The table shows the distribution of marks scored... | WAEC SSCE - Further Mathematics - 2016

Question 1 Report

The table shows the distribution of marks scored by some students in a test.

Marks	1-10	11-20	21-30	31-40	41-50	51-60	61-70	71-80	81-90	91-100
No. of students	3	17	41	85	97	115	101	64	21	6

(a)(i) Construct a cumulative frequency table for the distribution ; (ii) Draw a cumulative frequency curve for the distribution.

(b) Use the curve to estimate the :

(i) number of students who scored marks between 32 and 74 ; (ii) pass mark, if 18% of the students failed ; (iii) lowest mark for distinction, if 8% of the students passed with distinction.

Answer

(a)(i) Cumulative frequency table

The cumulative frequency is obtained by adding each class frequency to the running total. The curve is plotted against the upper class boundary of each class, so these are tabulated as well. The total number of students is \(N = 550\).

Marks	Class boundaries	Frequency \(f\)	Cumulative frequency \(\sum f\)
1 - 10	0.5 - 10.5	3	3
11 - 20	10.5 - 20.5	17	20
21 - 30	20.5 - 30.5	41	61
31 - 40	30.5 - 40.5	85	146
41 - 50	40.5 - 50.5	97	243
51 - 60	50.5 - 60.5	115	358
61 - 70	60.5 - 70.5	101	459
71 - 80	70.5 - 80.5	64	523
81 - 90	80.5 - 90.5	21	544
91 - 100	90.5 - 100.5	6	550

(a)(ii) Cumulative frequency curve (ogive)

Plot the cumulative frequency (vertical axis) against the upper class boundary (horizontal axis), starting the curve from \((0.5,\,0)\), and join the points with a smooth S-shaped curve.

graph — Ogive of cumulative frequency against upper class boundary; N = 550. Readings: CF(32) approx 74, CF(74) approx 481, mark at CF=99 approx 35, mark at CF=506 approx 78.

(b) Estimates read from the curve

(i) Number of students scoring between 32 and 74. Read the cumulative frequency at each mark from the ogive (dashed guide lines):

\[\text{CF}(32)\approx 61+\frac{32-30.5}{10}\times 85 \approx 74\]\[\text{CF}(74)\approx 459+\frac{74-70.5}{10}\times 64 \approx 481\]

Number scoring between 32 and 74:

\[481-74 = 407\ \text{students (approximately)}.\]

(ii) Pass mark, if 18% failed. The number who failed is

\[18\%\ \text{of}\ 550 = \frac{18}{100}\times 550 = 99.\]

The pass mark is the mark below which 99 students lie, i.e. the mark corresponding to \(\text{CF}=99\). Reading horizontally from \(99\) on the vertical axis to the curve and down to the mark axis:

\[\text{pass mark} \approx 30.5+\frac{99-61}{85}\times 10 \approx 35.\]

(iii) Lowest mark for distinction, if 8% passed with distinction. The distinction group is the top 8%, so the number below the distinction mark is

\[550-\Big(8\%\ \text{of}\ 550\Big)=550-44=506.\]

Read the mark corresponding to \(\text{CF}=506\):

\[\text{lowest distinction mark} \approx 70.5+\frac{506-459}{64}\times 10 \approx 78.\]

Read lesson note on Statistics (WAEC)

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