Maths · Statistics
Viewing as
Teacher view: every answer and mark scheme set out in full.
Histograms and Sampling
Using frequency density to draw and read histograms, and taking random, stratified and capture-recapture samples.
Learning Objectives
- 1Draw and interpret histograms with unequal class widths using frequency density.
- 2Find frequencies from a histogram, and estimate the number of values in part of a class.
- 3Explain the difference between a population and a sample, and why samples can be biased.
- 4Take a stratified sample, and use the capture-recapture method to estimate a population.
When the classes are not equal
In a bar chart the height shows the frequency, but when the classes have different widths that is no longer fair, because a wide class naturally holds more values. A histogram solves this by making the area of each bar show the frequency, so the height shows the frequency density. This lesson also looks at how data is collected, since a conclusion is only as good as the sample behind it. Both topics are Higher tier content, and they are mostly about careful working with a few simple formulae.
Frequency density and histograms
In a histogram the area of a bar is the frequency.
-
The formula
Frequency density \(= \dfrac{\text{frequency}}{\text{class width}}\).
-
Frequency from a bar
Frequency \(=\) frequency density \(\times\) class width, which is the area of the bar.
-
The axes
The vertical axis is labelled frequency density, and there are no gaps between the bars.
-
Why it works
A class twice as wide gets a bar half as high for the same frequency, so the areas stay in proportion.
A histogram
The green numbers are the frequencies. The class from 10 to 30 has width 20 and frequency density 3, so its frequency is \(3 \times 20 = 60\). The bar from 0 to 10 has width 10 and height 1.5, so its frequency is 15.
Reading the histogram
- Total frequency \(15 + 60 + 40 + 30 = 145\).
- Modal class The tallest bar is 30 to 40, with the highest frequency density.
- Part of a class An estimate of the number aged 30 to 35 is half the class, \(0.5 \times 40 = 20\), assuming the values are spread evenly.
- Drawing Work out the frequency density for each class, then draw bars of that height.
Drawing the bars
The table shows the lengths of 60 worms. For the class \(0 < l \leq 5\) the frequency is 10, for \(5 < l \leq 15\) it is 30, and for \(15 < l \leq 20\) it is 20. Work out the frequency density for each class.
Show the solutionHide the solution
- 1 First class The width is 5, so \(\dfrac{10}{5} = 2\).
- 2 Second class The width is 10, so \(\dfrac{30}{10} = 3\).
- 3 Third class The width is 5, so \(\dfrac{20}{5} = 4\).
- 4 Draw Bars of heights 2, 3 and 4 over the three classes, with no gaps.
Answer2, 3 and 4
Populations and samples
The population is everyone or everything you want to know about, and a sample is the part you actually measure.
-
Why sample
Asking everyone is slow or expensive, so a sample is used.
-
Bias
A sample is biased if some members of the population are more likely to be chosen than others, such as asking only people in a gym about exercise.
-
Random sample
Every member has an equal chance of being chosen, for example by using random numbers.
-
Sample size
A larger sample is generally more reliable, but takes more effort.
Stratified sampling
In a stratified sample, each group (stratum) is represented in proportion to its size.
-
The formula
Number in the sample from a group \(= \dfrac{\text{size of the group}}{\text{size of the population}} \times \text{sample size}\).
-
Example
A school has 600 pupils, 240 of them in Year 10. A sample of 50 should include \(\dfrac{240}{600} \times 50 = 20\) pupils from Year 10.
-
Check
The numbers from each group should add up to the sample size.
-
Choose randomly within each group
Each group's sample should still be chosen at random.
Estimating a population (capture-recapture)
To estimate how many animals are in a population, mark some and see how many marked ones turn up later.
-
Method
Capture and mark \(M\) animals, release them, then capture a second sample of \(n\) animals and count the \(m\) that are marked.
-
Assumption
The proportion marked in the second sample matches the proportion in the whole population: \(\dfrac{m}{n} = \dfrac{M}{N}\).
