Past papers › WAEC · 2012 · May/June · General Maths · Paper 2 › Question 10 Question WAEC General Maths 2012 Theory Statistics: data & averages Dispersion & cumulative frequency Statistics: data & averages, Dispersion & cumulative frequency
WAEC 2012 · Paper 2 · Q10
Marks
60–64
65–69
70–74
75–79
80–84
85–89
90–94
95–99
Frequency
2
3
6
11
8
7
2
1
The table shows the distribution of marks scored by students in an examination. Calculate, correct to 2 decimal places, the:
(a) (b) standard deviation of the distribution.
Worked solution (try it first) Use the class marks
62 , 67 , 72 , … , 97 62, 67, 72, \ldots, 97 62 , 67 , 72 , … , 97 .
The numbers are large, so an assumed mean
A = 77 A = 77 A = 77 keeps the arithmetic small:
d = x − 77 d = x - 77 d = x − 77 .
Marks
f f f x x x d d d f d fd f d f d 2 fd^2 f d 2
60–64
2
62
− 15 -15 − 15 − 30 -30 − 30 450
65–69
3
67
− 10 -10 − 10 − 30 -30 − 30 300
70–74
6
72
− 5 -5 − 5 − 30 -30 − 30 150
75–79
11
77
0
0
0
80–84
8
82
5
40
200
85–89
7
87
10
70
700
90–94
2
92
15
30
450
95–99
1
97
20
20
400
Total
40
70
2650
(a) Mean
= A + ∑ f d ∑ f = A + \frac{\sum fd}{\sum f} = A + ∑ f ∑ f d = 77 + 70 40 = 77 + \frac{70}{40} = 77 + 40 70 = 77 + 1.75 = 77 + 1.75 = 77 + 1.75 (b) Standard deviation
= ∑ f d 2 ∑ f − ( ∑ f d ∑ f ) 2 = \sqrt{\frac{\sum fd^2}{\sum f} - \left(\frac{\sum fd}{\sum f}\right)^2} = ∑ f ∑ f d 2 − ( ∑ f ∑ f d ) 2 = 2650 40 − 1.75 2 = \sqrt{\frac{2650}{40} - 1.75^2} = 40 2650 − 1.7 5 2 = 66.25 − 3.0625 = \sqrt{66.25 - 3.0625} = 66.25 − 3.0625 = 63.1875 = \sqrt{63.1875} = 63.1875 Watch out
With an assumed mean, subtract ( ∑ f d ∑ f ) 2 \left(\frac{\sum fd}{\sum f}\right)^2 ( ∑ f ∑ f d ) 2 x ˉ 2 \bar x^2 x ˉ 2 Use the class marks (60 + 64 2 = 62 \frac{60 + 64}{2} = 62 2 60 + 64 = 62 x x x Report a problem with this question