Marks 10 20 30 40 50 60 70 80 90 Frequency 1 1 x 5... | WAEC SSCE - General Mathematics - 2018

Question 1 Report

Marks	10	20	30	40	50	60	70	80	90
Frequency	1	1	x	5	y	1	4	3	1

The frequency distribution shows the marks distribution of a class of 30 students in an examination.

The mean mark of the distribution is 52.

(a) Find the values of x and y.

(b) Construct a group frequency distribution table starting with a lower class limit of 1 and class interval of 10.

(d) Use the histogram to estimate the mode.

Answer

Answer Details

(a) Determination of \(x\) and \(y\)

Since there are 30 students,

\[16+x+y=30 \quad\Rightarrow\quad x+y=14\tag{1}\]

Also,

\[52=\frac{900+30x+50y}{30}\]

\[1560=900+30x+50y\]

\[3x+5y=66.\tag{2}\]

From (1), \(x=14-y\). Hence,

\[3(14-y)+5y=66\]

\[42+2y=66\Rightarrow y=12.\]

Therefore, \(x=14-12=2\).

\(\boxed{x=2,\ y=12}\)

(b) Grouped frequency distribution

(c) Histogram

(d) Mode

The modal class is \(41-50\). Reading the peak position from the histogram by the usual intersecting-lines method gives a mode of approximately

\[\boxed{44\text{ marks}}\]

Marks 10 20 30 40 50 60 70 80 90 Frequency 1 1 x 5 y 1 4 3 1 The frequency distribution shows the marks distribution of a class of 30 students in an examina...