IB Math AI HL Topic 4.3: Mean, median, and mode | Notes & Flashcards | Aimnova

Key concepts in Mean, median, and mode

Key Idea: Topic 4.3 focuses on measuring the centre and spread of a dataset. The mean, median, and mode each describe the 'typical' value differently. Standard deviation measures how far values spread from the mean. Understanding how these statistics change when you transform the data (e.g., add a constant, multiply) is a key skill tested in exams.

✅ Measures of central tendency

✅ Measures of spread

Example: Grouped data mean: Class intervals: 0–10 (f=5), 10–20 (f=8), 20–30 (f=7). Midpoints: 5, 15, 25. x̄ = (5×5 + 8×15 + 7×25) / (5+8+7) = (25 + 120 + 175) / 20 = 320/20 = 16 Transformation: If x̄ = 50 and σ = 8, and each value is scaled by 1.1 then increased by 5: New mean = 50×1.1 + 5 = 60 New σ = 8×1.1 = 8.8

Use the GDC for standard deviation on real data — the manual formula is slow and error-prone. For skewed distributions: mean is pulled toward the tail, median stays near the centre. Use median when data is skewed or has outliers.

Paper 2 (GDC allowed): Enter all values in a list. Run 1-Var Stats to get x̄, σₓ, Sₓ, Q1, Q3 all at once. Write down the values you use. Paper 1: You may be asked to calculate the mean from a frequency table by hand — show the Σ(f×x) calculation step explicitly.

IB-style question [6 marks]

A courier company records the delivery times (minutes) of 30 parcels in the grouped table below. Time: [0,10) freq 4, [10,20) freq 9, [20,30) freq 12, [30,40) freq 5. (a) Use the midpoints of each class to estimate the mean delivery time. (b) The standard deviation of the delivery times is found to be 8.4 minutes. The company calculates a delivery charge using charge = 0.5 × time + 2. Find the mean and standard deviation of the delivery charges.

Step by step:

(a) For grouped data, estimate the mean using the class midpoints as the representative x-values.
$\bar{x} = \frac{\sum f x}{\sum f}$
The midpoints are 5, 15, 25, 35. Compute Σfx and Σf.
$\sum f x = 4(5) + 9(15) + 12(25) + 5(35) = 20 + 135 + 300 + 175 = 630$
Divide by the total frequency (30) to get the estimated mean.
$\bar{x} = \frac{630}{30} = 21\ \text{minutes}$
(b) The charge is a linear transformation Y = 0.5X + 2, so the mean uses the full rule.
$\bar{y} = 0.5(21) + 2 = 10.5 + 2 = 12.5$
The standard deviation scales by the multiplier only — the +2 shift does not affect spread.
$\sigma_y = |0.5| \times 8.4 = 4.2$

Final answer:

(a) Estimated mean = 21 minutes. (b) Mean charge = $12.50, standard deviation = $4.20.

Key concepts in Mean, median, and mode

Key Idea: Topic 4.3 focuses on measuring the centre and spread of a dataset. The mean, median, and mode each describe the 'typical' value differently. Standard deviation measures how far values spread from the mean. Understanding how these statistics change when you transform the data (e.g., add a constant, multiply) is a key skill tested in exams.

✅ Measures of central tendency

✅ Measures of spread

Example: Grouped data mean: Class intervals: 0–10 (f=5), 10–20 (f=8), 20–30 (f=7). Midpoints: 5, 15, 25. x̄ = (5×5 + 8×15 + 7×25) / (5+8+7) = (25 + 120 + 175) / 20 = 320/20 = 16 Transformation: If x̄ = 50 and σ = 8, and each value is scaled by 1.1 then increased by 5: New mean = 50×1.1 + 5 = 60 New σ = 8×1.1 = 8.8

Use the GDC for standard deviation on real data — the manual formula is slow and error-prone. For skewed distributions: mean is pulled toward the tail, median stays near the centre. Use median when data is skewed or has outliers.

Paper 2 (GDC allowed): Enter all values in a list. Run 1-Var Stats to get x̄, σₓ, Sₓ, Q1, Q3 all at once. Write down the values you use. Paper 1: You may be asked to calculate the mean from a frequency table by hand — show the Σ(f×x) calculation step explicitly.

IB-style question [6 marks]

Step by step:

(a) For grouped data, estimate the mean using the class midpoints as the representative x-values.
$\bar{x} = \frac{\sum f x}{\sum f}$
The midpoints are 5, 15, 25, 35. Compute Σfx and Σf.
$\sum f x = 4(5) + 9(15) + 12(25) + 5(35) = 20 + 135 + 300 + 175 = 630$
Divide by the total frequency (30) to get the estimated mean.
$\bar{x} = \frac{630}{30} = 21\ \text{minutes}$
(b) The charge is a linear transformation Y = 0.5X + 2, so the mean uses the full rule.
$\bar{y} = 0.5(21) + 2 = 10.5 + 2 = 12.5$
The standard deviation scales by the multiplier only — the +2 shift does not affect spread.
$\sigma_y = |0.5| \times 8.4 = 4.2$

Final answer:

(a) Estimated mean = 21 minutes. (b) Mean charge = $12.50, standard deviation = $4.20.

IB Math AI HL — Mean, median, and mode

Key concepts in Mean, median, and mode

✅ Measures of central tendency

✅ Measures of spread

IB-style question [6 marks]

What you'll learn in Topic 4.3

Study resources — 4.3 Mean, median, and mode

Mean (Average)

Median

Mode

Range and Standard Deviation

Effect of Transformations

Ready to study Mean, median, and mode?

Ready to practice?

IB Math AI HL — Mean, median, and mode

Key concepts in Mean, median, and mode

✅ Measures of central tendency

✅ Measures of spread

IB-style question [6 marks]

What you'll learn in Topic 4.3

Study resources — 4.3 Mean, median, and mode

Mean (Average)

Median

Mode

Range and Standard Deviation

Effect of Transformations

Ready to study Mean, median, and mode?

Ready to practice?