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Maths · Further statistics

Interpreting histograms

Read frequencies from histograms using area, estimate the number in part of a class, and estimate the median and mean.

  • Higher
  • 6 key terms
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Warm-up

Answer each one, then check.

  1. 1

    What is frequency density?

    Show answerHide answer

    Frequency divided by class width

  2. 2

    Work out \(20 \times 1.5\).

    Show answerHide answer

    \(30\)

  3. 3

    What is the midpoint of the class \(20 < x \le 40\)?

    Show answerHide answer

    \(30\)

  4. 4

    How do you estimate a mean from grouped data?

    Show answerHide answer

    Add up midpoint times frequency, divide by total frequency

  5. 5

    What is the median position for 124 values?

    Show answerHide answer

    The 62nd value

Learning Objectives

  1. 1Find frequencies from a histogram using area.
  2. 2Estimate the number in part of a class.
  3. 3Estimate the median from a histogram.
  4. 4Estimate the mean from a histogram.

READING A HISTOGRAM

Frequency = frequency density \(\times\) class width. That is the area of the bar.

To estimate a frequency for part of a class, use the width of just that part.

Frequency from a Bar

A histogram bar for the class \(10 < t \le 30\) has height 1.5. Find the frequency.

Show the solutionHide the solution
  1. 1 Class width \(30 - 10 = 20\)
  2. 2 Multiply \(20 \times 1.5 = 30\)

AnswerThe frequency is 30.

Part of a Class

In the class \(20 < x \le 40\) the frequency density is 2.2. Estimate the number of values between 25 and 40.

Show the solutionHide the solution
  1. 1 Width of the part \(40 - 25 = 15\)
  2. 2 Multiply \(15 \times 2.2 = 33\)

AnswerAbout 33 values.

Estimating the Median

Frequencies of 16, 30, 44, 24 and 10 are in classes \(0-10\), \(10-20\), \(20-40\), \(40-70\), \(70-90\) (total 124). The frequency density of \(20-40\) is 2.2. Estimate the median.

Show the solutionHide the solution
  1. 1 Position \(124 \div 2 = 62\)
  2. 2 Running total \(16 + 30 = 46\), so the median is in \(20-40\)
  3. 3 Values needed in this class \(62 - 46 = 16\)
  4. 4 Width needed \(16 \div 2.2 = 7.27\)
  5. 5 Median \(20 + 7.27 = 27.3\)

AnswerThe median is about 27.3.

Estimating the Mean

Use the same data to estimate the mean.

Show the solutionHide the solution
  1. 1 Midpoints \(5,\ 15,\ 30,\ 55,\ 80\)
  2. 2 Multiply \(80 + 450 + 1320 + 1320 + 800 = 3970\)
  3. 3 Divide by 124 \(3970 \div 124 = 32.0\)

AnswerThe mean is about 32.0.

Reading Tips

Keep the method tidy.

  • Add a frequency column

    Work out each bar's frequency first.

  • Running totals

    Use them to find the median class.

  • Median

    Interpolate inside the median class.

  • Estimates

    Values from grouped data are only estimates.

Read the Bars

A histogram has bars: \(0-10\) height 1.6, \(10-20\) height 3, \(20-40\) height 2.2, \(40-70\) height 0.8, \(70-90\) height 0.5. (a) Find each frequency. (b) Find the total. (c) Estimate the number aged 30 to 40.

1. Width times height.

2. Add them up.

A good answer shows: (a) 16, 30, 44, 24, 10. (b) 124. (c) \(10 \times 2.2 = 22\).

Can I...?

  1. 1Find frequency from area.
  2. 2Find the total frequency.
  3. 3Estimate part of a class.
  4. 4Find the median class.
  5. 5Interpolate the median.
  6. 6Work out midpoints.
  7. 7Estimate the mean.
  8. 8Explain why answers are estimates.

Summary & Exam Focus

  • Frequency \(=\) frequency density \(\times\) class width.
  • Median: find the class, then interpolate.
  • Mean: \(\dfrac{\sum (\text{midpoint} \times \text{frequency})}{\text{total frequency}}\).
  • Grouped answers are estimates.

