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IB Mathematics: Applications and Interpretation SL · 13.3: Statistics · 13.3.5

Histograms and cumulative frequency graphs Problem set B Worked solutions

Skills

  1. Draw a histogram for each set of data.

    1. (a) The times taken by 50 pupils to solve a puzzle: 0–3 min: 5, 3–6 min: 12, 6–9 min: 17, 9–12 min: 11, 12–15 min: 5.

      • The classes are all 3 wide, so each height is the frequency.
      • Heights 5, 12, 17, 11, 5; label both axes.

    2. (b) The ages of 60 people at a sports club: 10–15 years: 9, 15–20 years: 21, 20–30 years: 20, 30–50 years: 10.

      • The classes are unequal, so plot frequency density = frequency ÷ class width.
      • 10–15: 9 ÷ 5 = 1.8
      • 15–20: 21 ÷ 5 = 4.2
      • 20–30: 20 ÷ 10 = 2
      • 30–50: 10 ÷ 20 = 0.5

  2. The histogram shows the lengths of 80 fish.

    051015202581216202428Length (cm)Frequency

    1. (a) Estimate the number of fish shorter than 14 cm.

      • 8–12: whole bar, 10
      • 12–16: 2 of 4 units, \(\frac{2}{4}\times 24 = 12\)
      • Total: 10 + 12 = 22

    2. (b) Estimate the number of fish between 17 and 26 cm long.

      • 16–20: 3 of 4 units, \(\frac{3}{4}\times 20 = 15\)
      • 20–24: whole bar, 16
      • 24–28: 2 of 4 units, \(\frac{2}{4}\times 10 = 5\)
      • Total: 15 + 16 + 5 = 36

  3. The table shows the heights of 50 plants.

    Height (cm)10–2020–3030–4040–5050–60Frequency41119124

    1. (a) Complete a cumulative frequency table.

      • Running totals: 4, 15, 34, 46, 50
      • The last total is 50, the total frequency.

    2. (b) Draw the cumulative frequency graph.

      • Plot against the upper boundaries 20, 30, 40, 50, 60.
      • Start at (10, 0) and finish at (60, 50).

  4. The graph shows the journey times of 40 commuters.

    1015202530354045505560510152025303540Time (min)Cumulative frequency

    1. (a) Estimate the median.

      • n = 40. Go across from 20, then down: 27

    2. (b) Estimate the interquartile range.

      • Across from 10 and 30: \(Q_1 = 22\), \(Q_3 = 35\)
      • IQR \(= 35 - 22 = 13\)

    3. (c) Estimate the 90th percentile.

      • 90% of 40 is 36. Across from 36, then down: 45

    4. (d) Estimate how many commuters were less than 30 minutes.

      • Up from 30, then across: 26

    5. (e) Estimate the percentage of commuters that were more than 50 minutes.

      • Up from 50, across: 38
      • 40 − 38 = 2
      • \(\frac{2}{40}\times 100 = 5\%\)

  5. The numbers of goals scored by 12 teams in a season were: 31, 24, 28, 35, 40, 27, 33, 38, 29, 36, 25, 32.

    1. (a) Find the median and the quartiles.

      • Ordered: 24, 25, 27, 28, 29, 31, 32, 33, 35, 36, 38, 40
      • Median 31.5; the lower half has median \(Q_1 = 27.5\), the upper half \(Q_3 = 35.5\)

    2. (b) Find the interquartile range.

      • \(35.5 - 27.5 = 8\)

    3. (c) Show that there are no outliers, and draw a box plot.

      • \(1.5\times 8 = 12\). Fences: \(27.5 - 12 = 15.5\) and \(35.5 + 12 = 47.5\)
      • All values lie between 24 and 40, so there are no outliers
      • Box plot: whiskers 24 and 40, box 27.5 to 35.5, line at 31.5

  6. The lengths, in centimetres, of 10 leaves were: 45, 52, 3, 41, 48, 44, 38, 50, 42, 47.

    1. (a) Find the median, the quartiles and the interquartile range.

      • Ordered: 3, 38, 41, 42, 44, 45, 47, 48, 50, 52
      • From the calculator: median 44.5, \(Q_1 = 41\), \(Q_3 = 48\)
      • IQR \(= 48 - 41 = 7\)

    2. (b) Test whether there are any outliers.

      • \(1.5\times 7 = 10.5\)
      • Lower fence \(41 - 10.5 = 30.5\). Upper fence \(48 + 10.5 = 58.5\)
      • 3 is outside the fences, so it is an outlier

    3. (c) Draw a box plot.

      • Whiskers end at the most extreme values that are not outliers: 38 and 52
      • Mark 3 with a cross

  7. A data set has: smallest value 15, second smallest 19, lower quartile 30, median 38, upper quartile 46, second largest 57, largest 82.

    1. (a) Determine whether the data set has any outliers.

      • IQR \(= 46 - 30 = 16\), and \(1.5\times 16 = 24\)
      • Lower fence \(30 - 24 = 6\). Upper fence \(46 + 24 = 70\)
      • Outliers: 82

    2. (b) State where the whiskers of the box plot end.

      • The whiskers go to the smallest and largest values that are not outliers: 15 and 57

In context

  1. The histogram shows the lengths of the fish caught in a competition.

    0510152025303540203040506070Length (cm)Frequency

    1. (a) How many were measured?

