IB Maths AA HL Topic 4 — Statistics & Probability Paper 1 & 2 ~7 min read

Histograms

A frequency histogram is a bar chart for grouped continuous data — bars sit side-by-side with no gaps, heights show class frequencies, widths show class intervals. The shape tells you a lot: where the data clusters, whether it’s symmetric, skewed, or potentially normal.

📘 What you need to know

Anatomy of a histogram

0 5 10 15 20 frequency data values (continuous) → a b c d e f modal class
Frequency histogram: equal class widths, no gaps between bars. The tallest bar is the modal class.

Reading shape from a histogram

ShapeVisualImplication
Symmetric / bell-shapedpeaks in centre, tails decay equallycould be modelled by a normal distribution
Positively skewed (right-skewed)tall bars on the left, long tail on the rightmedian < mean; not normal
Negatively skewed (left-skewed)tall bars on the right, long tail on the leftmedian > mean; not normal
Bimodaltwo distinct peakslikely two underlying populations mixed together
“Could be normal” requires both: (i) approximately symmetric, (ii) bell-shaped (high in middle, decaying outwards). Symmetric + flat is uniform, not normal.

🧭 Recipe — draw and interpret a histogram

  1. Set up axes: x = data variable (continuous, with units); y = frequency.
  2. Mark class boundaries on the x-axis using an even scale.
  3. Draw bars for each class — no gaps, height = class frequency.
  4. Identify the modal class (tallest bar).
  5. Comment on shape: symmetric, skewed, bell-shaped, bimodal.
  6. For estimates of mean: use mid-interval values × frequencies, sum, divide by total.

Worked examples

WE 1

Draw a frequency histogram

The table shows the times in minutes (t) that 50 students spent on homework one evening.

Time (min)0 ≤ t < 2020 ≤ t < 4040 ≤ t < 6060 ≤ t < 8080 ≤ t < 100
Frequency6121895

(a) Describe the histogram you would draw. (b) State the modal class.

(a) Histogram description x-axis: time in minutes, scale 0 to 100, no gaps y-axis: frequency, scale 0 to 18 (or 20) 5 bars of equal width 20, heights 6, 12, 18, 9, 5 Bars touch each other (no gaps) (b) Modal class = tallest bar freq 18 → modal class is 40 ≤ t < 60 Modal class: 40 ≤ t < 60 shape rises to peak at 40-60 then falls — slight right-skew (tail extends further right)
WE 2

Modal class and estimated mean

The weights of 40 apples (in grams) are summarised below.

Weight (g)80 ≤ w < 100100 ≤ w < 120120 ≤ w < 140140 ≤ w < 160160 ≤ w < 180
Frequency4913104

(a) State the modal class. (b) Estimate the mean weight.

(a) Modal class — tallest bar freq 13 → 120 ≤ w < 140 (b) Mid-interval values 90, 110, 130, 150, 170 Σfx 4(90) + 9(110) + 13(130) + 10(150) + 4(170) = 360 + 990 + 1690 + 1500 + 680 = 5220 Estimated mean = 5220/40 = 130.5 g Modal class: 120 ≤ w < 140; estimated mean = 130.5 g distribution is roughly symmetric — mean lies near the middle of the modal class
WE 3

Comment on the shape of a histogram

The daily social media usage (in hours) of 47 students is shown below.

Hours0 ≤ h < 22 ≤ h < 44 ≤ h < 66 ≤ h < 88 ≤ h < 10
Frequency1814942

(a) State the modal class. (b) Comment on the shape of the distribution.

(a) Modal class freq 18 → 0 ≤ h < 2 (b) Shape Frequencies decrease as hours increase: 18, 14, 9, 4, 2 Tallest bar on the left; long tail extends to the right Modal class: 0 ≤ h < 2; positively skewed (right-skewed) distribution positive skew → mean > median; the data is NOT a good candidate for a normal distribution
WE 4

Could the data be modelled by a normal distribution?

The birth weights (in kg) of 54 newborns are recorded.

Weight (kg)2.0 ≤ w < 2.52.5 ≤ w < 3.03.0 ≤ w < 3.53.5 ≤ w < 4.04.0 ≤ w < 4.5
Frequency51220125

Comment on whether a normal distribution would be a suitable model for this data.

Step 1: Look at frequencies 5, 12, 20, 12, 5 — symmetric around the central class 3.0–3.5 Step 2: Check shape Peaks in middle, decays evenly to both tails → bell-shaped Step 3: Conclusion Yes — symmetric and bell-shaped, so a normal distribution is a suitable model “yes” requires BOTH symmetry AND bell shape; either alone is not enough
WE 5

Find a missing frequency from a histogram

A histogram has the following class frequencies, where one is unknown.

Class20 ≤ m < 4040 ≤ m < 6060 ≤ m < 8080 ≤ m < 100100 ≤ m < 120
Frequency614x83

The total frequency is 50. Find x and state the modal class.

Step 1: Sum of all frequencies = 50 6 + 14 + x + 8 + 3 = 50 31 + x = 50 → x = 19 Step 2: Identify modal class Frequencies: 6, 14, 19, 8, 3 → highest is 19 x = 19; modal class: 60 ≤ m < 80 always identify modal class AFTER finding the missing frequency — the unknown class might be the mode
WE 6

Use a histogram to estimate counts above a threshold

Using the apple-weight data from WE 2 (n = 40), find: (a) the number of apples weighing more than 140 g; (b) the percentage of apples weighing 120 g or less.

(a) Apples > 140 g — sum classes 140-160 and 160-180 10 + 4 = 14 (b) Apples ≤ 120 g — sum classes 80-100 and 100-120 4 + 9 = 13 Percentage: 13/40 × 100 = 32.5% (a) 14 apples; (b) 32.5% when the threshold falls exactly at a class boundary, just sum the relevant whole classes

💡 Top tips

⚠ Common mistakes

Final note in this section: Interpreting Data. With all the tools assembled — averages, dispersion, box plots, cumulative graphs, histograms — the question becomes which one to use. The answer depends on whether outliers are present, whether the data is symmetric, and what claim you’re trying to support. Comparing two distributions in context is the single most-tested skill in this section.

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