population statistics and deviation from mean Suppose $X \sim N (50, 20.5)$ represents a population. If $X_1, X_2, …, X_n$ is a random sample from this population, find $n$ such that $\mathbb{P}(X_1 + X_2 + … + X_n > 2000) = 0.95 $
 A: We assume that $20.5$ is the standard deviation of our normals. The random variable $W=X_1+X_2+\cdots +X_n$ is then normal, mean $50n$, standard deviation $20.5\sqrt{n}$. 
The probability that $W\gt 2000$ is equal to the probability that $Z\gt \frac{2000-50n}{20.5\sqrt{n}}$, where $Z$ is standard normal. 
Note that there is $5\%$ probability in the left tail of the standard normal if we take $z\approx -1.645$. 
So we end up looking at the equation
$$\frac{2000-50n}{20.5\sqrt{n}}=-1.645.$$
This simplifies to (approximately)
$$50n-33.7225\sqrt{n}-2000=0.$$
We have reached  a quadratic equation in $\sqrt{n}$. Solve.  
A: So, by the theorem on sums of normal distribution, if $Y_n = X_1 + X_2 + ... +X_n$ then we have that $Y_n \sim N(50n, 20.5n)$ (I can show you a proof for this if you'd like but hopefully you could do this on your own? Or at least your teacher/professor told you this)
Let $Z \sim N(0,1)$ then we have that $\mathbb{P} (Y_n >2000) = \mathbb{P} (Z > \frac{2000 -50n} { \sqrt{20.5n}}) = 1 - \mathbb{P} (Z \leqslant \frac{2000 -50n} { \sqrt{20.5n}})$ And we want this to be equal to $0.95$.
So we get (by looking up in normal tables) $\frac{2000 -50n} { \sqrt{20.5n}} = \Phi ^{-1}(1- 0.95)  = \Phi^{-1} (0.05) \approx -1.645$
And now we have the equation $2000 - 50n = -1.645\sqrt{20.5n}$ (note n need not be an integer, we will just round up to the nearest integer for our final solution)
I just chucked this equation into Wolfram Alpha to solve it, but if you wanted to you would square both sides of the equation and solve the quadratic you get, and you check which root really is a solution to the equation (the only positive one I imagine though I don't actually know) and we get n = 40.9533, so we round up and get n = 41 as our solution.
As a quick sanity check we note tat $Y_{41} \sim N(2050, 840.5)$ then our standard deviation is $\sqrt{20.5*41} \approx 28.9$ and then $\mathbb{P} (Y_{41} (= X_1 + X_2 + ... + X_{41}) > 2000) = 1 -  \mathbb{P} (Y_{41} \leqslant 2000) \approx 1 - \mathbb{P} (Z \leqslant \frac{2000 - 2050}  {28.9}) = 1 - \mathbb{P} (Z \leqslant -1.725) >0.95$
So we are correct, and $n = 41$ is the answer.