      • 10 + 25 + 40 + 20 + 5 = 100

    2. (b) Estimate the number between 36 and 55 cm.

      • 30–40: 4 of 10 units, \(\frac{4}{10}\times 25 = 10\)
      • 40–50: whole bar, 40
      • 50–60: 5 of 10 units, \(\frac{5}{10}\times 20 = 10\)
      • Total: 10 + 40 + 10 = 60

    3. (c) Estimate this as a percentage of the total.

      • \(\frac{60}{100}\times 100 = 60\%\)

  2. The histogram shows the times, in minutes, that 60 customers spent in a shop.

    051015202581216202428Time (min)Frequency

    1. (a) Write the data as a grouped frequency table.

      • Equal widths, so each bar height is a frequency: 8–12: 6, 12–16: 15, 16–20: 24, 20–24: 10, 24–28: 5

    2. (b) Estimate the mean.

      • Midpoints: 10, 14, 18, 22, 26
      • From the calculator, midpoints with frequencies: \(\bar x = 17.5\)
      • By hand: \(\sum fx = 1052\) and \(1052\div 60 = 17.5333\)

  3. The cumulative frequency table shows the ages, in years, of 80 people on a bus.

    Age (years)≤ 10≤ 20≤ 30≤ 40≤ 50Cumulative frequency725527080

    1. (a) Draw a histogram of the data.

      • Subtract: 7 − 0 = 7, 25 − 7 = 18, 52 − 25 = 27, 70 − 52 = 18, 80 − 70 = 10
      • Classes 0–10, 10–20, 20–30, 30–40, 40–50 with those frequencies; equal widths, so bars of those heights

    2. (b) Estimate the mean.

      • Midpoints: 5, 15, 25, 35, 45
      • \(\sum fx = 2060\), \(n = 80\)
      • \(\bar x = 25.8\)

  4. The graph shows the masses, in grams, of 80 eggs.

    1015202530354045505560612182430364248546066727884Mass (g)Cumulative frequency

    1. (a) Estimate the median.

      • n = 80. Across from 40, then down: 35

    2. (b) Estimate the lower and upper quartiles.

      • Across from 20: \(Q_1 = 28\). Across from 60: \(Q_3 = 42\)

    3. (c) Draw a box plot.

      • Smallest 10 and largest 60 from the ends of the graph
      • Five numbers: 10, 28, 35, 42, 60
      • Box from the lower to the upper quartile, a line at the median, whiskers to the ends

Stretch

  1. The table shows the times, in minutes, that 60 pupils took to cycle to school (group A). For group B the five-number summary was 15, 28, 41, 52, 60.

    Minutes10–2020–3030–4040–5050–60Frequency510251010

    1. (a) Complete a cumulative frequency table and draw the graph.

      • Running totals: 5, 15, 40, 50, 60
      • Plot against the upper boundaries 20, 30, 40, 50, 60, starting at (10, 0)

    2. (b) Estimate the median and the quartiles.

      • n = 60: across from 15, 30 and 45
      • \(Q_1 = 30\), median \(= 36\), \(Q_3 = 45\)

    3. (c) Draw a box plot for the data.

      • Smallest 10, largest 60
      • Box plot: 10, 30, 36, 45, 60

    4. (d) Compare the two groups. Refer to the median and the interquartile range.

      • Medians: group A 36, group B 41. The second group has the higher median, so its journey times are typically greater.
      • IQRs: group A \(45 - 30 = 15\), group B \(52 - 28 = 24\). The second group has the larger IQR, so its journey times are more spread out.

  2. Three histograms show the masses (in grams) of 50 items. Graphs A, B and C are the cumulative frequency graphs of the same data. Match each histogram to its graph. Histogram 1: frequencies 5, 12, 18, 10, 5 for classes 0–20, 20–40, 40–60, 60–80, 80–100. Histogram 2: frequencies 20, 14, 9, 5, 2 for classes 0–20, 20–40, 40–60, 60–80, 80–100. Histogram 3: frequencies 2, 5, 9, 14, 20 for classes 0–20, 20–40, 40–60, 60–80, 80–100.

    204060801005101520253035404550Mass (g)Cumulative frequencyABC
    • Histogram 1 rises, then falls: the graph is an S shape, steepest in the middle: C.
    • Histogram 2 has its tallest bars first: the graph climbs steeply, then flattens: A.
    • Histogram 3 has its tallest bars last: the graph is flat at first, then climbs steeply: B.

  3. The diagram shows three cumulative frequency graphs, A, B and C, for the lengths (in centimetres) of three batches of 80 plants. Three box plots P, Q and R show the same data in some order. Match each box plot to its graph. The five-number summaries are: Box plot P: 0, 25, 33, 38, 50; Box plot Q: 0, 20, 25, 30, 50; Box plot R: 0, 12, 17, 25, 50.

    5101520253035404550612182430364248546066727884Length (cm)Cumulative frequencyABC
    • Graph A: median 25, \(Q_1 = 20\), \(Q_3 = 30\) (read at n/4, n/2, 3n/4)
    • Graph B: median 17, \(Q_1 = 12\), \(Q_3 = 25\) (read at n/4, n/2, 3n/4)
    • Graph C: median 33, \(Q_1 = 25\), \(Q_3 = 38\) (read at n/4, n/2, 3n/4)
    • Match each graph's quartiles to the box plot with the same median and box.

jmaths.xyz · AI SL · 13.3.5 Histograms and cumulative frequency graphs