Chapter 13 — Statistics

When a large number of observations is collected, such as the marks of 200 students or the daily wages of 500 workers, it is convenient to group them into class intervals of equal width, and to record only how many observations fall in each class. The result is a grouped frequency distribution. For example, marks out of 100 may be grouped as 0–10, 10–20, …, 90–100, with the number of students in each class as its frequency f.

In the exclusive form of classes such as 10–20, 20–30, the upper limit of one class is the lower limit of the next, and an observation equal to the common value, say 20, is put in the higher class 20–30. The lower limit and upper limit of a class are its two ends, and the class size or width h is their difference, 10 here.

Once the data are grouped, the individual values are lost. To compute anything we assume that all the observations in a class are concentrated at its middle value, the class mark:

Class mark x_i = (upper limit + lower limit)/2.

The class mark of 10–20 is 15, of 20–30 is 25, and so on. The class marks of consecutive classes of equal width differ by h.

Some data are given in the inclusive form, such as 1–10, 11–20, 21–30, where both limits belong to the class and there is a gap of 1 between the upper limit of one class and the lower limit of the next. For the median and mode formulas, and for ogives, such classes must first be converted to exclusive form by subtracting half the gap from each lower limit and adding it to each upper limit: 1–10 becomes 0.5–10.5, 11–20 becomes 10.5–20.5, and so on. The class marks do not change, so the mean can be found without conversion.

The total frequency is n = Σf_i, the sum of all the frequencies, where Σ (sigma) denotes summation. The symbols used throughout the chapter are: x_i the class mark of the i-th class, f_i its frequency, n = Σf_i, h the class size, and a an assumed mean.

Worked example. The classes 0–20, 20–40, 40–60, 60–80, 80–100 have class marks 10, 30, 50, 70, 90, width h = 20. The inclusive classes 11–20, 21–30, 31–40 become 10.5–20.5, 20.5–30.5, 30.5–40.5 with class marks 15.5, 25.5, 35.5 and width 10.

Data may also be given as a less-than or more-than cumulative table, which must be converted back to ordinary frequencies before the mean or mode is found; this is discussed with the median.

The mean of ungrouped observations x₁, x₂, …, x_n is their sum divided by n. If the value x_i occurs f_i times, the sum is Σf_ix_i and the number of observations is Σf_i, so

Mean x̄ = Σf_ix_i / Σf_i.

For grouped data we take x_i to be the class mark of the i-th class, on the assumption that the f_i observations in that class are all at its centre. This is the direct method.

Steps. Write the class marks x_i; multiply each by its frequency to get f_ix_i; add the f_i column and the f_ix_i column; divide.

Worked example. The marks of 30 students are distributed as follows.

Σf = 30, Σfx = 1860. Mean = 1860/30 = 62.

Worked example. Daily wages of 50 workers.

Σfx = 27260, n = 50, mean = ₹545.20.

The mean obtained from grouped data is only approximate, since the actual values are replaced by class marks; the finer the grouping, the closer it is to the true mean of the raw data. In the examination, however, the grouped mean is the answer wanted.

The direct method is straightforward but the products f_ix_i can be large and tedious. The next two methods reduce the size of the numbers without changing the result. A common one-mark question asks which method is easiest; the answer depends on the data, and all three are acceptable unless a specific method is demanded.

Finding a missing frequency. If the mean is given and one frequency is unknown, write the mean formula with the unknown and solve. For example, if classes 0–20, 20–40, 40–60, 60–80 have frequencies 7, p, 10, 9 and the mean is 40, then Σfx = 70 + 30p + 500 + 630 = 1200 + 30p and n = 26 + p; (1200 + 30p)/(26 + p) = 40 gives 1200 + 30p = 1040 + 40p, so 10p = 160 and p = 16.

The products f_ix_i are large because the x_i are large. If we subtract a fixed number a from every class mark, the products become smaller, and the mean of the original data is recovered by adding a back. The number a is the assumed mean; any value can be used, but choosing one of the class marks, usually the middle one, makes the deviations small and symmetric.

Let d_i = x_i − a be the deviation of the i-th class mark from a. Then x_i = a + d_i, so Σf_ix_i = Σf_i(a + d_i) = aΣf_i + Σf_id_i. Dividing by n = Σf_i,

x̄ = a + Σf_id_i / Σf_i.

