EST I HS Math · Unit 2 · Data Analysis & Probability
Lesson 2.1 · Mean, Median, Mode & Spread
On the paper: Data Analysis & Probability is 27–32% of the paper; averages alone are 2–4 questions every year
Big idea. Every average question is a totals question in disguise. Mean × count = total; work with the total and the question solves itself. Median and mode need one thing: sort first.
1The five numbers 📊
🍼 Baby steps
- 1
5, 15, 2, 18, 7, 3
The 2026 sample (Q15): the list with x found from "the sum is 50" → x = 3.
- 2
sorted: 2, 3, 5, 7, 15, 18
Always sort first. Median, quartiles and range all need the order.
- 3
mean = 50 ÷ 6 = 8.33
Sum ÷ count.
- 4
median = (5 + 7) ÷ 2 = 6
Even count → average the two middle values. Odd count → the middle one.
- 5
Q1 = median of {2, 3, 5} = 3; Q3 = median of {7, 15, 18} = 15; IQR = 12
Split the sorted list in half (even count: clean halves), take each half’s median. That IS the sample question.
- 6
range = 18 − 2 = 16; mode = none (no repeat)
Range is max − min. Mode is the most frequent value — the EST tables always have one.
🎛 Play with it · eight numbers, every statistic
Edit any value. Then push x8 towards 100 with the slider and watch mean vs median.
sorted
2, 3, 5, 7, 9, 11, 15, 18
n · sum
8 · 70
mean = sum ÷ n
8.75
median (middle)
8
mode
none
range = max − min
16
Q1 · Q3 · IQR
4 · 13 · 9
standard deviation
5.31
Push x8 to 100: the mean, range and SD jump; the median and mode barely move. That is why the EST asks "which statistic changes the most / least".
2Totals, not averages 🧮
🍼 Baby steps
- 1
mean of x, y, z, t is 11.5 → total = 46
June 2021. Mean × count = total. This one line replaces all algebra.
- 2
mean of x, y, z, t, q is 11.8 → total = 59
11.8 × 5.
- 3
q = 59 − 46 = 13
The new value is the difference of the totals.
- 4
Brad: 3 tests avg 76 → 228; 4 tests avg 80 → 320; needs 92
October 2020, same move.
- 5
one value goes up by 12 among 6 → mean goes up by 12 ÷ 6 = 2
2026 sample Q43: Giza 30 → 42 in a list of six stores. Δmean = Δvalue ÷ n. No need to recompute anything.
3Frequency tables and weighted averages 📋
🍼 Baby steps
- 1
pens 1, 2, 3, 5 × employees 23, 17, 25, 10
2026 sample Q32. n = 75 employees.
- 2
total pens = 23 + 34 + 75 + 50 = 182 → mean = 182 ÷ 75 = 2.43
Multiply value × frequency, add, divide by the total frequency.
- 3
median = 38th employee: 23 own 1, 40 own ≤2 → median 2
Cumulate frequencies until you pass n/2 = 37.5.
- 4
mode = 3 (largest frequency 25) → product 2.43 × 2 × 3 = 14.56
Mode is the value with the biggest frequency, not the biggest frequency itself.
- 5
25 students: boys avg 12, girls avg 14, class avg 13.2
December 2021. Totals: 12b + 14g = 13.2 × 25 = 330 and b + g = 25 → g = 15.
Trap · mode vs frequency
4Spread: range, IQR, standard deviation 📏
🍼 Baby steps
- 1
range = max − min
December 2020: list 5 … 40 has range 35 already; no x can make the range 29. Smaller than the current range is impossible.
- 2
standard deviation: how far values sit from the mean, on average
You will never compute it by hand on the EST. You compare it: smaller SD = more consistent (June 2021: Tyler, 0.75).
- 3
add 5 to every value → mean + 5, SD unchanged
March 2021. Shifting the whole list moves the centre but not the spread.
- 4
multiply every value by 2 → mean × 2, SD × 2
Scaling stretches the spread too.
- 5
one outlier → mean, range, SD jump; median, mode barely move
The playground above shows it. "Which statistic is most affected" → mean/range; "least affected" → median/mode.
5Real EST questions, solved twice 📄
The sum of 5, 15, 2, 18, 7 and x is 50. What is the value of the interquartile range?
- A.12
- B.13
Q3 − Q1 with Q1 = 2 (took the minimum instead of the lower median).
- C.15
That is Q3 itself.
- D.16
The range (18 − 2), not the IQR.
Safe way · ~60 s
- x = 50 − 47 = 3.
- Sort: 2, 3, 5, 7, 15, 18.
- Lower half 2, 3, 5 → Q1 = 3. Upper half 7, 15, 18 → Q3 = 15. IQR = 12.
