# S3 Variance question - linear combinationWatch

This discussion is closed.
#1
The following question from the new specimen on the OCR website:

Boxes of matches contain 50 matches. Full boxes have mean mass 20.0 grams and standard deviation
0.4 grams. Empty boxes have mean mass 12.5 grams and standard deviation 0.2 grams. Stating any
assumptions that you need to make, calculate the mean and standard deviation of the mass of a match.

Mass is easy enough - 0.15g

Now for standar deviation

Letting M represent a match and B an empty box

Var(50M+B) = 50Â²Var(M) + Var(B)
0.4Â² = 50Â²Var(M) + 0.2Â²
is how I think it should be worked out, using Var(aX+bY )=aÂ²Var(X )+bÂ²Var(Y)

But why does the mark scheme give 0.4Â² = 0.2Â² + 50σÂ², ie not 50Â²?
0
13 years ago
#2
You have taken one match and multiplied its mass by 50.

That gives a different distribution to adding the masses of 50 different matches.

The rule you need is:

Var( X(i)) = Var(X(i))

for independent random variables X(1), X(2), X(3), ... .
0
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