This is the assumed mean method. The mean of the deviations, d̄ = Σf_id_i/n, is added to the assumed mean.

Steps. Choose a; compute d_i = x_i − a (some negative, some positive); compute f_id_i; add; divide by n; add a.

Worked example. Same marks data with a = 47.5.

Σfd = 435, n = 30, mean = 47.5 + 435/30 = 47.5 + 14.5 = 62, the same as by the direct method.

Worked example. Number of plants in 20 houses: classes 0–2, 2–4, 4–6, 6–8, 8–10, 10–12, 12–14 with frequencies 1, 2, 1, 5, 6, 2, 3. Class marks 1, 3, 5, 7, 9, 11, 13. With a = 7, d = −6, −4, −2, 0, 2, 4, 6 and fd = −6, −8, −2, 0, 12, 8, 18, Σfd = 22, mean = 7 + 22/20 = 8.1 plants.

Why any a works. The derivation used only algebra, not a special choice of a. A different a changes every d_i by the same amount and changes Σf_id_i/n by exactly the opposite amount, so a + d̄ is unchanged. Choosing a near the centre merely keeps the numbers small and the negative and positive deviations balanced.

Worked example. Daily expenditure on food of 25 households: classes 100–150, 150–200, 200–250, 250–300, 300–350 with frequencies 4, 5, 12, 2, 2. Class marks 125, 175, 225, 275, 325; a = 225; d = −100, −50, 0, 50, 100; fd = −400, −250, 0, 100, 200; Σfd = −350; mean = 225 − 350/25 = 225 − 14 = ₹211.

The assumed mean method is the method of choice when the class marks are large but the class size is not a convenient divisor; when it is, the step-deviation method goes one step further.

When the classes have equal width h, all the deviations d_i = x_i − a are multiples of h, since the class marks are equally spaced. Dividing by h gives small integers u_i = (x_i − a)/h, usually …, −2, −1, 0, 1, 2, …, and the arithmetic becomes trivial.

Since d_i = hu_i, Σf_id_i = hΣf_iu_i, and the assumed mean formula becomes

x̄ = a + h × (Σf_iu_i / Σf_i), with u_i = (x_i − a)/h.

This is the step-deviation method. The quantity Σf_iu_i/n is the mean of the u-values, ū, so x̄ = a + hū.

Steps. Choose a (a central class mark); compute u_i = (x_i − a)/h; compute f_iu_i; add; divide by n; multiply by h; add a.

Worked example. The marks data again, with a = 47.5 and h = 15.

Σfu = 29, n = 30, mean = 47.5 + 15 × 29/30 = 47.5 + 14.5 = 62. All three methods agree.

Worked example. Percentage of female teachers in 35 cities: classes 15–25, 25–35, …, 75–85 with frequencies 6, 11, 7, 4, 4, 2, 1. Class marks 20, 30, 40, 50, 60, 70, 80; a = 50, h = 10; u = −3, −2, −1, 0, 1, 2, 3; fu = −18, −22, −7, 0, 4, 4, 3; Σfu = −36; mean = 50 + 10 × (−36/35) = 50 − 10.29 = 39.71.

Worked example. Wickets taken by 45 bowlers: classes 20–60, 60–100, 100–150, 150–250, 250–350, 350–450 with frequencies 7, 5, 16, 12, 2, 3. The classes are unequal, but the step-deviation method still works with any convenient h; take a = 200 and h = 20: class marks 40, 80, 125, 200, 300, 400; u = −8, −6, −3.75, 0, 5, 10; fu = −56, −30, −60, 0, 10, 30; Σfu = −106; mean = 200 + 20 × (−106/45) = 200 − 47.11 = 152.89 wickets.

Comparison of the three methods. All give the same value because each is an algebraic rearrangement of Σfx/n. The direct method needs no choice; the assumed mean method reduces the size of the numbers; the step-deviation method reduces them further when a common factor h is available. If the class marks are small or the frequencies are small, the direct method is quickest; if the class marks are large and equally spaced, the step-deviation method is quickest.

Remember to multiply Σfu/n by h before adding a; forgetting the factor h is the most frequent error in this method.

