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Grade 11 · Statistics
Statistics: Spread, Ogives & Box Plots (Grade 11)
MEMORANDUM
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  / 31

Answer all questions. Show all your working — marks are awarded for method as well as the final answer. Teacher copy: accept any correct equivalent method.

  1. 1
    The mean of 64, 71, 58, 67 and x is 66. Solve for x.
    (3)
    (64 + 71 + 58 + 67 + x)/5 = 66 (1)
    260 + x = 66 × 5 = 330 (1)
    x = 330 − 260 = 70 (1)
    ✓✓✓ (3)
  2. 2
    For the box-and-whisker diagram below, write down the five-number summary.
    (3)
    minQ₁MQ₃max12243342550102030405060
    min = 12, Q₁ = 24, median = 33, Q₃ = 42, max = 55✓✓✓ (3)
  3. 3
    Calculate the interquartile range from the diagram below.
    (2)
    minQ₁MQ₃max12243342550102030405060
    42 − 24 = 18✓✓ (2)
  4. 4
    Calculate the mean of 12 ; 15 ; 18 ; 18 ; 21 ; 24 ; 27 ; 30 ; 33.
    (2)
    198/9 = 22,00✓✓ (2)
  5. 5
    Calculate the standard deviation of 12 ; 15 ; 18 ; 18 ; 21 ; 24 ; 27 ; 30 ; 33.
    (3)
    variance = 44,00 → σ = 6,63✓✓✓ (3)
  6. 6
    How many data values lie within one standard deviation of the mean for the data above?
    (2)
    Interval [15,37 ; 28,63] contains 5 values✓✓ (2)
  7. 7
    Use the ogive below to estimate the median mark.
    (2)
    markscumulative frequency102030405051015202530354045
    The 20th of 40 values → ≈ 25 marks✓✓ (2)
  8. 8
    Use the ogive below to write down the total number of learners.
    (1)
    markscumulative frequency102030405051015202530354045
    40✓ (1)
  9. 9
    Determine the outlier boundaries for the box plot below.
    (3)
    minQ₁MQ₃max12243342550102030405060
    IQR = 18, 1,5 × 18 = 27 → below −3 or above 69; there are no outliers✓✓✓ (3)
  10. 10
    Estimate the mean of: 0–10 (f = 4), 10–20 (f = 9), 20–30 (f = 12), 30–40 (f = 5).
    (4)
    Midpoints 5, 15, 25, 35 ✓; ∑(x·f) = 20 + 135 + 300 + 175 = 630 ✓; ∑f = 30 ✓; mean = 630 ÷ 30 = 21 ✓(4)
  11. 11
    Why can the mean of grouped data only be ESTIMATED?
    (2)
    The individual values are lost once the data is grouped ✓; the midpoint is used to stand in for every value in the interval, which is an approximation ✓(2)
  12. 12
    At which point of each interval must a frequency polygon be plotted, and why?
    (2)
    At the midpoint ✓, because the midpoint represents the interval — plotting at a boundary shifts the whole distribution ✓(2)
  13. 13
    State ONE difference between a frequency polygon and an ogive.
    (2)
    A frequency polygon plots frequency and is drawn with straight lines between midpoints ✓; an ogive plots CUMULATIVE frequency and is used to read off the median and quartiles ✓(2)
TOTAL: 31 marks
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