# Distribution of a sum of normal distributions?

$X$ = weight of a small bag of crisps has Normal distribution with mean = $35.5$ and $var = 0.8$ .

$Y$ = weight of a large bag of crisps has Normal distribution with mean = $152$ and $var = 3.2$

$$X \sim N(35.5, 0.8)$$ $$Y \sim N(152, 3.2)$$

What is the distribution of $Y - 4X$ ?

The answer given in the book is mean = $10$ and $var = 20.5$

I can see that the mean would be $10 \space (152 - 4 \times 35.5 = 10)$. But how do I get the variance?

What is the variance of $4X$ in this case, I think, it should be $4^2 \times Var(X) = 12.8$ and then var of $Y-4X$ would give $12.8+3.2 = 16$

so I get $Y-4X$ has normal distribution $N \sim (10,16)$.

Additional details: Due to a comment by another user, I've got a couple of questions to extend this.

Suppose I want to find the probability that a randomly selected large bag of crisps is more than four times the weight of a randomly selected small bag, would I use $W \sim N(10,16)$ and find $P(W>0)$

Also, if the question instead was, find the probability that a randomly selected large bag of crisps is more than the sum of 4 randomly selected small bags, how would this change my anaswer?

• With the numbers as given, variance is $16$. But if you are calculating some probabilities involving $4$ small bags, you may be looking at the wrong random variable. Jun 25, 2014 at 14:43
• Thanks for your answer. I'm not quite sure what you mean. Suppose I want to find the probability that a randomly selected large bag of crisps is more than four times the weight of a randomly selected small bag, wouldn't I use W~N(10,16) and find P(W>0) Jun 25, 2014 at 15:01
• I'm curious how anyone got a variance of $20.5$ out of that in the first place! Jun 25, 2014 at 15:08
• @DavidK I've asked a couple of others as well and it seems that the correct answer is indeed 16 and it's just a mistake on the book. Jun 25, 2014 at 15:12
• @DavidK I've added a a few more details to the original question. Would you please take a look at it. Really appreciate your help. Jun 25, 2014 at 15:13

So I think we all agree the book gave a wrong answer.

For the extended questions, yes, I believe $P(W > 0)$ where $W = Y - 4X \sim N(10,16)$ would tell you the probability that the large bag will be more than four times the weight of the (single) small bag, both selected at random.

If you instead select four small bags, it seems to me a reasonable model is that their weights are iid random variables $X_i$, $i = 1, 2, 3, 4$, all with distribution $X_i = N(35.5,0.8)$. In that case, the difference between the weight of the large bag and the total weight of all the small bags is $Y - \sum X_i$, and the variance of that is $Var(Y) + 4 \,Var(X_1),$ that is, $6.4$ rather than $16$.

• Thanks very much for your answer. This answers another question that has been bothering me, i.e. when to add variances vs taking n^2 times the variance Jun 25, 2014 at 16:50

We have that $-X\sim N(-\mu,\sigma)$, so regarding the variance we can indeed just sum up to get

$V(Y-4X)=V(Y)+V(4X)=V(Y)+16V(X)=3.2+16\cdot 0.8=16$.

That the variance from the "negative" X adds positively is from the fact that the Normal distribution is symmetric.

Probably not relevant for the original poster of the question but it may be of help for some future visitors. The answer given by the book for the variance is indeed correct. The variance of the lineair combination of normally distributed variables Y - 4X is calculated as follows:

V = 3.2^2 * ( (-4)^2 + 0.8^2 ) = 20.48