The mode of ungrouped data is the value that occurs most often. For grouped data the individual values are unknown, so we can only say which class has the greatest frequency, the modal class, and then estimate where within it the mode lies.

The estimate uses the frequencies of the classes on either side. If the class before the modal class has a larger frequency than the class after it, the mode is pulled towards the lower end of the modal class, and vice versa. The formula that expresses this is

Mode = l + [(f₁ − f₀) / (2f₁ − f₀ − f₂)] × h,

where l = lower limit of the modal class, h = its class size, f₁ = frequency of the modal class, f₀ = frequency of the class preceding it, f₂ = frequency of the class succeeding it.

The fraction (f₁ − f₀)/[(f₁ − f₀) + (f₁ − f₂)] lies between 0 and 1, so the mode lies inside the modal class. If f₀ = f₂ the mode is at the centre; if f₀ = f₁ the mode is at the lower limit.

Worked example. Family sizes of 20 families: classes 1–3, 3–5, 5–7, 7–9, 9–11 with frequencies 7, 8, 2, 2, 1. The modal class is 3–5 (largest frequency 8). l = 3, h = 2, f₁ = 8, f₀ = 7, f₂ = 2. Mode = 3 + [(8 − 7)/(16 − 7 − 2)] × 2 = 3 + (1/7) × 2 = 3.286. The most common family size is about 3.3. The mean of the same data is 4.4, larger because the few large families pull it up.

Worked example. Marks of 30 students: classes 10–25, 25–40, 40–55, 55–70, 70–85, 85–100 with frequencies 2, 3, 7, 6, 6, 6. Modal class 40–55; l = 40, h = 15, f₁ = 7, f₀ = 3, f₂ = 6. Mode = 40 + [(7 − 3)/(14 − 3 − 6)] × 15 = 40 + (4/5) × 15 = 52. The mean was 62. The mode tells us the marks most students scored near; the mean gives the average.

Worked example. Lifetimes of 225 electrical components: classes 0–20, 20–40, 40–60, 60–80, 80–100, 100–120 with frequencies 10, 35, 52, 61, 38, 29. Modal class 60–80; l = 60, h = 20, f₁ = 61, f₀ = 52, f₂ = 38. Mode = 60 + [9/(122 − 52 − 38)] × 20 = 60 + (9/32) × 20 = 60 + 5.625 = 65.625 hours.

Worked example. Runs scored by top batsmen in one-day cricket: classes 3000–4000, 4000–5000, …, 10000–11000 with frequencies 4, 18, 9, 7, 6, 3, 1, 1. Modal class 4000–5000; mode = 4000 + [(18 − 4)/(36 − 4 − 9)] × 1000 = 4000 + (14/23) × 1000 = 4608.7 runs.

When the modal class is the first class, f₀ = 0; when it is the last, f₂ = 0. If two non-adjacent classes share the highest frequency, the distribution is bimodal and the formula is applied to each. Inclusive classes must be converted to exclusive form before l and h are read.

The median of ungrouped data is the middle value when the observations are arranged in order: for n odd it is the (n + 1)/2-th observation; for n even it is the average of the n/2-th and (n/2 + 1)-th. For grouped data we locate the class in which the middle observation falls, and to do this we need the cumulative frequency.

The cumulative frequency (cf) of a class is the number of observations less than its upper limit, obtained by adding the frequencies of all classes up to and including it. The cf of the last class is n. A table of classes with their cumulative frequencies is a cumulative frequency distribution of the less-than type.

Worked example. Marks of 53 students.

Here n = 53 and n/2 = 26.5. The median class is the class whose cumulative frequency is the first to reach or exceed n/2: the cf 22 of 50–60 is less than 26.5, and the cf 29 of 60–70 is greater, so the median class is 60–70. The 26.5-th observation lies in this class.

For grouped data we always use n/2, not (n + 1)/2, because the median is estimated by interpolation within the class, as the next topic shows, and the distinction between odd and even n disappears.

Reading cf tables given in the question. Sometimes the data are given as a less-than table: number of students scoring less than 10, less than 20, and so on. These are cumulative frequencies; the ordinary frequencies are recovered by subtraction: f of 10–20 = cf(less than 20) − cf(less than 10). A more-than table lists the number of observations greater than or equal to each lower limit; the frequency of a class is the difference of consecutive entries, and the entry for the first class is n.

