most.

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• A data set can have two modes (bimodal) if two values share the highest frequency, or no mode if every value occurs equally often. SSC questions are usually set so that there is one clear mode.
• In a frequency table, the mode is the value with the highest frequency. Don't confuse it with the largest value.
• Finding k from a given mode. If the mode is given as 13, then 13 must occur more times than any other value. Count how often each value already appears, and choose k so that the unknown term becomes 13 (or makes 13 the clear winner).

Mode of grouped data. The class with the highest frequency is the modal class. Then

where l = lower limit of the modal class, f₁ = its frequency, f₀ = frequency of the class before it, f₂ = frequency of the class after it and h = class width. The mode is pulled towards whichever neighbouring class is heavier.

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Range is the simplest measure of spread:

Range = Largest value − Smallest value

For grouped data, range = upper limit of the last class − lower limit of the first class. The coefficient of range is H − LH + L. It has no unit, so it can compare two data sets measured in different units.

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Many questions combine these, for example find 2 × median + mode − range or the mean of the median and the mode. There is nothing new in them. Sort once, and read the median, mode and range from the same sorted list.

In a perfectly symmetrical distribution (a neat bell shape), the mean, median and mode all sit at the same point: Mean = Median = Mode.

Most real data is a little lopsided, because a few very high (or very low) values pull the mean to one side. Such data is called skewed. Karl Pearson observed that for a moderately skewed (moderately asymmetrical) distribution the three measures keep a steady pattern. The median lies between the mean and the mode, and the gap between the mean and the mode is about three times the gap between the mean and the median. As a formula:

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Mode ≈ 3 Median − 2 Mean

This is called the empirical relation. Empirical means it comes from observation, not from a proof, so it is an approximation. In SSC questions, however, calculate with it as if it were exact (and read the answer as an estimate) whenever the question says moderately skewed or moderately asymmetrical, or gives two of the three measures and asks for the third.

Rearrange it as needed:

• Mean = 3 Median − Mode2
• Median = Mode + 2 Mean3
• Mean − Mode = 3(Mean − Median)

In Radhika's class the mean height is 140 cm and the median is 138 cm. Then mode = 3 × 138 − 2 × 140 = 414 − 280 = 134 cm. Look at the order: mean 140 > median 138 > mode 134. This is the pattern of a positively skewed distribution, where a few tall students pull the mean up. For negative skew the order reverses: mean < median < mode.

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Ratio questions. If the ratio of two measures is given, write them as, say, 7k and 6k, find the third in terms of k with the formula, and then form the required ratio. The k cancels.

Range uses only two values. To measure spread properly, we look at how far every value is from the centre. That distance is the deviation: d = x − x̄.

But we already know that Σ(x − x̄) = 0, because the positive and negative deviations cancel. So we need a way to stop the cancelling. There are two ways, and they give the two important measures.

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1. Ignore the sign: Mean Deviation (MD).
MD about the mean = Σ|x − x̄|n, and MD about the median = Σ|x − M|n.
The coefficient of mean deviation = MDmean (or MDmedian, if the MD was taken about the median). It has no unit.

2. Square the deviations: Variance and Standard Deviation (SD).
Variance σ² = Σ(x − x̄)²n, and standard deviation σ = √variance.
A shorter form of the same thing is σ² = Σx²n − x̄², that is, the mean of the squares minus the square of the mean.

Take Radhika's numbers 2, 4, 6, 8, 10. Mean = 6, and the deviations are −4, −2, 0, 2, 4.

• MD = 4 + 2 + 0 + 2 + 45 = 125 = 2.4
• Squares: 16, 4, 0, 4, 16, with sum 40. Variance = 405 = 8, and SD = √8 ≈ 2.83

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Population or sample. The formula above (divide by n) gives the population variance. When the question says sample variance or sample standard deviation, divide by (n − 1) instead: s² = Σ(x − x̄)²n − 1. For the five numbers above, the sample variance = 404 = 10. Look for the word sample in the question; it decides whether you divide by n or by n − 1.

Coefficient of Variation (CV) = σx̄ × 100. It expresses the SD as a percentage of the mean, so it can compare the spread of data sets that have different means, or even different units. A lower CV means more consistent data.

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Facts to remember.

• SD is never negative, and SD = 0 only when all the values are equal.
• Σ(x − x̄) = 0, and Σ(x − A)² is the least when A is the mean.
• Mean deviation is the least when taken about the median.
• The variance of the first n natural numbers is n² − 112. For n terms of an AP with common difference d, variance = n² − 112 × d².

Working back from the mean and the SD. Many questions do not give the data at all. They give n, the mean and the SD, then change something: one value was read wrongly, or a new value joins the set. Recover the two sums first:

• Σx = n × x̄
• Σx² = n(σ² + x̄²), which is just σ² = Σx²n − x̄² turned round.

Correct both sums (remove the wrong value and add the right one, in Σx and in Σx²), then find the new mean and the new variance from the corrected sums. Take Radhika's 8 readings with mean 5 and SD 2. Then Σx = 40 and Σx² = 8 × (4 + 25) = 232. Never correct the SD directly; always correct Σx and Σx².

This type has no new formula, only one clear idea. Some measures tell you where the data is, and some tell you how spread out it is. When you shift every value by the same amount, the data moves but its spread does not change. When you stretch every value, both change.