-
Formula
\(N = \dfrac{M \times n}{m}\).
-
Example
50 fish are marked. A second sample of 40 has 8 marked. The population is estimated as \(\dfrac{50 \times 40}{8} = 250\).
Capture-recapture
A scientist catches 30 birds, marks them and releases them. A week later she catches 60 birds, and 9 of them are marked. Estimate the number of birds in the population.
Show the solutionHide the solution
- 1 Use the proportions \(\dfrac{9}{60} = \dfrac{30}{N}\).
- 2 Rearrange \(N = \dfrac{30 \times 60}{9}\).
- 3 Calculate \(\dfrac{1800}{9} = 200\).
- 4 Comment It is an estimate, and it assumes the birds mix evenly and none leave or join.
Answer200
Test yourself
-
1
What is frequency density?
Show answerHide answer
Frequency divided by class width.
-
2
What does the area of a bar in a histogram show?
Show answerHide answer
The frequency.
-
3
What makes a sample biased?
Show answerHide answer
Some members of the population are more likely to be chosen than others.
-
4
How many Year 10 pupils are in a stratified sample of 50 if Year 10 is 240 of 600?
Show answerHide answer
20.
-
5
What is the capture-recapture formula?
Show answerHide answer
\(N = \dfrac{M \times n}{m}\).
Exam technique: histograms and sampling
Show each calculation separately.
-
Label the vertical axis
Frequency density, not frequency.
-
Work out the width of each class
Use the class boundaries, not the number of values.
-
Use the area for estimates
Part of a bar is the same fraction of the frequency.
-
State the assumptions
In capture-recapture, say the population has not changed and the marked animals have mixed in.
Summary and exam focus
- In a histogram, frequency density is frequency divided by class width, and the area of the bar is the frequency.
- A random sample gives every member an equal chance, and a biased sample does not.
- In a stratified sample, each group is represented in proportion to its size.
- Capture-recapture estimates a population as \(N = \dfrac{M \times n}{m}\).
Exam focus
A school has 600 pupils, and 240 of them are in Year 10. A stratified sample of 50 pupils is taken. How many Year 10 pupils should be in the sample? (2 marks) (2 marks)
The fraction of the school in Year 10 is \(\dfrac{240}{600} = \dfrac{2}{5}\), so \(\dfrac{2}{5} \times 50 = 20\) pupils. Show the fraction before multiplying, as it earns the method mark.
Key terms
The words this lesson expects you to use. Each one is linked from the first place it appears above.
- Histogram
- A chart with touching bars where the area of each bar shows the frequency.
- Frequency density
- Frequency divided by class width, used as the height of a histogram bar.
- Class width
- The size of a class, found by subtracting its lower boundary from its upper boundary.
- Population
- The whole group that is being studied.
- Sample
- A part of the population that is selected for study.
- Bias
- A tendency for a sample to favour some outcomes over others.
- Random sample
- A sample where every member of the population has an equal chance of being chosen.
- Stratified sample
- A sample where each group is represented in proportion to its size.
- Capture-recapture
- A method of estimating the size of a population by marking and recapturing members.
Questions and answers
15 questions set on this lesson, with the mark schemes and model answers open.
The table shows the ages of some people. For the class \(0 < a \leq 10\) the frequency is 5, for \(10 < a \leq 30\) it is 30, and for \(30 < a \leq 50\) it is 60. Calculate the frequency density for each class. [3 marks]
Mark scheme — 3 marks available
- Class widths 10, 20 and 20 used — M1
- At least two frequency densities correct — A1
- 0.5, 1.5 and 3 — A1
Model answer
The widths are 10, 20 and 20. The frequency densities are \(\dfrac{5}{10} = 0.5\), \(\dfrac{30}{20} = 1.5\) and \(\dfrac{60}{20} = 3\).