Exam focus

The histogram shows the ages of visitors to a museum. Work out an estimate for the number of visitors aged 25 to 40. (3 marks) (3 marks)

Use the class width of only the part you need.

Key terms

The vocabulary this lesson expects you to use. Each one is linked from the first place it appears above.

Frequency density
Height of a histogram bar.
Estimate
An approximate answer from grouped data.
Midpoint
The middle of a class.
Median class
The class that contains the median.
Interpolate
Find a value part of the way through a class.
Area
Width times height.

Practice questions

Have a go at each one before you open its answer.

  1. Question 1 Use the histogram 4 marks

    The histogram gives information about the ages of visitors to a museum. (a) Work out the number of visitors aged 10 to 20. (b) Work out an estimate for the number of visitors aged 25 to 40.

    A histogram of visitors' ages with unequal class widths.
    Show answerHide answer

    Model answer

    (a) \(10 \times 3 = 30\). (b) \(15 \times 2.2 = 33\).

    Mark scheme

    • \(10 \times 3\) — M1
    • 30 — A1
    • \(15 \times 2.2\) — M1
    • 33 — A1
  2. Question 2 Work out 2 marks

    Use the same histogram to work out the total number of visitors.

    Show answerHide answer

    Model answer

    \(16 + 30 + 44 + 24 + 10 = 124\)

    Mark scheme

    • All five frequencies — M1
    • 124 — A1
  3. Question 3 Estimate 3 marks

    Use the same histogram to find an estimate for the median age.

    Show answerHide answer

    Model answer

    Median position 62; running total 46 after age 20; \(62 - 46 = 16\); \(16 \div 2.2 = 7.27\); median \(\approx 27.3\).

    Mark scheme

    • Median position 62 — B1
    • Interpolates in the class 20 to 40 — M1
    • 27.3 (accept 27 to 27.5) — A1
  4. Question 4 Estimate 3 marks

    Use the same histogram to find an estimate for the mean age of the visitors.

    Show answerHide answer

    Model answer

    \(\dfrac{16 \times 5 + 30 \times 15 + 44 \times 30 + 24 \times 55 + 10 \times 80}{124} = \dfrac{3970}{124} = 32.0\)

    Mark scheme

    • Midpoints times frequencies — M1
    • 3970 — A1
    • 32.0 — A1
  5. Question 5 Work out 2 marks

    A histogram bar for a class of width 20 has height 1.4. Work out the frequency of the class.

    Show answerHide answer

    Model answer

    \(20 \times 1.4 = 28\)

    Mark scheme

    • Uses \(20 \times 1.4\) — M1
    • 28 — A1
  6. Question 6 Explain 2 marks

    Explain why the answer to part (b) of the first question is only an estimate.

    Show answerHide answer

    Model answer

    The data are grouped, so we do not know how the values are spread inside the class. We assumed they are evenly spread.

    Mark scheme

    • Data grouped — M1
    • Assumes even spread — C1

Quick check

  1. To find a frequency from a histogram bar, you...

    1. ARead the height only
    2. BMultiply height by width
    3. CDivide height by width
    4. DAdd height and width
    Show answerHide answer

    B: Multiply height by width

    Multiply the height by the width.

  2. A bar has width 10 and height 3. Frequency...

    1. A3
    2. B13
    3. C30
    4. D300
    Show answerHide answer

    C: 30

    \(10 \times 3 = 30\).

  3. To estimate the number in 25 to 40 from a class 20 to 40, use the width...

    1. A15
    2. B20
    3. C25
    4. D40
    Show answerHide answer

    A: 15

    15, the width of the part you need.

  4. For 124 values, the median is the...

    1. A31st
    2. B42nd
    3. C50th
    4. D62nd
    Show answerHide answer

    D: 62nd

    The 62nd value.

  5. The midpoint of the class 40 to 70 is...

    1. A50
    2. B55
    3. C60
    4. D30
    Show answerHide answer

    B: 55

    \((40 + 70) \div 2 = 55\).

  6. Why are mean values from a histogram estimates?

    1. AThe data are grouped
    2. BThe bars touch
    3. CThe heights are large
    4. DThe axes are labelled
    Show answerHide answer

    A: The data are grouped

    We use the midpoint of each class, not the actual values.

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