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Question
The frequency distribution table below shows the height of 50 students of grade 10.
| Heights (in cm) | 138 | 139 | 140 | 141 | 142 |
| Frequency | 6 | 11 | 16 | 10 | 7 |
Find the median, the upper quartile and the lower quartile of the heights.
Solution
| Height (in cm) | Frequency (f) | Cumulative frequency |
| 138 | 6 | 6 |
| 139 | 11 | 17 |
| 140 | 16 | 33 |
| 141 | 10 | 43 |
| 142 | 7 | 50 |
No. of terms = 50
The mean of 25th and 26th term is the median
25th and 26th terms lay under 140 and 140
median = `(140 + 140)/2 = 140`
Hence, Median = 140
Upper Quartile (Q3) = `(n xx 3)/4 = (50 xx 3)/4` = 37.5th term = 141
Lower Quartile (Q1) = `n/4 = 50/4` = 12.5th term = 139
Upper Quartile = 141 and Lower Quartile = 139
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