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

Histograms and cumulative frequency graphs Problem set A Worked solutions

Skills

  1. Draw a histogram for each set of data.

    1. (a) The masses of 40 parcels: 0–5 kg: 4, 5–10 kg: 9, 10–15 kg: 15, 15–20 kg: 8, 20–25 kg: 4.

      • The classes are all 5 wide, so each height is the frequency.
      • Heights 4, 9, 15, 8, 4; label both axes.

    2. (b) The lengths of 70 journeys: 0–5 km: 10, 5–10 km: 15, 10–20 km: 24, 20–40 km: 21.

      • The classes are unequal, so plot frequency density = frequency ÷ class width.
      • 0–5: 10 ÷ 5 = 2
      • 5–10: 15 ÷ 5 = 3
      • 10–20: 24 ÷ 10 = 2.4
      • 20–40: 21 ÷ 20 = 1.05

  2. The histogram shows the lengths of 60 phone calls.

    051015200510152025Length (min)Frequency

    1. (a) Estimate the number of calls shorter than 7 minutes.

      • 0–5: whole bar, 5
      • 5–10: 2 of 5 units, \(\frac{2}{5}\times 15 = 6\)
      • Total: 5 + 6 = 11

    2. (b) Estimate the number of calls between 12 and 22 minutes long.

      • 10–15: 3 of 5 units, \(\frac{3}{5}\times 20 = 12\)
      • 15–20: whole bar, 10
      • 20–25: 2 of 5 units, \(\frac{2}{5}\times 10 = 4\)
      • Total: 12 + 10 + 4 = 26

  3. The table shows the lengths of 40 trout.

    Length (cm)20–2525–3030–3535–4040–45Frequency3714106

    1. (a) Complete a cumulative frequency table.

      • Running totals: 3, 10, 24, 34, 40
      • The last total is 40, the total frequency.

    2. (b) Draw the cumulative frequency graph.

      • Plot against the upper boundaries 25, 30, 35, 40, 45.
      • Start at (20, 0) and finish at (45, 40).

  4. The graph shows the masses of 50 potatoes.

    510152025303540455055605101520253035404550Mass (g)Cumulative frequency

    1. (a) Estimate the median.

      • n = 50. Go across from 25, then down: 24

    2. (b) Estimate the interquartile range.

      • Across from 12.5 and 37.5: \(Q_1 = 15\), \(Q_3 = 31\)
      • IQR \(= 31 - 15 = 16\)

    3. (c) Estimate the 90th percentile.

      • 90% of 50 is 45. Across from 45, then down: 46

    4. (d) Estimate how many potatoes were less than 20 g.

      • Up from 20, then across: 17

    5. (e) Estimate the percentage of potatoes that were more than 40 g.

      • Up from 40, across: 42
      • 50 − 42 = 8
      • \(\frac{8}{50}\times 100 = 16\%\)

  5. The scores of 11 players in a game were: 14, 22, 9, 17, 25, 19, 12, 21, 16, 28, 18.

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

      • Ordered: 9, 12, 14, 16, 17, 18, 19, 21, 22, 25, 28
      • Median 18; with the median left out of the halves when n is odd: \(Q_1 = 14\), \(Q_3 = 22\)

    2. (b) Find the interquartile range.

      • \(22 - 14 = 8\)

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

      • \(1.5\times 8 = 12\). Fences: \(14 - 12 = 2\) and \(22 + 12 = 34\)
      • All values lie between 9 and 28, so there are no outliers
      • Box plot: whiskers 9 and 28, box 14 to 22, line at 18

  6. The waiting times, in minutes, of 11 buses were: 7, 9, 8, 13, 5, 31, 10, 12, 8, 14, 11.

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

      • Ordered: 5, 7, 8, 8, 9, 10, 11, 12, 13, 14, 31
      • From the calculator: median 10, \(Q_1 = 8\), \(Q_3 = 13\)
      • IQR \(= 13 - 8 = 5\)

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

      • \(1.5\times 5 = 7.5\)
      • Lower fence \(8 - 7.5 = 0.5\). Upper fence \(13 + 7.5 = 20.5\)
      • 31 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: 5 and 14
      • Mark 31 with a cross

  7. A data set has: smallest value 4, second smallest 18, lower quartile 24, median 29, upper quartile 35, second largest 50, largest 71.

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

      • IQR \(= 35 - 24 = 11\), and \(1.5\times 11 = 16.5\)
      • Lower fence \(24 - 16.5 = 7.5\). Upper fence \(35 + 16.5 = 51.5\)
      • Outliers: 4, 71

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

      • The whiskers go to the smallest and largest values that are not outliers: 18 and 50

In context

  1. The histogram shows the ages of the visitors to a library one day.

    0102030405060102030405060Age (years)Frequency

    1. (a) How many were measured?

