A-level · Data presentation

Histograms and frequency polygonsExam-style questions

Why a histogram for continuous data plots frequency DENSITY, not frequency, so that the AREA of each bar is proportional to the frequency — and how that one rule lets you find a missing frequency, a missing class width, or read a frequency off a drawn bar. Plus the frequency polygon, which joins class midpoints.

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16 questions · A-level level · mark scheme and worked solution on each · no sign-up to try

DifficultyAllLevel 2Level 3Level 4Level 5
  1. 1
    The ages, xx years, of the 200 members of a bowls club are summarised in the table below.
    Age (xx years)Frequency
    20⩽x<4020 \leqslant x < 4016
    40⩽x<5540 \leqslant x < 5524
    55⩽x<6555 \leqslant x < 6540
    65⩽x<7065 \leqslant x < 7032
    70⩽x<7570 \leqslant x < 7548
    75⩽x<9075 \leqslant x < 9040
    The data are to be shown on a histogram. On this histogram, the bar for the class 55⩽x<6555 \leqslant x < 65 has a height of 2 cm.
    (a) Find the height of the tallest bar on this histogram.

    (3 marks)
    (b) Estimate the number of members of the club who are aged between 60 and 78 years.

    (2 marks)
    (c) Use linear interpolation to estimate the median age of the members.

    (2 marks)

    The upper quartile of the ages is 73.96 years, correct to 4 significant figures.
    (d) Use linear interpolation to estimate

    (i) the lower quartile,

    (ii) the interquartile range

    of the ages.

    (4 marks)
    (e) Describe the skewness of these data, giving a reason for your answer.

    (2 marks)

    The club secretary thinks that the ages of the members could be modelled by a normal distribution.
    (f) Using your answer to part (e), comment on the secretary's view.

    (1 mark)
    Level 5 · 14 marksTry it
  2. 2
    The diameters, in millimetres, of a random sample of 240 oranges from a large consignment were measured. The histogram summarises the results.556065707580859095246810121416Diameter (mm)Frequency densityAn orange is selected at random from the 240, and XX denotes its diameter in millimetres.
    (a) Show that P(80<X<85)=0.146\mathrm{P}(80 < X < 85) = 0.146, correct to 3 significant figures.

    (2 marks)

    Ella proposes the model X∼N(75,25)X \sim \mathrm{N}(75, 25).
    (b) (i) Explain why a normal distribution may be a sensible model for XX.

    (ii) Explain how Ella could have chosen the values 75 and 25 from the histogram.

    (3 marks)
    (c) Use Ella's model to find P(80<X<85)\mathrm{P}(80 < X < 85).

    (1 mark)

    Marcus prefers to estimate the mean and the standard deviation of the 240 diameters from the class midpoints, obtaining mm mm and ss mm respectively. His model is X∼N(m,s2)X \sim \mathrm{N}(m, s^2).
    (d) Use Marcus's model to find P(80<X<85)\mathrm{P}(80 < X < 85).

    (4 marks)

    The table compares probabilities found from the histogram and from the two models for three other intervals. Two of the entries, pp and qq, have been left blank.
    X<70X < 7070<X<8070 < X < 80X>85X > 85
    Histogram0.1670.6460.0417
    Ella's modelpp0.6830.0228
    Marcus's model0.204qq0.0542
    (e) (i) Find the values of pp and qq.

    (ii) Comment on how well each model fits the data, referring to the values in the table and to the probabilities you found in parts (a),
    (c) and (d).

    (4 marks)
    Level 5 · 14 marksTry it