Chapter 9.3 — Exercise 9.2 — Mean, Median and Mode

The Arithmetic Mean, Two Ways

For raw, ungrouped data, the mean is simply the sum of all observations divided by how many there are. For data already organized into a frequency table, with values x₁, x₂, ... and matching frequencies f₁, f₂, ..., each value is first weighted by how often it occurs:

Weights of parcels handled at a transport office illustrate this directly. Weight values of 50, 65, 75, 90, 110, 120 kg occur with frequencies 25, 34, 38, 40, 47, 16 respectively — 200 parcels in total. Multiplying each weight by its own frequency and summing gives Σfᵢxᵢ = 17000, so the mean weight is 17000 ÷ 200 = 85 kg. This is the identical raw-data formula from above, just computed more efficiently: instead of adding 200 individual weights one at a time, each distinct weight value is added exactly as many times as its frequency demands, in a single multiplication rather than dozens of repeated additions.

Solving Backward From a Known Mean

A frequency distribution has values 2, 4, 6, 8, 10, 12 with frequencies 4, 7, 10, 16, k, 3 — and the mean is given as 7.2, with k unknown. Building Σfᵢxᵢ = 8 + 28 + 60 + 128 + 10k + 36 = 260 + 10k, and Σfᵢ = 40 + k, the mean formula gives 7.2 = (260 + 10k) / (40 + k). Cross-multiplying, 7.2(40 + k) = 260 + 10k, so 288 + 7.2k = 260 + 10k, giving 288 − 260 = 10k − 7.2k, so 28 = 2.8k and k = 10.

This problem runs the mean formula in reverse: instead of computing a mean from known frequencies, a known mean is used to recover a missing frequency. The algebra is ordinary equation-solving once Σfᵢxᵢ and Σfᵢ are written out symbolically in terms of k — the only genuinely new step is remembering that k appears in both the numerator (as 10k, since it's paired with the value 10) and the denominator (as plain k) simultaneously.

The Median: the Middle Value, Precisely Defined

The median is the middle observation once data is sorted in order. For n observations with n odd, the median is the ((n+1)/2)th value; with n even, it's the average of the (n/2)th and (n/2 + 1)th values. Sorting 75, 21, 56, 36, 81, 05, 42 gives 5, 21, 36, 42, 56, 75, 81 — seven values, so the median is the 4th, which is 42. Adding one more score, 68, gives eight sorted values, 5, 21, 36, 42, 56, 68, 75, 81 — now the median is the average of the 4th and 5th values, (42 + 56) / 2 = 49.

For a frequency distribution rather than a short raw list, the median is located using a running cumulative frequency column instead of physically counting through sorted values one at a time. Employee wages of ₹7500 through ₹11000, with frequencies 4, 18, 30, 20, 15, 8, 5, build a cumulative total reaching N = 100; since N/2 = 50, the median sits at the value where the cumulative frequency first reaches or passes 50 — the ₹8500 row, where the cumulative total climbs from 22 to 52. Median wage = ₹8500.

Comparing Two Data Sets by Their Medians

Both medians are found the identical way — locate the row where the cumulative frequency first reaches or exceeds N/2 — applied separately to the boys' and girls' height tables. The comparison itself only becomes meaningful once both medians are computed on their own terms: with 37 boys, N/2 = 18.5 lands inside the 147 cm row's cumulative range; with only 29 girls, N/2 = 14.5 lands inside the 152 cm row instead. The girls' median height comes out taller here — a genuine finding about this particular data set, not a general rule about the two groups, since a different school's data could easily reverse it.

Mode, Bimodal Data, and a Data Set Showing All Three at Once

The mode is simply the value occurring most often. Shoe sizes 6, 7, 8, 9, 10, 6, 7, 10, 7, 6, 7, 9, 7, 6 tally to frequencies 4, 5, 1, 2, 1 — size 7, appearing 5 times, is the mode. When two values tie for the highest frequency, the data is bimodal (two modes); three-way ties make it trimodal; and if every value occurs equally often, there is no meaningful mode at all.

Centuries scored by cricketers, across five centuries-counts (5, 10, 15, 20, 25) with frequencies 56, 23, 39, 13, 8, showcase all three measures on one data set at once: the mean works out to 1555 ÷ 139 ≈ 11.18, the median (at cumulative position N/2 = 69.5) falls in the 10-row, giving median = 10, and the mode is 5, since 56 is the highest frequency in the table. All three numbers are legitimate "typical value" answers for the same 139 cricketers, and all three disagree — a genuine illustration of why a single data set can be honestly summarized several different ways depending on which question is actually being asked.

Working Backward From a Given Mean

Three students have a mean weight of 40 kg; Ranga weighs 46 kg, and Rahim and Reshma weigh the same unknown amount, x kg. Since the mean is the total divided by the count, (x + x + 46) / 3 = 40, giving 2x + 46 = 120, so 2x = 74 and x = 37 — Rahim (and Reshma) each weigh 37 kg.

A related problem chains three separate means together: four unknown numbers, with the first equal to 2, the mean of the first two equal to 4, the mean of the first three equal to 9, and the mean of all four equal to 15. Each stated mean converts into one linear equation, solved in sequence: (2 + a)/2 = 4 gives a = 6; (2 + 6 + b)/3 = 9 gives b = 19; (2 + 6 + 19 + c)/4 = 15 gives c = 33. Both problems reduce a "mean" statement into a solvable equation the same way — total divided by count equals the given mean — the only difference being how many unknowns and how many chained means need untangling before every value is found.

From Central Tendency to Measuring Solids

This chapter closes here, having moved data all the way from raw, unsorted numbers through frequency tables to a single representative statistic. Chapter 10, Surface Areas and Volumes leaves data analysis behind entirely, turning next to measuring three-dimensional solids — a genuinely different branch of mathematics, though it leans on the same careful, step-by-step arithmetic built up across this chapter's worked problems.