Use graph sheet for this question. Take 2 cm = 10 students along one axis and 2 cm = 5 cm of height along the other axis.
The following table shows the height of 80 students in a class.
| Height (in cm) | No. of students |
|---|---|
| 150 – 155 | 8 |
| 155 – 160 | 15 |
| 160 – 165 | 24 |
| 165 – 170 | 17 |
| 170 – 175 | 11 |
| 175 – 180 | 5 |
Draw the cumulative frequency curve for the above distribution and find the:
(a) median height.
(b) lower – quartile height.
(c) number of students whose height is more than 173 cm.
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Forming the cumulative frequency table :
| Height (in cm) | No. of students | Cumulative frequency (cf) |
|---|---|---|
| 150 – 155 | 8 | 8 |
| 155 – 160 | 15 | 23 |
| 160 – 165 | 24 | 47 |
| 165 – 170 | 17 | 64 |
| 170 – 175 | 11 | 75 |
| 175 – 180 | 5 | 80 |
Steps :
Take 2 cm = 10 students on the y-axis and 2 cm = 5 cm of height on the x-axis.
Plot the points (150, 0), (155, 8), (160, 23), (165, 47), (170, 64), (175, 75) and (180, 80).
Join the points with a free hand smooth curve to obtain the ogive.
(a) Here, $n$ = 80.
Median = $\frac{n}{2} = \frac{80}{2}$ = 40th term.
Through the point A = 40 on the y-axis, draw a horizontal line to meet the curve at B, then a vertical line down to meet the x-axis at point C = 163.5.
Hence, the median height = 163.5 cm.
(b) Lower quartile = $\frac{n}{4} = \frac{80}{4}$ = 20th term.
Through the point D = 20 on the y-axis, draw a horizontal line to meet the curve at E, then a vertical line down to meet the x-axis at point F = 159.
Hence, the lower-quartile height = 159 cm.
(c) Through the point G = 173 on the x-axis, draw a vertical line to meet the curve at H, then a horizontal line to meet the y-axis at point I = 71.
Thus, 71 students have a height less than or equal to 173 cm..
∴ Number of students with height more than 173 cm = 80 − 71 = 9.
Hence, the number of students whose height is more than 173 cm = 9.