Worked example. Given less-than cumulative frequencies 4, 9, 22, 30, 36, 40 for less than 10, 20, 30, 40, 50, 60 respectively, the frequencies of 0–10, …, 50–60 are 4, 5, 13, 8, 6, 4 and n = 40.

Worked example. Given more-than frequencies: 100 or more: 60; 200 or more: 52; 300 or more: 40; 400 or more: 22; 500 or more: 10; 600 or more: 0. The frequencies of 100–200, …, 500–600 are 8, 12, 18, 12, 10 and n = 60.

Always add a cf column to the table before finding the median, and mark the median class clearly; the examiner looks for it.

Having found the median class, we estimate where within it the n/2-th observation lies, assuming the observations in the class are spread evenly over its width. If cf observations lie below the class and the class holds f observations over a width h, then the n/2-th observation is (n/2 − cf) observations into the class, and each observation occupies h/f of the width. So

Median = l + [(n/2 − cf) / f] × h,

where l = lower limit of the median class, n = total frequency, cf = cumulative frequency of the class preceding the median class, f = frequency of the median class, h = class size.

Worked example. Marks of 53 students (previous topic). Median class 60–70; l = 60, n/2 = 26.5, cf = 22 (the cf of 50–60), f = 7, h = 10. Median = 60 + [(26.5 − 22)/7] × 10 = 60 + (4.5/7) × 10 = 60 + 6.43 = 66.43. So about half the students scored below 66.4.

Worked example. Heights of 51 girls: less-than table 140: 4, 145: 11, 150: 29, 155: 40, 160: 46, 165: 51. Frequencies of 135–140, 140–145, 145–150, 150–155, 155–160, 160–165 are 4, 7, 18, 11, 6, 5. n/2 = 25.5; cf 11 < 25.5 ≤ 29, so the median class is 145–150; l = 145, cf = 11, f = 18, h = 5. Median = 145 + [(25.5 − 11)/18] × 5 = 145 + (14.5/18) × 5 = 145 + 4.03 = 149.03 cm.

Worked example. Lifetimes of 400 lamps: classes 1500–2000, …, 4500–5000 with frequencies 14, 56, 60, 86, 74, 62, 48. cf: 14, 70, 130, 216, 290, 352, 400. n/2 = 200; median class 3000–3500 (cf 216 is the first ≥ 200); l = 3000, cf = 130, f = 86, h = 500. Median = 3000 + [(200 − 130)/86] × 500 = 3000 + 406.98 = 3406.98 hours.

Worked example. Length of leaves, inclusive classes 118–126, 127–135, …, 172–180 with frequencies 3, 5, 9, 12, 5, 4, 2. Convert to 117.5–126.5, 126.5–135.5, …, 171.5–180.5 (h = 9). cf: 3, 8, 17, 29, 34, 38, 40; n/2 = 20; median class 144.5–153.5; l = 144.5, cf = 17, f = 12. Median = 144.5 + [(20 − 17)/12] × 9 = 144.5 + 2.25 = 146.75 mm.

Finding missing frequencies from the median. If the median is given and one or two frequencies are unknown, the median formula gives one equation and the total n gives another. Example: classes 0–10 to 50–60 with frequencies 5, x, 20, 15, y, 5, n = 60 and median 28.5. Then x + y = 15. The median class is 20–30 (since 28.5 lies there); cf before it = 5 + x; 28.5 = 20 + [(30 − 5 − x)/20] × 10, so 8.5 = (25 − x)/2, x = 8 and y = 7.

Write l, n/2, cf, f, h explicitly before substituting; this earns method marks even if the arithmetic slips.

The three measures of central tendency answer different questions, and the examination asks you to say which is appropriate and why.

Mean. The mean uses every observation, so it is the best measure when the total matters: total marks, total production, total expenditure. It is the balance point of the data. Its weakness is that extreme values pull it strongly: in a village where nine families earn ₹5000 a month and one earns ₹5,00,000, the mean income of ₹54,500 describes nobody.

Median. The median is the middle value; half the observations are below it and half above. It is unaffected by how extreme the extreme values are, so it is the right measure for typical income, typical house price, or typical marks when a few scores are very different from the rest. In the village example the median is ₹5000, a fair description of a typical family.

