# Probability that a box of $3$ liters of wine is sufficient?

Maybe I'm a bit tired now or I just don't know this but here is the question:

You want to invite people on a party. Assume that 10 people came to the party and everyone drinks wine with a quantity that is normal distributed with mean $25$cl and standard deviation $10$cl. What is the probability that a box of $3$ liters of wine will suffice?

Attempt: $3$ liters $=$ $300$ cl. Let the total wine being drunk be $D$. Since there are $10$ their total mean is $250$ cl.

For the total variance, I know that $k\cdot N(\mu,\sigma^2)=N(k\cdot250,k^2\cdot10^2)$, so $k=10$ gives the convoluted distribution $N(250,\color{red}{10000}).$

So now we can compute

$$P(D\le300)=\Phi\left(\frac{300-250}{100}\right)=\Phi\left(0.5\right)=0.69.$$

However, the correct answer seems to be $0.943$ because the variance is not my red $10000$ but it's $1000$. Why did I get wrong variance?

• The wiki page suggests that you just sum the variances. This would give $N(250k, 10^2 k)$. Would this match your expected result? – rwbogl Aug 20 '18 at 23:02
• How would it give that? You have multiplied $k$ with $10^2$. In my case there is no other random variable so there is not a convolution going on in that sense you suggested I believe? Look here: math.stackexchange.com/questions/1865323/… – Parseval Aug 20 '18 at 23:15
• If the ten people are independent normals, wouldn't the amount they drink be the sum of ten normals rather than the product of one normal by 10? If I remember correctly, this is different from multiplying a single normal by a constant. (If $X \sim Y$, the variable $X + Y$ is different from $2X$.) – rwbogl Aug 20 '18 at 23:20
• Having just read your link, I think that the difference is what I mentioned about being the sum of different, independent normals. At least, that's the only difference I can think of off the top of my head. – rwbogl Aug 20 '18 at 23:23
• Ahh...you're right. I did not consider them as random variables only as 10 as a constant. Ofcourse now it made sense and I also get the correct answer. Thanks! – Parseval Aug 20 '18 at 23:25

In particular, if $X \sim N(\mu, \sigma^2)$, then, as you mentioned, $$k X \sim N(k \mu, (k\sigma)^2).$$ However, if $X_1, \dots, X_k$ are $k$ independently distributed normals, say $X_i \sim N(\mu, \sigma^2)$, then $$\sum_{i = 1}^k X_i \sim N(k \mu, k\sigma^2).$$ This follows from the usual sum of independent normals.