⏱ 90-second way · ~25 s
- x = 3, sort, split into two halves of three, take the middles: 3 and 15. 15 − 3 = 12.
Answer: A
pens: 1, 2, 3, 5 · employees: 23, 17, 25, 10
A survey recorded the number of pens owned by employees in a company (table above). What is the product of the mean, median and mode of the number of pens?
- A.10.25
Mean × median × 2 — mode taken as 2 (the median) by mistake.
- B.13.50
Uses mean 2.25 (divided by 4 categories, not 75 employees).
- C.14.50
Rounded the mean to 2.42 too early.
- D.14.56
Safe way · ~75 s
- n = 75. Total pens = 23 + 34 + 75 + 50 = 182 → mean 2.4267.
- Cumulative: 23, 40 → the 38th employee owns 2 → median 2.
- Mode = 3 (frequency 25). Product = 2.4267 × 2 × 3 = 14.56.
⏱ 90-second way · ~30 s
- Median 2 and mode 3 are read off the table in five seconds; only the mean needs the calculator: 182 ÷ 75 × 6.
Answer: D
The mean of x, y, z and t is 11.5. The mean of x, y, z, t and q is 11.8. What is the value of q?
- A.11.8
The new mean, not the new value.
- B.12.1
Adding the two means’ difference (0.3) to 11.8 — no.
- C.13
- D.14.5
11.5 + 3, a guess.
Safe way · ~40 s
- x + y + z + t = 4 × 11.5 = 46.
- x + y + z + t + q = 5 × 11.8 = 59.
- q = 59 − 46 = 13.
⏱ 90-second way · ~10 s
- 5 × 11.8 − 4 × 11.5 = 59 − 46 = 13. Two multiplications, one subtraction.
Answer: C
staff: Cairo 33, Alexandria 37, Port Said 19, Giza 30, Aswan 11, Luxor 10
A company has 6 stores (table above). If the Giza store increases its staff to 42, which is true about the new average number of staff members?
- A.The average will decrease.
A value went UP; the mean cannot go down.
- B.The average will increase by 2.
- C.The average will increase by 4.
Δ ÷ 3 instead of Δ ÷ 6.
- D.The average will not change.
Only if the total were unchanged.
Safe way · ~50 s
- Old total 140 → mean 23.33. New total 152 → mean 25.33.
- Difference 2.
⏱ 90-second way · ~10 s
- One value rises by 12 among 6 values → the mean rises by 12 ÷ 6 = 2. Never add the list.
Answer: B
Rounds 1–5 — Diego 7, 5, 8, 9, 4 · Mark 3, 10, 9, 5, 8 · Tyler 8, 7, 6, 7, 8 · Toni 10, 4, 9, 10, 7 · SD: 1.85, 2.61, 0.75, 2.28
The winner is the player with the highest average. Who won — and which player was the most consistent?
- A.Toni won; Tyler was most consistent
- B.Tyler won; Toni was most consistent
Tyler has the SMALLEST spread, not the highest total (36 vs Toni’s 40).
- C.Toni won; Diego was most consistent
Diego’s SD 1.85 is not the smallest.
- D.Mark won; Tyler was most consistent
Mark totals 35, below Toni’s 40.
Safe way · ~50 s
- Totals: Diego 33, Mark 35, Tyler 36, Toni 40 → Toni (same count, so compare totals, not means).
- Smallest SD = most consistent = Tyler (0.75).
⏱ 90-second way · ~15 s
- Same number of rounds → highest total wins: Toni. "Consistent" = smallest SD: Tyler, already printed in the table.
Answer: A
⏱ The 90-second lane
One question = 90 seconds. These are the moves that finish this topic in one line.
- 1Mean × n = total. Every "find the missing value / the fourth test / the new member" is total₂ − total₁.
- 2One value changes by Δ → the mean changes by Δ ÷ n. Nothing else needs recomputing.
- 3Same count? Compare totals, skip the division.
- 4Median: sort, then middle (odd) or average of the two middles (even). Frequency table: cumulate until you pass n/2.
- 5Q1 and Q3 are the medians of the two halves; IQR = Q3 − Q1. Range = max − min.
- 6Mode = the value with the largest frequency — the value, not the frequency.
- 7Add c to every value → mean + c, median + c, SD and range unchanged. Multiply by k → everything × k.
- 8Outlier: mean, range, SD react; median, mode do not. "Most consistent" = smallest SD.
- 9Combined averages: (n₁·m₁ + n₂·m₂) ÷ (n₁ + n₂). Never average the averages.
Take with you
- ▸Averages are totals divided by counts — work with the totals.
- ▸Order the data before any median or quartile.
- ▸Spread words: range (max − min), IQR (middle half), SD (typical distance from the mean, compared, never computed by hand).
- ▸Shifts move the centre; scaling moves everything; outliers move the mean but not the median.