Mode. The mode is the most frequent value. It is the measure a shopkeeper wants: the shirt size or shoe size sold most, the most common family size, the most frequent number of wickets. It is the only measure meaningful for non-numerical data, such as the most popular subject.

Empirical relation. For a distribution that is moderately skewed, the three measures are connected approximately by

3 Median = Mode + 2 Mean, i.e. Mode = 3 Median − 2 Mean.

For a symmetric distribution the three coincide. The relation lets you estimate one measure when the other two are known and is tested in one-mark questions: if mean = 24 and median = 26, then mode = 78 − 48 = 30.

Worked example. For the marks data of 30 students: mean = 62, mode = 52, and the median (computed from cf 2, 5, 12, 18, 24, 30 with n/2 = 15, median class 55–70, cf = 12, f = 6, h = 15) = 55 + (3/6) × 15 = 62.5. Check: 3 × 62.5 = 187.5 and 52 + 124 = 176; the relation holds only roughly here because the distribution is not smooth.

Worked example. Lifetimes of 225 components: mode 65.625; mean by the step-deviation method with a = 70, h = 20: u = −3, −2, −1, 0, 1, 2 and fu = −30, −70, −52, 0, 38, 58, Σfu = −56, mean = 70 − 20 × 56/225 = 65.02; median: cf 10, 45, 97, 158, 196, 225, n/2 = 112.5, median class 60–80, median = 60 + (15.5/61) × 20 = 65.08. All three are close, so the distribution is nearly symmetric.

Interpreting a comparison. If the mean exceeds the median, the data have a tail towards high values (a few very large observations); if the median exceeds the mean, the tail is towards low values. When a question gives both the mean and the median of two sets, it usually wants this kind of remark.

The standard textbook remark to reproduce: the mean is the most reliable when data are evenly spread; the median is preferred when extreme values are present; the mode is preferred when the most typical item is wanted.

A cumulative frequency distribution can be shown graphically by an ogive (pronounced o-jive), a cumulative frequency curve. For the less-than type, the points plotted are (upper limit of class, cumulative frequency), and they are joined by a smooth freehand curve. The curve starts at the lower limit of the first class with cumulative frequency 0 and rises to n at the upper limit of the last class.

Steps.

• Prepare the less-than cf table: for each class, its upper limit and the total frequency up to it.
• Take upper limits on the x-axis and cumulative frequencies on the y-axis, with suitable scales.
• Plot the points (upper limit, cf) and join them by a smooth curve. Also plot (lower limit of first class, 0).

Worked example. Daily income of 50 workers: classes 100–120, 120–140, 140–160, 160–180, 180–200 with frequencies 12, 14, 8, 6, 10. Less-than table: less than 120: 12; less than 140: 26; less than 160: 34; less than 180: 40; less than 200: 50. Plot (120, 12), (140, 26), (160, 34), (180, 40), (200, 50), starting from (100, 0), and join.

Reading the median from the ogive. Locate n/2 on the y-axis, draw a horizontal line to meet the ogive, and from that point drop a perpendicular to the x-axis; the foot is the median. For the workers, n/2 = 25; the horizontal at 25 meets the curve just below the point (140, 26); the perpendicular falls at about 138.6, and the formula confirms: median class 120–140, l = 120, cf = 12, f = 14, h = 20, median = 120 + (13/14) × 20 = 138.57.

Reading other values. The ogive also answers questions like how many workers earn less than 150: read the y-value of the curve at x = 150, about 30. And the quartiles: the value below which a quarter of the data lie is read at y = n/4.

Worked example. Weights of 35 students recorded as less-than 38: 0, 40: 3, 42: 5, 44: 9, 46: 14, 48: 28, 50: 32, 52: 35. Plotting these gives the ogive directly. n/2 = 17.5; the horizontal at 17.5 meets the curve between (46, 14) and (48, 28); the median is about 46.5 kg. By formula, the frequencies are 0, 3, 2, 4, 5, 14, 4, 3; median class 46–48; median = 46 + (3.5/14) × 2 = 46.5 kg, in agreement.

The ogive is an increasing curve; it never comes down. Its steepness over a class shows the class's frequency: a steep rise means many observations there. Label both axes with their quantities and units and give the graph a title; these earn presentation marks.