The histogram shows the ages of some people at a village event. (a) Calculate the number of people aged more than 40 and up to 60. [2 marks] (b) Calculate the number of people aged over 60. [2 marks]
Mark scheme — 4 marks available
- (a) \(2 \times 20\) — M1
- (a) 40 — A1
- (b) \(0.5 \times 40\) — M1
- (b) 20 — A1
Model answer
(a) The frequency is the area of the bar: \(2 \times 20 = 40\). (b) The bar from 60 to 100 has a frequency density of 0.5 and a width of 40, so \(0.5 \times 40 = 20\).
On a histogram the bar for the class \(20 < x \leq 40\) has a frequency density of 2.5. (a) Calculate the frequency of the class. [2 marks] The class \(40 < x \leq 50\) has a frequency of 30. (b) Calculate the height of its bar. [1 mark]
Mark scheme — 3 marks available
- (a) \(2.5 \times 20\) — M1
- (a) 50 — A1
- (b) 3 — B1
Model answer
(a) The width is 20, so the frequency is \(2.5 \times 20 = 50\). (b) The width is 10, so the frequency density is \(\dfrac{30}{10} = 3\).
A club has 90 juniors, 60 seniors and 30 veterans. A stratified sample of 36 members is taken. Calculate the number of members from each group in the sample. [3 marks]
Mark scheme — 3 marks available
- \(\dfrac{90}{180} \times 36\) or the total 180 seen — M1
- At least two numbers correct — A1
- 18, 12 and 6 — A1
Model answer
There are 180 members. Juniors: \(\dfrac{90}{180} \times 36 = 18\). Seniors: \(\dfrac{60}{180} \times 36 = 12\). Veterans: \(\dfrac{30}{180} \times 36 = 6\).
A farmer marks 25 sheep in a flock. Later he selects 50 sheep, and 5 of them are marked. (a) Calculate an estimate for the number of sheep in the flock. [2 marks] (b) State one assumption you have made. [1 mark]
Mark scheme — 3 marks available
- (a) \(\dfrac{25 \times 50}{5}\) — M1
- (a) 250 — A1
- (b) The marked sheep have mixed evenly, or the population has not changed — B1
Model answer
(a) \(\dfrac{5}{50} = \dfrac{25}{N}\), so \(N = \dfrac{25 \times 50}{5} = 250\). (b) The marked sheep have mixed evenly with the others, and the size of the flock has not changed.
There are 800 students in a school. Describe how to take a random sample of 20 of them. [2 marks]
Mark scheme — 2 marks available
- Numbers each of the students — B1
- Uses random numbers to choose 20 students — B1
Model answer
Give each student a different number from 1 to 800. Use a random number generator to pick 20 different numbers, and choose the students with those numbers.
What is the formula for frequency density?
Why: Frequency density \(= \dfrac{\text{frequency}}{\text{class width}}\).
What does the area of a bar in a histogram show?
Why: Area \(=\) frequency density \(\times\) class width \(=\) frequency.
A histogram bar has class width 20 and frequency density 3. What is the frequency?
Why: \(3 \times 20 = 60\).
A class of width 10 has frequency 30. What is its frequency density?
Why: \(\dfrac{30}{10} = 3\).
A survey about exercise only asks people leaving a gym. What is wrong with the sample?
Why: People at a gym are not typical of the whole population.
A school has 600 pupils, 240 in Year 10. A stratified sample of 50 is taken. How many Year 10 pupils are in the sample?
Why: \(\dfrac{240}{600} \times 50 = 20\).
50 fish are marked and released. A second sample of 40 fish contains 8 marked fish. What is the estimate of the population?
Why: \(\dfrac{50 \times 40}{8} = 250\).
A histogram bar from 30 to 40 has frequency 40. Estimate the number of values from 30 to 35.
Why: 35 is halfway through the class, so about half the frequency: 20.
Why does a larger sample usually give a better estimate?
Why: Bigger samples are less affected by chance.