The ages of a group of athletes are as follows:... | WAEC SSCE - General Mathematics - 2019

Question 1 Report

The ages of a group of athletes are as follows: 18, 16. 18,20, 17, 16, 19, 17, 18, 17 and 13. (a) Find the range of the distribution.

(b) Draw a frequency distribution table for the data.

(d) Calculate, correct to two decimal places, the;

(i) mean age:

(ii) standard deviation.

Answer

Arranging the ages in ascending order gives:

\[13,\ 16,\ 16,\ 17,\ 17,\ 17,\ 18,\ 18,\ 18,\ 19,\ 20\]

(a) Range

\[\text{Range}=20-13=7\text{ years}.\]

(b) Frequency distribution table

Age, \(x\)	Tally	Frequency, \(f\)	Cumulative frequency	\(fx\)	\(fx^2\)
13	I	1	1	13	169
16	II	2	3	32	512
17	III	3	6	51	867
18	III	3	9	54	972
19	I	1	10	19	361
20	I	1	11	20	400
Total		\(\sum f=11\)		\(\sum fx=189\)	\(\sum fx^2=3281\)

(c) Median age

There are \(11\) observations. Thus the median is the \(\frac{11+1}{2}=6\)th observation.

\[\text{Median}=17\text{ years}.\]

(d)(i) Mean age

\[\bar{x}=\frac{\sum fx}{\sum f}=\frac{189}{11}=17.1818\ldots\]

\[\boxed{\text{Mean age}=17.18\text{ years}}\]

(d)(ii) Standard deviation

\[\sigma=\sqrt{\frac{\sum fx^2}{\sum f}-\left(\frac{\sum fx}{\sum f}\right)^2}\]

\[\sigma=\sqrt{\frac{3281}{11}-\left(\frac{189}{11}\right)^2}=\sqrt{3.0579\ldots}=1.7487\ldots\]

\[\boxed{\text{Standard deviation}=1.75\text{ years}}\]

Read lesson note on Measures Of Dispersion (JAMB)

Read lesson note on Statistics (WAEC)

Answer Details