Past papers › WAEC · 2020 · Private · General Maths · Paper 2 › Question 3 Question WAEC General Maths 2020 Theory Statistics: data & averages Dispersion & cumulative frequency Statistics: data & averages, Dispersion & cumulative frequency
The ages of some lecturers are 42, 54, 50, 54, 50, 42, 46, 46, 48 and 48. Calculate the:
(a) (b) Worked solution (try it first) (a) The ages add up to
42 + 54 + 50 + 54 + 50 + 42 + 46 + 46 + 48 + 48 = 480 42 + 54 + 50 + 54 + 50 + 42 + 46 + 46 + 48 + 48 = 480 42 + 54 + 50 + 54 + 50 + 42 + 46 + 46 + 48 + 48 = 480 , and there are 10, so the mean is
480 10 = 48 \frac{480}{10} = 48 10 480 = 48 years.
(b) Set out a table.
Each age appears twice, so
f = 2 f = 2 f = 2 for each:
x x x 42
46
48
50
54
Total
f f f 2
2
2
2
2
10
x − 48 x - 48 x − 48 − 6 -6 − 6 − 2 -2 − 2 0
2
6
f ( x − 48 ) 2 f(x - 48)^2 f ( x − 48 ) 2 72
8
0
8
72
160
Standard deviation
= 160 10 = 16 = 4 = \sqrt{\frac{160}{10}} = \sqrt{16} = 4 = 10 160 = 16 = 4 years.
Watch out
Show the table of deviations and their squares; a bare answer with no table loses the method marks. Take the square root at the end: the variance is 16, the standard deviation is 4. Report a problem with this question