Page 73 - DMGT209_QUANTITATIVE_TECHNIQUES_II

Quantitative Techniques-II

Notes Solution:


Class interval Sample A Midpoint x x 1000 Fu fu
2
u =
200
700 – 900 10 800 – 1 – 10 10
900 – 1100 16 1000 0 0 0
1100 – 1300 26 1200 1 26 26
1300 – 1500 8 1400 2 16 32
60 32 68

32
u = = 0.533
A 60
= 1000 + 200
xA
 = 1000 + 200 (0.533) = 1106.67
xA
1 68 2
2
s 2 = = fu – ( u ) =  (0.533)
u N 60
= 1.133 – 0.2809
s 2 u = 0.8524

s = 0.9233
u
s = 200 × 0.9233 = 184.66
x
σ A
 CV for sample A = ×100
x A

184.66
= 100  16.68 %

1106.67
Class interval Sample B Midpoint x x 1000 fu fu 2

u =
200
700 – 900 3 800 – 1 – 3 3
900 – 1100 42 1000 0 0 0
1100 – 1300 12 1200 1 12 12
1300 – 1500 3 1400 2 6 12
60 15 27

15

V = 60 0.25
x B = 1000 + 200 V
= 1000 + 58

 x B = 1058

1 27
s 2 = fv – ( V ) = – (0.25) 2
2
2
v N 60
= 0.45 – 0.0625

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