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November 28th, 2016, 04:34 AM   #1
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Talking Linear combination of expectancy and variance

6 The weights, in kilograms, of men and women have the distributions N(78,7^2) and N(66, 5^2) respectively.

(i) The maximum load that a certain cable car can carry safely is 1200kg.
If 9 randomly chosen men and 7 randomly chosen women enter the cable car, find the probability that the cable car can operate safely. [5]

(i do understand that there's a mark scheme for this A level problem, but what I don't get is why the variance doesn't need the 7 and 9 to be squared. Can anyone explain to me when to square or not square.)

I've posted this in the high school forum, just realized that this is heavily anchored in advanced statistics. sorry for the repost XD
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December 17th, 2016, 07:52 PM   #2
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What do you mean by 'the variance doesn't need the 7 and 9 to be squared'? The variance in this case, assuming independence, should be $\displaystyle 9 \times 7^2 + 7 \times 5^2$, which you can derive easily with the normal MGF.

Edit: I think I understand now what you're getting at - you don't understand why 7 and 9 aren't squared.

Recall that the MGF of a normal distribution is $\displaystyle \exp{(\mu t + \frac{1}{2} \sigma^2 t^2)}$ and that $\displaystyle M_{X+Y}(t) = M_X(t) M_Y(t)$. Then, if the weight of the ith man is $\displaystyle X_i$ and that of the ith woman is $\displaystyle Y_i$, we have $\displaystyle M_{\sum_{i=1}^9 X_i + \sum_{j=1}^7 Y_i}(t) = \Pi_{i=1}^9 \exp{(78t + \frac{1}{2}\times 7^2 t^2)}\Pi_{i=1}^7 \exp{(66t + \frac{1}{2}\times 5^2 t^2)} = (\exp{(78t + \frac{1}{2}\times 7^2 t^2)})^7 (\exp{(66t + \frac{1}{2}\times 5^2 t^2)})^9 = \exp{((9 \times 78 + 7 \times 66)t + \frac{1}{2}\times (9 \times 7^2 + 7 \times 5^2) t^2)}$.
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Last edited by 123qwerty; December 17th, 2016 at 08:49 PM.
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December 17th, 2016, 08:32 PM   #3
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December 17th, 2016, 08:51 PM   #4
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