Mathematics · AQA 7357 · Topic L

Data presentation and interpretation

Showing and summarising data: diagrams, averages, spread, correlation and regression, outliers and cleaning data.

  • 4 sub-topics
  • Taught in Year 12
  • 55 practice questions
  • 4 animated lessons
Practice questions (55) Lessons Worksheets All AQA Mathematics topics

What the specification says

Every sub-topic AQA can examine in this topic, with its reference number from the official specification. The Y12 and Y13 tags show when each sub-topic is usually taught.

  1. L1Diagrams for single-variable data, including histogramsY12
  2. L2Scatter diagrams, regression and correlationY12
  3. L3Averages, spread and standard deviationY12
  4. L4Outliers and cleaning dataY12

Try 3 questions

Easy, medium and hard questions from this topic, each with a worked solutions and the mark for every step. All 55 questions in this topic, free, no account needed.

Data presentation and interpretation

Easy

Averages · Year 12 · 3 marks

3, 7, 7, 8, 10, 123,\ 7,\ 7,\ 8,\ 10,\ 12
  1. Find the mean, median and mode.
    (3)
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  1. Add them up and divide by how many there are.Mean =476=7.83= \dfrac{47}{6} = 7.83
  2. The middle of six values is halfway between the 3rd and 4th.Median =7+82=7.5= \dfrac{7 + 8}{2} = 7.5
  3. The most common value.Mode =7= 7

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Data presentation and interpretation

Medium

Grouped data · Year 12 · 6 marks

Times, tt minutes: 0≤t<100 \le t < 10: 5; 10≤t<2010 \le t < 20: 12; 20≤t<3020 \le t < 30: 8.
  1. (a)
    Estimate the mean.
    (3)
  2. (b)
    Use linear interpolation to estimate the median.
    (3)
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Part (a)3 marks

  1. With grouped data, use each class's midpoint.Midpoints 5, 15, 25
  2. 5(5)+12(15)+8(25)25\dfrac{5(5) + 12(15) + 8(25)}{25}=40525{}= \dfrac{405}{25}
  3. =16.2= 16.2 minutes

Part (b)3 marks

  1. 5 values come before this class.The median is the 12.5th value, in the class 10≤t<2010 \le t < 20
  2. It is 7.5 of the 12 values into the class.1010+12.5−512×10{}+ \dfrac{12.5 - 5}{12} \times 10
  3. =16.25= 16.25 minutes

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Data presentation and interpretation

Hard

Combining two groups · Year 12 · 6 marks

Group A: 10 values, mean 12, standard deviation 2. Group B: 15 values, mean 14, standard deviation 3.
  1. Find the mean and standard deviation of all 25 values together.
    (6)

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  1. Total of each group = number × mean.Mean =10(12)+15(14)25= \dfrac{10(12) + 15(14)}{25}=13.2{}= 13.2
  2. From σ2\sigma^2=∑x2n{}= \dfrac{\sum x^2}{n}−xˉ2{}- \bar{x}^2.∑xA2=10(22+122)=1480\sum x_A^2 = 10(2^2 + 12^2) = 1480
  3. ∑xB2=15(32+142)=3075\sum x_B^2 = 15(3^2 + 14^2) = 3075
  4. σ2\sigma^2=455525−13.22{}= \dfrac{4555}{25} - 13.2^2=7.96{}= 7.96
  5. σ=2.82\sigma = 2.82

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