Weighted Mean of College A
\(\overline { x } w=\frac { \sum _{ i=1 }^{ n }{ { W }_{ i }{ X }_{ i } } }{ \sum _{ i=1 }^{ n }{ { W }_{ i } } } \) = 910000/227=4008.8
Weighted Mean of College B = 931000/238 = 3911.764
Simple Mean of College A \(\sum\)X/N = 12000/3 = 4000
Simple Mean College B \(\sum\)X/N = 12000/3 = 4000
In this situation, weighted mean is better because it gives a better analysis.

\(\overline { x } w=\frac { \sum _{ i=1 }^{ n }{ { W }_{ i }{ X }_{ i } } }{ \sum _{ i=1 }^{ n }{ { W }_{ i } } } \)
1620/70 = 95.29

Old \(\sum\)FX = 100 x 2000 = 200000
New \(\sum\)FX = 200000 - 10000 = 190000
New \(\sum\)F = 99
New Mean = 190000/99 = Rs. 1919.19

Old \(\sum\) FX = 50 x 42 = 2100
New \(\sum\)FX = 2100 + (50 x 2) = 2200
New Mean = 2200/50 = 44

Old \(\sum\)FX = 40 x 25 = 1000
New \(\sum\)FX = 1000-30 = 970
New \(\sum\)F = 39
New Mean = 970/39 = 24.87

Method I:Incorrect Mean = 100
N = 20
Incorrect \(\sum \)FX = 20 x 100 = 2000
Correct \(\sum\)FX = 2000 - (120+80)+(60+40)
Correct\(\sum\)FX = 1900
Correct Mean = 1900/20 = 95
Method II: Difference between Incorrect and correct items = - (120 + 80) + (60 + 40) = -100
Amount to be deducted = -100/20 = 5
Correct mean = 100 - 5 = 95.

\(\overline { x } 12=\frac { { N }_{ 1 }\overline { x } 1+{ N }_{ 2 }\overline { x } 2 }{ { N }_{ 1 }+{ N }_{ 2 } } \)
65 = \(\frac { (62+50)+(Y\times 30) }{ 80 } \)
30Y = 2494
Y = 2490/30 = 83

Combined Mean
\(\overline { x } 12=\frac { { N }_{ 1 }\overline { x1 } +{ N }_{ 2 }\overline { x2 } }{ { N }_{ 1 }+{ N }_{ 2 } } \)
(a) The mean will be equal to 
X = (5 x 6 + 5 x 12)/(5 + 5) = 9
(b) The mean of the combined set would be
X = (4 x 6 + 8 x 12)/(4+ 8) = 10.

 \(\overline { x } =\frac { 2460 }{ 60 } =41\)

Mean
\(\overline { x } =\frac { \sum { x } }{ n } \)
140/7 = 20
Let us verify \(\sum _{ i=1 }^{ n }{ ({ x }_{ i }-\overline { x } ) } \) = 0