The data shows the marks obtained by students in a... | WAEC SSCE - General Mathematics - 2019

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

(i) median

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

Answer

(a) Frequency distribution table. Tallying the 50 marks into class intervals of width 10 gives:

Class	Frequency	Cumulative frequency (less than upper boundary)
0 - 9	1	1
10 - 19	3	4
20 - 29	3	7
30 - 39	6	13
40 - 49	8	21
50 - 59	12	33
60 - 69	6	39
70 - 79	3	42
80 - 89	4	46
90 - 99	4	50

(b) Cumulative frequency curve. Plot cumulative frequency against the upper class boundaries (9.5, 19.5, 29.5, ..., 99.5) and join the points with a smooth ogive.

(c)(i) Median. With \(N=50\), read across from \(\frac{N}{2}=25\). By interpolation in the 50 - 59 class (boundaries 49.5 - 59.5, frequency 12, cumulative 21 before it):

\[\text{Median}=49.5+\frac{25-21}{12}\times10 = 49.5+3.33 \approx 53.\]

(c)(ii) Percentage scoring at least 66 marks. Read the cumulative frequency at 66. In the 60 - 69 class, cumulative at 66 \(=33+\frac{66-59.5}{10}\times6 \approx 36.9\). So about \(50-37=13\) students scored at least 66.

\[\frac{13}{50}\times100\% \approx 26\%.\]

Read lesson note on Representation Of Data (JAMB)

Read lesson note on Inequalities (JAMB)

Read lesson note on Lengths And Perimeters (WAEC)

Answer Details

(a) Frequency distribution table. Tallying the 50 marks into class intervals of width 10 gives:

Class	Frequency	Cumulative frequency (less than upper boundary)
0 - 9	1	1
10 - 19	3	4
20 - 29	3	7
30 - 39	6	13
40 - 49	8	21
50 - 59	12	33
60 - 69	6	39
70 - 79	3	42
80 - 89	4	46
90 - 99	4	50

(b) Cumulative frequency curve. Plot cumulative frequency against the upper class boundaries (9.5, 19.5, 29.5, ..., 99.5) and join the points with a smooth ogive.

(c)(i) Median. With \(N=50\), read across from \(\frac{N}{2}=25\). By interpolation in the 50 - 59 class (boundaries 49.5 - 59.5, frequency 12, cumulative 21 before it):

\[\text{Median}=49.5+\frac{25-21}{12}\times10 = 49.5+3.33 \approx 53.\]

\[\frac{13}{50}\times100\% \approx 26\%.\]

Read lesson note on Representation Of Data (JAMB)

Read lesson note on Inequalities (JAMB)

Read lesson note on Lengths And Perimeters (WAEC)

Representation Of Data

Inequalities

Lengths And Perimeters

View Answer

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 ...

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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

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

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 ...

Related Questions

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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