(a) The table shows the distribution of marks scored by some candidates in an examination.
| Marks | 11 – 20 | 21 – 30 | 31 – 40 | 41 – 50 | 51 – 60 | 61 – 70 | 71 – 80 | 81 – 90 |
91 – 100 |
| Num of candidates | 5 | 39 | 14 | 40 | 57 | 25 | 11 | 8 | 1 |
Construct a cumulative frequency table for the distribution.
(b) The table shows the distribution of marks scored by some candidates in an examination.
| Marks | 11 – 20 | 21 – 30 | 31 – 40 | 41 – 50 | 51 – 60 | 61 – 70 | 71 – 80 | 81 – 90 |
91 – 100 |
| Num of candidates | 5 | 39 | 14 | 40 | 57 | 25 | 11 | 8 | 1 |
Draw a cumulative frequency curve for the distribution.
(ci) Use the curve to estimate the:
number of candidates who scored marks between 24 and 58 ;
(cii) Use the curve to estimate the:
lowest mark for distinction, if 12% of the candidates passed with distinction.
Explanation
(a)
| Marks | Class Boundaries | Frequency | Cumulative Frequency |
| 11 - 20 | 10.5 - 20.5 | 5 | 5 |
| 21 - 30 | 20.5 - 30.5 | 39 | 5+39=44 |
| 31 - 40 | 30.5 - 40.5 | 14 | 44+14=58 |
| 41 - 50 | 40.5 - 50.5 | 40 | 58+40=98 |
| 51 - 60 | 50.5 - 60.5 | 57 | 98+57=155 |
| 61 - 70 | 60.5 - 70.5 | 25 | 155+25=180 |
| 71 - 80 | 70.5 - 80.5 | 11 | 180+11=191 |
| 81 - 90 | 80.5 - 90.5 | 8 | 191+8=199 |
| 91 - 100 | 90.5 - 100.5 | 1 | 199+1=200 |
(b)
(ci) The number of candidates who scored marks between 24 and 58 = 146 - 14 = 132
(cii) if 12% of the candidates passed with distinction then 88% did not pass with distinction
⇒\(88% of 200 = \frac{88}{100}\times 200=176\)
From the cumulative frequency curve, 176 corresponds to 69 marks
∴ The lowest mark for distinction = 69 marks