The more-than ogive plots the cumulative frequency of the more-than type: for each class, the number of observations greater than or equal to its lower limit. The points are (lower limit, more-than cf), starting at (lower limit of first class, n) and falling to (lower limit of last class, frequency of last class), and then to (upper limit of last class, 0). The curve is decreasing.

Worked example. Daily income of 50 workers again. More-than table: 100 or more: 50; 120 or more: 38; 140 or more: 24; 160 or more: 16; 180 or more: 10; 200 or more: 0. Plot (100, 50), (120, 38), (140, 24), (160, 16), (180, 10), (200, 0) and join.

Median from both ogives. Draw the less-than and more-than ogives on the same axes. They intersect at one point. The x-coordinate of the point of intersection is the median. Reason: at the median, the number of observations below it equals the number above it, each n/2, so both curves have the same height n/2 there. For the workers, the two curves cross at about (138.6, 25), giving median ≈ 138.6, matching the formula value 138.57.

Worked example. Production yield of 100 farms: classes 50–55, 55–60, 60–65, 65–70, 70–75, 75–80 with frequencies 2, 8, 12, 24, 38, 16. Less-than cf at 55, 60, 65, 70, 75, 80: 2, 10, 22, 46, 84, 100. More-than cf at 50, 55, 60, 65, 70, 75: 100, 98, 90, 78, 54, 16. The curves cross near x = 70.5; by formula, n/2 = 50, median class 70–75, cf = 46, f = 38, median = 70 + (4/38) × 5 = 70.53 kg per hectare.

Worked example. Given a more-than table, draw the more-than ogive and find the median: monthly consumption in units, more than or equal to 65: 68; 85: 64; 105: 59; 125: 46; 145: 26; 165: 12; 185: 4; 205: 0. n = 68, n/2 = 34. On the more-than ogive, the horizontal at 34 meets the curve between (125, 46) and (145, 26); the foot of the perpendicular is about 137. By formula, frequencies are 4, 5, 13, 20, 14, 8, 4; less-than cf 4, 9, 22, 42, 56, 64, 68; median class 125–145, cf = 22, f = 20, h = 20; median = 125 + (12/20) × 20 = 137 units.

Converting between the tables. The frequency of a class = (more-than cf at its lower limit) − (more-than cf at the next lower limit) = (less-than cf at its upper limit) − (less-than cf at the previous upper limit). Any one of the three tables determines the other two.

In the examination, when asked to find the median graphically, draw the ogive(s) to scale, show the construction lines, state the median read from the graph, and, if asked, verify with the formula.

Examination data are not always tidy. Here is how each irregular case is handled.

Inclusive classes. Classes such as 1–10, 11–20 leave gaps. For the mean, use the class marks directly (5.5, 15.5, …). For the median, mode and ogives, convert to exclusive classes 0.5–10.5, 10.5–20.5, … by subtracting and adding half the gap; the class size is then the difference of the new limits (10) and l is the new lower limit.

Unequal class sizes. The mean is found by any of the three methods; in the step-deviation method h may be any convenient number, and the u-values need not be integers. The median formula uses h of the median class alone, so unequal classes cause no difficulty. The mode formula also uses h of the modal class; but note that with unequal classes the modal class should strictly be the class with the greatest frequency density (frequency ÷ width). At this level the class with the greatest frequency is taken.

Worked example. Number of wickets taken by 45 bowlers in classes 20–60, 60–100, 100–150, 150–250, 250–350, 350–450 with frequencies 7, 5, 16, 12, 2, 3. Mode: modal class 100–150, l = 100, h = 50, f₁ = 16, f₀ = 5, f₂ = 12; mode = 100 + [11/(32 − 5 − 12)] × 50 = 100 + (11/15) × 50 = 136.67 wickets. Median: cf 7, 12, 28, 40, 42, 45; n/2 = 22.5; median class 100–150, cf = 12, f = 16, h = 50; median = 100 + (10.5/16) × 50 = 132.81. Mean (found earlier) = 152.89.

Open-ended classes. A first class such as below 10 or a last class such as 60 and above has no definite class mark, so the mean cannot be found exactly; the median and mode can, provided the median or modal class is not the open one. If the mean is asked, assume the open class has the same width as its neighbour.

