The table below shows the mark distribution of... | WAEC SSCE - General Mathematics - 1997 (Essay)

Question 1 Report

The table below shows the mark distribution of candidates in an aptitude test for selection into the public service.

Marks (in %)	Freq
44 - 46	2
47 - 49	5
50 - 52	11
53 - 55	20
56 - 61	42
62 - 64	46
65 - 67	36
68 - 70	9
71 - 73	3

(a) Make a cumulative frequency for the distribution

(b) Draw the cumulative frequency curve.

(d) The cut-off mark was 63%. What percentage of the candidates was selected?

Answer

(a) Cumulative frequency table

Total frequency, \(N=174\).

Marks (%)	Class boundaries	Frequency	Cumulative frequency
44–46	43.5–46.5	2	2
47–49	46.5–49.5	5	7
50–52	49.5–52.5	11	18
53–55	52.5–55.5	20	38
56–61	55.5–61.5	42	80
62–64	61.5–64.5	46	126
65–67	64.5–67.5	36	162
68–70	67.5–70.5	9	171
71–73	70.5–73.5	3	174

(b) Cumulative frequency curve (ogive)

Plot the upper class boundaries against the cumulative frequencies, beginning with \((43.5,0)\), and join the points with a smooth curve.

graph — Ogive plotted using upper class boundaries and cumulative frequencies. The guide readings show a median of about 62% and a cumulative frequency of about 103 at 63%.

(c) Median mark

The median corresponds to \(\frac{N}{2}=\frac{174}{2}=87\) on the cumulative-frequency axis. From the ogive, this gives a mark of approximately \(\boxed{62\%}\).

(d) Percentage selected

At a mark of \(63\%\), the cumulative frequency read from the curve is approximately \(103\). Hence, the number selected is

\[174-103=71.\]

\[\text{Percentage selected}=\frac{71}{174}\times100\%=40.8\%\approx\boxed{41\%}.\]

Read lesson note on Loci (WAEC)

Read lesson note on Probability (WAEC)

Read lesson note on Euclidean Geometry (JAMB)

Read lesson note on Surds (radicals) (WAEC)

Answer Details

(a) Cumulative frequency table

Total frequency, \(N=174\).

Marks (%)	Class boundaries	Frequency	Cumulative frequency
44–46	43.5–46.5	2	2
47–49	46.5–49.5	5	7
50–52	49.5–52.5	11	18
53–55	52.5–55.5	20	38
56–61	55.5–61.5	42	80
62–64	61.5–64.5	46	126
65–67	64.5–67.5	36	162
68–70	67.5–70.5	9	171
71–73	70.5–73.5	3	174

(b) Cumulative frequency curve (ogive)

Plot the upper class boundaries against the cumulative frequencies, beginning with \((43.5,0)\), and join the points with a smooth curve.

(c) Median mark

The median corresponds to \(\frac{N}{2}=\frac{174}{2}=87\) on the cumulative-frequency axis. From the ogive, this gives a mark of approximately \(\boxed{62\%}\).

(d) Percentage selected

At a mark of \(63\%\), the cumulative frequency read from the curve is approximately \(103\). Hence, the number selected is

\[174-103=71.\]

\[\text{Percentage selected}=\frac{71}{174}\times100\%=40.8\%\approx\boxed{41\%}.\]

Read lesson note on Loci (WAEC)

Read lesson note on Probability (WAEC)

Read lesson note on Euclidean Geometry (JAMB)

Read lesson note on Surds (radicals) (WAEC)

Loci

Probability

Euclidean Geometry

Surds (radicals)

View Answer

The table below shows the mark distribution of candidates in an aptitude test for selection into the public service. Marks (in %) Freq 44 - 46 2 47 - 49 5 5...

Download The App On Google Playstore

The table below shows the mark distribution of candidates in an aptitude test for selection into the public service. Marks (in %) Freq 44 - 46 2 47 - 49 5 5...

Related Questions

Download The App On Google Playstore