Question 13
The table shows the marks scored by some candidates in an examination.
Marks (%) 
09 
1019 
2029 
3039 
4049 
5059 
6069 
7079 
8089 
9099 
Frequency 
7 
11 
17 
20 
29 
34 
30 
25 
21 
6 
 Construct a cumulative frequency table for the distribution and draw a cumulative frequency curve.
 Use the curve to estimate, correct to one decimal [place, the:
(i) lowest mark for distinction if 5% of the candidates passed with distinction;
(ii) probability of selecting a candidate who scored at most 45%.
Observation
The Chief Examiner reported that this question was attempted by majority of the candidates. According to the report, majority of them were able to construct the cumulative frequency table, a little less were able to determine a suitable scale and drew the cumulative frequency curve correctly but majority of them were unable to read from the curve. Teachers were encouraged to put in more effort at correcting this trend so that their performance may improve in subsequent examinations. Candidates were expected to obtain the following cumulative frequency table:
Class interval 
CLASS BOUNDARIES 
FREQUENCY 
CUMULATIVE FREQUENCY 

LOWER 
UPPER 

0 – 9 
0.5 
9.5 
7 
7 
10 – 19 
9.5 
19.5 
11 
18 
20 – 29 
19.5 
29.5 
17 
35 
30 – 39 
29.5 
39.5 
20 
55 
40 – 49 
39.5 
49.5 
29 
84 
50 – 59 
49.5 
59.5 
34 
118 
60 – 69 
59.5 
69.5 
30 
148 
70  79 
69.5 
79.5 
25 
173 
80 – 89 
79.5 
89.5 
21 
194 
90  99 
89.5 
99.5 
6 
200 
Using the table and an appropriate scale, candidates would draw the cumulative frequency curve as follows:
In part (b), candidates were expected to show that if 5% of the candidates passed with distinction, then 95% of them did not pass with distinction. Therefore, candidates would calculate 95% of 200 which was 190. They would then trace 190 from the cumulative frequency axis to the marks axis and read off the point of interception. This was obtained as 87.5% (± 1%). In part (b)(ii), candidates were expected to trace the 45% mark to the cumulative frequency axis and read off the value which was 70. Therefore, the required probability was = 0.35 (± 0.01).
 SUMMARY OF EXAMINERS VIEW OF THE STANDARD OF THE PAPER AND CANDIDATES’ PERFORMANCE

STANDARD OF THE PAPER 
CANDIDATES’ PERFORMANCE 
HIGHER THAN PREVIOUS YEARS 
5 
12 
SAME AS PREVIOUS YEARS 
45 
12 
LOWER THAN PREVIOUS YEARS 
1 
21 
CANOT BE COMPARED 
 
 
NO COMMENTS 
3 
9 
 SUGGESTED CRITICAL GRADES
X − 70  100
Y − 50  59
Z − 40  49