Cumulative data given in the question. Convert to ordinary frequencies before finding the mean or mode; the median and ogives can be done directly.

Classes given out of order. Arrange them in increasing order before computing cf.

Worked example (inclusive). Ages of 100 policy holders given as below 20: 2, below 25: 6, below 30: 24, below 35: 45, below 40: 78, below 45: 89, below 50: 92, below 55: 98, below 60: 100. These are less-than cf's. Frequencies of 15–20, 20–25, …, 55–60: 2, 4, 18, 21, 33, 11, 3, 6, 2. n/2 = 50; median class 35–40 (cf 45 < 50 ≤ 78); l = 35, cf = 45, f = 33, h = 5; median = 35 + (5/33) × 5 = 35.76 years.

Worked example (letters per surname). Number of letters in 100 surnames: 1–4: 6, 4–7: 30, 7–10: 40, 10–13: 16, 13–16: 4, 16–19: 4. Median: cf 6, 36, 76, 92, 96, 100; n/2 = 50; median class 7–10; median = 7 + (14/40) × 3 = 8.05 letters. Mode: modal class 7–10, l = 7, f₁ = 40, f₀ = 30, f₂ = 16, h = 3; mode = 7 + [10/(80 − 30 − 16)] × 3 = 7 + (10/34) × 3 = 7.88. Mean by step-deviation with a = 11.5, h = 3: class marks 2.5, 5.5, 8.5, 11.5, 14.5, 17.5; u = −3, −2, −1, 0, 1, 2; fu = −18, −60, −40, 0, 4, 8; Σfu = −106; mean = 11.5 − 3.18 = 8.32 letters.

Whatever the irregularity, write the table in full, in exclusive increasing classes, with x, f and cf columns; then every formula applies.

A full-length statistics question usually asks for two of the three measures, or one measure plus an ogive, on a table of six to ten classes. Here is a complete model answer and the checklist the examiner uses.

Model question. The following table gives the distribution of the life time of 400 neon lamps. Find the median life time and the mode.

Median. n = 400, n/2 = 200. The cumulative frequency just greater than 200 is 216, of the class 3000–3500, which is the median class. l = 3000, cf = 130, f = 86, h = 500. Median = l + [(n/2 − cf)/f] × h = 3000 + [(200 − 130)/86] × 500 = 3000 + (70/86) × 500 = 3000 + 406.98 = 3406.98 hours.

Mode. The maximum frequency 86 belongs to the class 3000–3500, the modal class. l = 3000, f₁ = 86, f₀ = 60, f₂ = 74, h = 500. Mode = l + [(f₁ − f₀)/(2f₁ − f₀ − f₂)] × h = 3000 + [26/(172 − 60 − 74)] × 500 = 3000 + (26/38) × 500 = 3000 + 342.11 = 3342.11 hours.

Mean (if asked). Step-deviation with a = 3250, h = 500: class marks 1750, 2250, 2750, 3250, 3750, 4250, 4750; u = −3, −2, −1, 0, 1, 2, 3; fu = −42, −112, −60, 0, 74, 124, 144; Σfu = 128; mean = 3250 + 500 × 128/400 = 3250 + 160 = 3410 hours. Check with the empirical relation: 3 × 3407 = 10221 and 3342 + 2 × 3410 = 10162; close, as expected for a nearly symmetric distribution.

Examiner's checklist.

• Table redrawn with the extra columns needed (x, d or u, fd or fu, cf).
• Column totals shown.
• Median class or modal class identified in words, with n/2 or the maximum frequency stated.
• Formula written in symbols before substitution.
• Values of l, cf, f, h (or f₀, f₁, f₂) listed.
• Arithmetic to two decimal places, with units.
• For an ogive: axes labelled, scale stated, points plotted at the correct limits, smooth curve, construction lines for the median, and the answer read and stated.

Frequent errors. Using (n + 1)/2 instead of n/2 for grouped data; taking cf of the median class itself instead of the preceding class; using the upper limit as l; forgetting to multiply by h in the step-deviation method; plotting a less-than ogive against lower limits; not converting inclusive classes; mixing up f₀ and f₂.

Time management. Compute the cf column once and use it for both the median and the ogive. Choose the step-deviation method for the mean whenever the classes are equal, and check the sign of Σfu carefully.