Theory question 113
StatisticsThe table shows the frequency distribution of heights (in cm) of pupils in a certain school.
Heights (cm): 100−109, 110−119, 120−129, 130−139, 140−149, 150−159, 160−169
Frequency: 27, 58, 130, 105, 50, 25, 5
(a)(i) Construct a cumulative frequency table. (ii) Use the table to draw a cumulative frequency curve. (b) Using the curve, estimate the: (i) median height; (ii) interquartile range; (iii) percentage of students whose heights are at most 130 cm.
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(a)(i) Upper boundaries 109.5, 119.5, …, 169.5; cumulative frequencies 27, 85, 215, 320, 370, 395, 400.
(ii) Plot these (start at (99.5, 0)) and join with a smooth curve.
(b)(i) Median = 200th value ≈ 119.5 + (115/130) × 10 ≈ 128.4 cm.
(ii) Q₁ (100th) ≈ 120.7 cm, Q₃ (300th) ≈ 137.6 cm; IQR ≈ 16.9 cm.
(iii) About 220 pupils are at most 130 cm: 220/400 ≈ 55%.
(Values read from a graph may differ slightly.)
Read the median at half the total frequency on the cumulative curve.