1. Correct. The answer is true.

 

To simplify the calculations subtract, some number, e.g. 35, from all the data without affecting the values of variation. We then obtain the table 2.

 

3

4

5

4

2

4

3

3

4

6

5

5

 

a) The scenario (row) means for table 9.2 are given respectively by

 

`x1=(3+4+5+4)/4=4, `x2=(2+4+3+30/3=3, `x3=(4+6+5+5)/4=5.

Thus the mean prices, obtained by adding 35, are 39, 38 and 40 for scenarios A, B and C respectively.

 

b) `x=(3+4+5+4+2+4+3+3+4+6+5+5)/12=4

 

Thus the grand mean for the original set of data is 39.

 

c) Total variation = v= Σj,k(xj,k-`x)2=(3-4)2+(4-4)2+(5-4)2+(4-4)2+(2-4)2+(4-4)2+(3-4)2+(3-4)2+(4-4)2+(6-4)2+(5-4)2+(5-4)2=14

d) Total variation between scenarios=vb=j(`xj,-`x)2=4[(4-4)2+(3-4)2+(5-4)2]=8.

e) Total variation within scenarios=vw=v-vb=14-8=6

 

Another method:

vw = Σj,k(xj,k-`xj)2=(3-4)2+(4-4)2+(5-4)2+(4-4)2+(2-3)2+(4-3)2+(3-3)2+(3-3)2+(4-5)2+(6-5)2+(5-5)2+(5-5)2=6