      • 10 + 40 + 60 + 60 + 30 = 200

    2. (b) Estimate the number between 25 and 45 years.

      • 20–30: 5 of 10 units, \(\frac{5}{10}\times 40 = 20\)
      • 30–40: whole bar, 60
      • 40–50: 5 of 10 units, \(\frac{5}{10}\times 60 = 30\)
      • Total: 20 + 60 + 30 = 110

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

      • \(\frac{110}{200}\times 100 = 55\%\)

  2. The histogram shows the distances, in metres, that 100 children threw a ball.

    05101520253035020406080100Distance (m)Frequency

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

      • Equal widths, so each bar height is a frequency: 0–20: 12, 20–40: 28, 40–60: 34, 60–80: 18, 80–100: 8

    2. (b) Estimate the mean.

      • Midpoints: 10, 30, 50, 70, 90
      • From the calculator, midpoints with frequencies: \(\bar x = 46.4\)
      • By hand: \(\sum fx = 4640\) and \(4640\div 100 = 46.4\)

  3. The cumulative frequency table shows the times, in seconds, that 100 students took to solve a puzzle.

    Time (s)≤ 30≤ 60≤ 90≤ 120≤ 150Cumulative frequency9337393100

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

      • Subtract: 9 − 0 = 9, 33 − 9 = 24, 73 − 33 = 40, 93 − 73 = 20, 100 − 93 = 7
      • Classes 0–30, 30–60, 60–90, 90–120, 120–150 with those frequencies; equal widths, so bars of those heights

    2. (b) Estimate the mean.

      • Midpoints: 15, 45, 75, 105, 135
      • \(\sum fx = 7260\), \(n = 100\)
      • \(\bar x = 72.6\)

  4. The graph shows the time, in minutes, that 60 students spent reading last night.

    5101520253035404550556051015202530354045505560Time (min)Cumulative frequency

    1. (a) Estimate the median.

      • n = 60. Across from 30, then down: 25

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

      • Across from 15: \(Q_1 = 15\). Across from 45: \(Q_3 = 36\)

    3. (c) Draw a box plot.

      • Smallest 0 and largest 60 from the ends of the graph
      • Five numbers: 0, 15, 25, 36, 60
      • Box from the lower to the upper quartile, a line at the median, whiskers to the ends

Stretch

  1. The table shows the hours 40 students spent revising in one week (group A). For group B the five-number summary was 5, 22, 30, 41, 50.

    Hours0–1010–2020–3030–4040–50Frequency248206

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

      • Running totals: 2, 6, 14, 34, 40
      • Plot against the upper boundaries 10, 20, 30, 40, 50, starting at (0, 0)

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

      • n = 40: across from 10, 20 and 30
      • \(Q_1 = 25\), median \(= 33\), \(Q_3 = 38\)

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

      • Smallest 0, largest 50
      • Box plot: 0, 25, 33, 38, 50

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

      • Medians: group A 33, group B 30. The first group has the higher median, so its revision times are typically greater.
      • IQRs: group A \(38 - 25 = 13\), group B \(41 - 22 = 19\). The second group has the larger IQR, so its revision times are more spread out.

  2. Three histograms show the times (in minutes) for 40 people. Graphs A, B and C are the cumulative frequency graphs of the same data. Match each histogram to its graph. Histogram 1: frequencies 4, 9, 14, 9, 4 for classes 0–10, 10–20, 20–30, 30–40, 40–50. Histogram 2: frequencies 16, 11, 7, 4, 2 for classes 0–10, 10–20, 20–30, 30–40, 40–50. Histogram 3: frequencies 2, 4, 7, 11, 16 for classes 0–10, 10–20, 20–30, 30–40, 40–50.

    1020304050510152025303540Time (min)Cumulative frequencyABC
    • Histogram 1 rises, then falls: the cumulative frequency graph is steepest in the middle, an S shape: B.
    • Histogram 2 has its tallest bars first: the graph climbs steeply, then flattens: C.
    • Histogram 3 has its tallest bars last: the graph is flat at first and climbs steeply at the end: A.

  3. The diagram shows three cumulative frequency graphs, A, B and C, for the times (in minutes) of three groups of 60 people. 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, 12, 18, 30, 60; Box plot Q: 0, 30, 42, 48, 60; Box plot R: 0, 27, 35, 43, 60.

    5101520253035404550556051015202530354045505560Time (min)Cumulative frequencyABC
    • Graph A: median 35, \(Q_1 = 27\), \(Q_3 = 43\) (read at n/4, n/2, 3n/4)
    • Graph B: median 18, \(Q_1 = 12\), \(Q_3 = 30\) (read at n/4, n/2, 3n/4)
    • Graph C: median 42, \(Q_1 = 30\), \(Q_3 = 48\) (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