The data shows the marks obtained by students in a biology test 52 56 25 56 68 73 66 64 56 48 20 39 9 50 46 54 54 40 50 96 36 44 18 97 65 21 60 44 54 32 92 ...

Assessment: WAEC SSCE - General Mathematics - 2019 (Essay) Subject: General Mathematics

Question 1 Report

The data shows the marks obtained by students in a biology test

52 56 25 56 68 73 66 64 56 48
20 39 9 50 46 54 54 40 50 96
36 44 18 97 65 21 60 44 54 32
92 49 37 94 72 88 89 35 59 34
15 88 53 16 84 52 72 46 60 42

(a) Construct a frequency distribution table using the class interval 0 - 9, 10 - 19, 19, 20, 29...

(b) Draw a cumulative frequency curve for the distribution

(c) Use the graph to estimate the;

(i) median

(ii) Percentage of students who scored at least 66 marks, correct to the nearest whole number.

Answer Details

(a) Frequency distribution

Class intervalClass boundariesFrequency, \(f\)Cumulative frequency, \(cf\)
0 - 9-0.5 - 9.511
10 - 199.5 - 19.534
20 - 2919.5 - 29.537
30 - 3929.5 - 39.5613
40 - 4939.5 - 49.5821
50 - 5949.5 - 59.51233
60 - 6959.5 - 69.5639
70 - 7969.5 - 79.5342
80 - 8979.5 - 89.5446
90 - 9989.5 - 99.5450
Total50

(b) Cumulative frequency curve (less-than ogive)

The cumulative frequencies are plotted against the upper class boundaries, beginning with \((-0.5,0)\), and joined by a smooth curve.

graph
Less-than ogive plotted from the upper class boundaries and cumulative frequencies.

(c)(i) Median

There are \(50\) students, so the median corresponds to cumulative frequency \(\frac{50}{2}=25\). Reading from \(cf=25\) on the ogive gives a mark of approximately \(53\).

Therefore, the estimated median mark is \(53\) marks.

(c)(ii) Percentage scoring at least 66 marks

From the ogive, the cumulative frequency at \(66\) marks is approximately \(37\). Hence, the estimated number scoring at least \(66\) marks is

\[50-37=13.\]

\[\text{Percentage}=\frac{13}{50}\times100\%=26\%.\]

Therefore, the percentage is \(26\%\), correct to the nearest whole number.

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