Chapter 7.2 — Exercise 7.1 — Mean, Median and Mode

The Deviation Method — A Faster Route for Awkward Numbers

Rather than adding large or inconvenient numbers directly, pick any one value as an "assumed mean" (call it A), find how far every observation deviates from A, and add the mean of those deviations back onto A:

For 18, 14, 13, 15, 17, 19, taking A = 17: the deviations are +1, −3, −4, −2, 0, +2, summing to −6. Mean = 17 + (−6/6) = 17 − 1 = 16. Choosing A close to the middle of the data keeps the deviations small and easy to add — the arithmetic gets simpler even though the final answer is identical to adding all six numbers directly.

A natural question follows: does the choice of A actually matter? Testing the same twelve marks (4, 21, 13, 17, 5, 9, 10, 20, 19, 12, 20, 14) with A = 16 gives a mean of 13.67; testing with A = 12 instead gives 13.67 again — identical. The assumed mean is exactly that: assumed, and temporary. It cancels out completely once its own deviation is added back in, which is exactly why any convenient number can be used without affecting the final result.

A Property Worth Knowing on Its Own: Deviations Always Sum to Zero

For 5, 8, 10, 15, 22 (mean = 12), the deviations from the mean are −7, −4, −2, 3, 10 — and these sum to exactly 0. This isn't a coincidence specific to this data set: it's true of every data set's deviations from its own mean, since the mean is defined precisely as the balancing point where the total shortfall below it equals the total excess above it. Knowing this in advance turns "find the mean deviation given the sum of 20 deviations is 100" into pure division (100/20 = 5) — no data values needed at all.

Reconstructing a Mean From Partial Information

Several questions give a mean and ask you to work backward through a change to the data:

• Removing an observation: eight observations with mean 25 have sum 200; removing the value 11 leaves a sum of 189 across 7 observations, for a new mean of 27.
• Fixing a data-entry mistake: nine observations with a (wrongly calculated) mean of 38 have sum 342; if 27 was mistakenly recorded instead of 72, the correct sum is 342 − 27 + 72 = 387, giving a corrected mean of 387/9 = 43.
• A person leaving a group: 40 people with mean age 11 two years ago have a present mean age of 13 (everyone ages by the same 2 years, shifting the mean by 2). If the mean age of the remaining 39 people is now 12, the person who left has a present age of (40×13) − (39×12) = 520 − 468 = 52.

Every one of these relies on the same underlying fact: sum = mean × count. Once that's treated as the starting equation rather than the mean formula rearranged each time, all three problems reduce to ordinary arithmetic.

Finding an Unknown Piece of the Data

If ten students' marks have a mean of 15, and nine of those students' deviations from a tenth student's (Karishma's) marks are known (−8, −6, −3, −1, 0, 2, 3, 4, 6, summing to −3, plus Karishma's own deviation of 0 from herself), then 15 = Karishma's marks + (−3/10), giving Karishma's marks = 15.3. A related, more abstract version: if the sum of n deviations from 25 is 25, and the sum of the same n deviations from 35 is −25, setting the two resulting mean expressions equal (25 + 25/n = 35 − 25/n) gives n = 5, and a mean of 30.

The Same Shift-and-Scale Rules Apply to Groups of Observations

If x₁ through x₁₀ have mean 20, then x₁+4, x₂+8, x₃+12, ... , x₁₀+40 has a mean of 20 + (4+8+12+...+40)/10 = 20 + 220/10 = 42 — the added amounts themselves have their own mean (22) added straight onto the original.

Median Problems With an Unknown to Solve For

For 3.3, 3.5, 3.1, 3.7, 3.2, 3.8 (6 values, even count): ordered, this is 3.1, 3.2, 3.3, 3.5, 3.7, 3.8, and the median is the average of the 3rd and 4th values, (3.3+3.5)/2 = 3.4. When the median is already known and a value inside the data is unknown: for 10, 12, 14, x−3, x, x+2, 25 (7 values, already in ascending order) with median 15, the 4th value must equal 15, so x−3 = 15, giving x = 18.

Two Median Questions That Reward Careful Reasoning Over Calculation

Given six of nine integers — 3, 5, 5, 7, 8, 9 — what's the largest possible median of all nine? Since the median of 9 values is the 5th one in ascending order, and the three unknown values can be made larger than all six known ones without affecting which value lands in the 5th position, the median is fixed at the largest of the six known values that could occupy that slot: 8. A second one: if a set of 9 distinct observations has median 20, and the largest 4 values are each increased by 2, what happens to the median? The 5th value (the median) isn't among the largest 4, so it's completely untouched — the median stays 20.

Working With the Mode

For 10, 12, 11, 10, 15, 20, 19, 21, 11, 9, 10, the value 10 appears three times, more than any other — mode = 10. Like the mean, the mode shifts along with the data: if every score in a set decreases by 3, the mode decreases by exactly 3 as well, since the most frequent value is still the most frequent after the same shift is applied to everything. A genuinely different kind of mode question: among all the digits used to write every natural number from 1 to 100, which digit appears most often? Digit 1 appears 21 times (including both digits of 100), while 2 through 9 each appear 20 times and 0 appears 11 times — so the mode of this digit list is 1.

Adding Data to Hit a Target

Given 5, 28, 15, 10, 15, 8, 24 (mean 15, mode 15), what four extra numbers keep the mean and median unchanged while raising the mode to 16? At least three of the four new numbers must equal 16 to outnumber the existing pair of 15's. Letting the fourth new number be x, and requiring the mean of all eleven numbers to stay at 15: (105 + 48 + x)/11 = 15 gives x = 12 — so the four numbers added are 16, 16, 16 and 12.

Where This Leads

Every technique here — the deviation method especially — becomes essential once data is organised into class intervals rather than listed individually, which is exactly the shift Exercise 7.2 makes.