The table shows the frequency distribution of heights (in cm) of pupils in a certain school. Heights 100-109 110-119 120-129 130-139 140-149 150-159 160-169...
Question 1 Report
The table shows the frequency distribution of heights (in cm) of pupils in a certain school.
| Heights | 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) inter quartile range (iii) percentage of students whose heights are most 130cm.
Answer Details
(a)(i) Cumulative frequency table (cf against upper boundaries). \(N = 400\).
| Height | Freq | Upper boundary | Cumulative freq |
|---|---|---|---|
| 100 - 109 | 27 | 109.5 | 27 |
| 110 - 119 | 58 | 119.5 | 85 |
| 120 - 129 | 130 | 129.5 | 215 |
| 130 - 139 | 105 | 139.5 | 320 |
| 140 - 149 | 50 | 149.5 | 370 |
| 150 - 159 | 25 | 159.5 | 395 |
| 160 - 169 | 5 | 169.5 | 400 |
(a)(ii) Plot (109.5, 27), (119.5, 85), ... , (169.5, 400) and join with a smooth ogive.
(b)(i) Median at position \( \tfrac{400}{2} = 200\), in class 120 - 129 (\(L=119.5, \text{cf}=85, f=130\)):
\[ \text{Median} = 119.5 + \frac{200 - 85}{130}\times 10 = 119.5 + 8.85 \approx \mathbf{128.3 \text{ cm}} \](b)(ii) Interquartile range. \(Q_1\) at 100th (class 120 - 129):
\[ Q_1 = 119.5 + \frac{100 - 85}{130}\times 10 = 119.5 + 1.15 = 120.65 \]\(Q_3\) at 300th (class 130 - 139, \(L=129.5, \text{cf}=215, f=105\)):
\[ Q_3 = 129.5 + \frac{300 - 215}{105}\times 10 = 129.5 + 8.10 = 137.60 \] \[ \text{IQR} = Q_3 - Q_1 = 137.60 - 120.65 \approx \mathbf{17.0 \text{ cm}} \](b)(iii) Percentage with height at most 130 cm. Read cf at 130 (class 130 - 139):
\[ \text{cf}(130) = 215 + \frac{130 - 129.5}{10}\times 105 = 215 + 5.25 = 220.25 \] \[ \text{Percentage} = \frac{220.25}{400}\times 100 \approx \mathbf{55\%} \]Download The App On Google Playstore
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