Adding or subtracting a constant k to every value (change of origin).

• Mean, median, mode, quartiles: all increase (or decrease) by k.
• Range, mean deviation, quartile deviation, SD, variance: no change.

Picture the whole class getting 5 grace marks. Everyone moves up by 5, so the average moves up by 5. The gap between the topper and the last student stays the same, so the spread is unchanged.

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Multiplying every value by k (change of scale).

• Mean, median, mode, quartiles: all get multiplied by k.
• Range, mean deviation, QD, SD: get multiplied by |k|.
• Variance: gets multiplied by k².

Dividing by k is the same as multiplying by 1k.

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Both together: y = ax + b. New mean = a × old mean + b, and new SD = |a| × old SD. The b has no effect on the SD. New variance = a² × old variance.

Say Radhika's five test scores have mean 12 and SD 3. If the teacher adds 2 marks to every score, the new mean is 14 and the SD stays 3. If instead every score is doubled, the new mean is 24, the SD is 6, and the variance becomes 36 (it was 9).

When only some values change. The mean changes whenever the total changes, because it uses every value. The median, however, depends only on the middle position. If the values that change stay on the same side of the middle (for example, the largest few values become even larger), the median does not change at all.

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This type takes the idea of the median one step further.

The median cuts the sorted data into two halves. Quartiles cut it into four equal parts:

• Q₁ (lower quartile): the n + 14th value. A quarter of the data lies below it.
• Q₂ = the median: the n + 12th value.
• Q₃ (upper quartile): the 3(n + 1)4th value. Three quarters of the data lie below it.

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In the same way, percentiles cut the data into 100 parts: P_k = the k(n + 1)100th value. (Deciles cut it into 10 parts: D_k = the k(n + 1)10th value.) Textbooks use slightly different conventions for quartiles, but SSC answers follow this (n + 1) method.

When the position is not a whole number, go that fraction of the way between the two neighbouring values. Position 2.75 means 2nd value + 0.75 × (3rd value − 2nd value).

Take Radhika's data 4, 7, 9, 10, 12, 15, 18 (n = 7). Q₁ = the 84 = 2nd value = 7, and Q₃ = the 6th value = 15. If the data were 4, 7, 9, 10, 12, 15, 18, 21 (n = 8), Q₁ would be the 2.25th value = 7 + 0.25 × (9 − 7) = 7.5.

Measures built on quartiles.

• Inter-quartile range (IQR) = Q₃ − Q₁
• Quartile deviation (QD, the semi-inter-quartile range) = Q₃ − Q₁2
• Coefficient of QD = Q₃ − Q₁Q₃ + Q₁

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These measures ignore the lowest and the highest quarters, so a single extreme value (like 2 in a list of numbers near 50) does not disturb them.

Skewness tells you which way the data leans. Karl Pearson's coefficient of skewness is

Sk = Mean − ModeSD, or, when the mode is not clear, using the empirical relation, Sk = 3(Mean − Median)SD

• Sk = 0: symmetrical (mean = mode).
• Sk > 0: positively skewed (mean > mode, with a long tail of high values).
• Sk < 0: negatively skewed (mean < mode).

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Whenever the mean equals the mode, the coefficient of skewness is 0, and you don't even need the SD.

Look, every graph in this type is only a frequency table drawn as a picture. Your first job is always the same: read the picture back into a table. After that you already know the mean, median and mode formulas from the earlier types.

Histogram. Bars stand on the class intervals with no gaps between them, and the height of a bar is the frequency of that class (when all classes have the same width). The tallest bar is the modal class. The mode can even be found on the graph itself: join the top-left corner of the tallest bar to the top-left corner of the next bar, and its top-right corner to the top-right corner of the previous bar. The two lines cross right above the mode.

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In the figure, the modal class is 20–30 with f₁ = 18, f₀ = 10 and f₂ = 12. The formula gives 20 + 18 − 1036 − 10 − 12 × 10 = 20 + 8014 ≈ 25.7, the same point where the red lines cross.

Frequency polygon. Mark the point (class mark, frequency) for every class and join the points with straight lines. To close the shape, the polygon comes down to zero at the class marks of an empty class just before the first and just after the last. So the corners of the polygon give the values x and f directly: x̄ = ΣfxΣf with x = the class marks you read.

Cumulative frequency (cf). The less-than cf of a class is the number of observations below its upper limit; it keeps rising and ends at N. The more-than cf is the number of observations at or above a lower limit; it starts at N and falls to 0. To get back the frequency of a class, subtract two neighbouring cf values: f = cf(upper) − cf(lower) for a less-than table, f = cf(this) − cf(next) for a more-than table.

Ogive. Plot the cumulative frequencies against the class limits and join the points: the less-than ogive uses the upper limits and rises, the more-than ogive uses the lower limits and falls. The two curves cross at the point where cf = N2, so the x-value of the point where the two ogives meet is the median, and its y-value is N2.

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Here N = 40. The rising curve and the falling curve meet at cf = 20, and below that point the x-axis reads 25. The formula agrees: median = 20 + 20 − 1412 × 10 = 25. In the same way, Q₁ is read at cf = N4 and Q₃ at cf = 3N4 on the less-than ogive.