Open Math›Year 8›Data classification and visualisation›Histograms and frequency polygons

Histograms and frequency polygons

Read and interpret histograms, identify the modal class, calculate an approximate mean from grouped data using class midpoints, and describe the shape of a distribution.

Worked examples

Reading a frequency from a histogram

Straightforward

Problem

A histogram shows the ages of visitors to a wildlife park. The frequency table is:

0–10 years: 7, 10–20 years: 13, 20–30 years: 9, 30–40 years: 6, 40–50 years: 5

How many visitors were aged 10–20?

Locate the class interval 10–20 in the table.

The row for the 10–20 age group is the second entry in the table.

Read the frequency for that interval.

The frequency for the 10–20 age group is

13

Answer

13 visitors were aged 10–20.

Finding the modal class and approximate mean

Moderate

Problem

A frequency table shows the time (minutes) Year 8 students spend on social media each day:

0–30: 6, 30–60: 14, 60–90: 11, 90–120: 7, 120–150: 2

(a) State the modal class. (b) Calculate the approximate mean.

Find the class interval with the highest frequency.

Frequencies: 6, 14, 11, 7, 2. The highest is

14

, in the 30–60 interval.
Modal class = 30–60 minutes.

Find the midpoint of each class interval.

\dfrac{0+30}{2} = 15, \quad \dfrac{30+60}{2} = 45, \quad \dfrac{60+90}{2} = 75, \quad \dfrac{90+120}{2} = 105, \quad \dfrac{120+150}{2} = 135

Multiply each midpoint by its frequency and sum the products.

6 \times 15 + 14 \times 45 + 11 \times 75 + 7 \times 105 + 2 \times 135

= 90 + 630 + 825 + 735 + 270 = 2550

Divide the sum by the total frequency.

\text{Total frequency} = 6 + 14 + 11 + 7 + 2 = 40

\text{Approximate mean} = \dfrac{2550}{40} = 63.75

minutes

Answer

(a) Modal class = 30–60 minutes. (b) Approximate mean

= 63.75

minutes.

Describing distribution shape and comparing modal classes

Challenging

Problem

Two histograms show the number of items purchased by customers at two stores on one day.

Store A: 0–5: 2, 5–10: 5, 10–15: 14, 15–20: 9, 20–25: 2
Store B: 0–5: 15, 5–10: 11, 10–15: 6, 15–20: 3, 20–25: 1

(a) Describe the shape of each distribution. (b) Which store has a higher modal class?

Describe the shape of Store A's distribution.

Frequencies: 2, 5, 14, 9, 2. The distribution rises to a peak at 10–15 then falls, with roughly equal values on both sides. Store A is approximately symmetric.

Describe the shape of Store B's distribution.

Frequencies: 15, 11, 6, 3, 1. The distribution is highest on the left and decreases toward the right, with a tail extending to the right. Store B is positively skewed.

Identify the modal class for each store.

Store A: highest frequency is

14

, in the 10–15 interval. Modal class = 10–15.
Store B: highest frequency is

15

, in the 0–5 interval. Modal class = 0–5.

Compare the modal classes.

10–15 > 0–5, so Store A has the higher modal class.

Answer

(a) Store A is approximately symmetric; Store B is positively skewed. (b) Store A has a higher modal class (10–15).

Practise

Q1·Straightforward

A histogram shows the ages of visitors to a holiday camp. The frequencies for each class interval are:

10–12 years: 6, 12–14 years: 15, 14–16 years: 11, 16–18 years: 8

How many visitors are aged 12–14 years?

Explanation

The frequency for the 12–14 years class interval is 15, so 15 visitors are in that age group.

Q2·Straightforward

A histogram shows the masses of parcels delivered to an office. The frequencies are:

0–2 kg: 8, 2–4 kg: 13, 4–6 kg: 7, 6–8 kg: 2

How many parcels have a mass in the 4–6 kg interval?

Explanation

The frequency for the 4–6 kg class interval is 7, so 7 parcels have a mass in that range.

Q3·Straightforward

A histogram shows the time (minutes) customers spent in a shop. The frequencies are:

0–10: 4, 10–20: 11, 20–30: 18, 30–40: 9, 40–50: 3

How many customers spent 20–30 minutes in the shop?

Explanation

The frequency for the 20–30 minute interval is 18, so 18 customers spent that amount of time in the shop.

