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

You computed the probability by taking one normal and multiplying it by a constant 10. This is different from the sum of 10 identically distributed normals! See, for instance, this question.

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.

Here's an amusing way to think about the difference in this problem: If ten people drink their regular amount of wine, then you would expect them to all hover pretty close to their regular amount. If one person drinks ten times as much as they normally would, you might expect wild variation in the total when all is said and done.