Q4·Straightforward

A histogram shows the daily maximum temperatures recorded over 40 days. The frequencies are:

15–20°C: 5, 20–25°C: 12, 25–30°C: 16, 30–35°C: 7

How many days had a maximum temperature in the 25–30°C interval?

Explanation

The frequency for the 25–30°C class interval is 16, so 16 days had a maximum temperature in that range.

Q5·Moderate

A frequency table shows the number of hours students spent on a school project:

0–2 hours: 3, 2–4 hours: 11, 4–6 hours: 8, 6–8 hours: 3

Which class interval is the modal class?

0–2 hours2–4 hours4–6 hours6–8 hours

Explanation

The frequencies are 3, 11, 8, 3. The highest frequency is 11, which belongs to the 2–4 hours interval. This is the modal class.

Q6·Moderate

A frequency table shows the lengths of leaves collected from a garden:

3–5 cm: 4, 5–7 cm: 9, 7–9 cm: 14, 9–11 cm: 8, 11–13 cm: 5

Which is the modal class?

3–5 cm5–7 cm7–9 cm9–11 cm

Explanation

The frequencies are 4, 9, 14, 8, 5. The highest frequency is 14, in the 7–9 cm interval. This is the modal class.

Q7·Moderate

A frequency table shows the distances (km) students travel to school:

0–2 km: 5, 2–4 km: 12, 4–6 km: 8, 6–8 km: 5

Calculate the approximate mean distance using class midpoints. Give your answer to 2 decimal places.

Explanation

Midpoints: 1, 3, 5, 7. Total frequency

= 5 + 12 + 8 + 5 = 30

\text{Sum} = 5 \times 1 + 12 \times 3 + 8 \times 5 + 5 \times 7 = 5 + 36 + 40 + 35 = 116

\text{Approximate mean} = \dfrac{116}{30} \approx 3.87

Q8·Moderate

A frequency table shows the test scores of 25 students:

40–50: 2, 50–60: 5, 60–70: 9, 70–80: 6, 80–90: 3

Calculate the approximate mean score using class midpoints. Give your answer to 2 decimal places.

Explanation

Midpoints: 45, 55, 65, 75, 85. Total frequency

= 25

\text{Sum} = 2 \times 45 + 5 \times 55 + 9 \times 65 + 6 \times 75 + 3 \times 85

= 90 + 275 + 585 + 450 + 255 = 1655

\text{Approximate mean} = \dfrac{1655}{25} = 66.20

Q9·Challenging

A histogram shows daily rainfall (mm) recorded over 40 days. The frequencies are:

0–5 mm: 18, 5–10 mm: 10, 10–15 mm: 5, 15–20 mm: 4, 20–25 mm: 3

Which term best describes the shape of this distribution?

SymmetricPositively skewedNegatively skewedBimodal

Explanation

The frequencies decrease from left to right (18, 10, 5, 4, 3), with most data in the lower intervals and a tail extending to the right. This is a positively skewed (right-skewed) distribution.

Q10·Challenging

A histogram shows the points scored by a basketball team across 45 games. The frequencies are:

40–50: 1, 50–60: 2, 60–70: 6, 70–80: 15, 80–90: 21

Which term best describes the shape of this distribution?

SymmetricPositively skewedNegatively skewedBimodal

Explanation

The frequencies increase from left to right (1, 2, 6, 15, 21), with most data in the higher intervals and a tail extending to the left. This is a negatively skewed (left-skewed) distribution.

Q11·Challenging

A histogram shows the wait times (minutes) of customers at a bus stop. The frequencies are:

0–5: 4, 5–10: 10, 10–15: 14, 15–20: 10, 20–25: 4

Which term best describes the shape of this distribution?

Positively skewedNegatively skewedSymmetricBimodal

Explanation

The frequencies are 4, 10, 14, 10, 4 — they peak in the middle and decrease equally on both sides. This distribution is symmetric.

Q12·Challenging

Two classes sat the same test. Their score distributions are:

Class A — 40–50: 2, 50–60: 4, 60–70: 10, 70–80: 6, 80–90: 3
Class B — 40–50: 1, 50–60: 2, 60–70: 5, 70–80: 11, 80–90: 6

Which statement is correct?

Class A has a higher modal class than Class BClass B has a higher modal class than Class ABoth classes have the same modal classClass A has more students than Class B

Explanation

Class A: highest frequency is 10, in the 60–70 interval. Modal class = 60–70.
Class B: highest frequency is 11, in the 70–80 interval. Modal class = 70–80.
Since 70–80 > 60–70, Class B has a higher modal class than Class A.