My text book has the following question
A highway department has enough salt to handle a total of 80 inches of snowfall. Suppose the daily amount of snow has a mean of 1.5 inches and a standard deviation of 0.3 inches. Approximate the probability that the salt on hand will suffice for the next 50 days
I looked up the solution as follows but I don't understand some steps
Let $X_i$ be a random variable for snow falling on the $i^{th}$ day.
Given,
$$
E[X_i]=1.5 \;inches
$$
$$
SD[Xi]=0.3 \;inches
$$
Let X be the amount of snow expected in 50 days
$$
X=\sum_{i=1}^{50} X_i
$$
Then,
$$
E[X] = \sum_{i=1}^{50} E(X_i)=50(1.5)=75
$$
step 2
$$
Var[X] = \sum_{i=1}^{50} (SD[X_i])^2=50(0.09) = 4.5
$$
By central limit theorem, $X \sim \mathcal{N}(E[X_i],Var[X])$
Using general conversion formula
$$
Z=\frac{X-\mu_{X}}{\sigma_{X}}
$$
We have
step 3
$$
P(X\leq 80) = P\left(\frac{X -75}{\sqrt{4.5}} < \frac{80 -75}{\sqrt{4.5}}\right)
$$
step 4
$$
\approx P(Z \leq 2.3570) =\phi(2.36) =0.99079
$$
I am wanting to know the following;
Is Poisson involved here?
What is the technical name for X? "The expected value"?
Why do we sum the variance?
As far as I can tell there are no Poisson distributions involved in this problem.
$X$ is the random variable that represents the total amount of snowfall in 50 days. The expected value is $E[X]$.
You can sum the variances because it is assumed that $X_1, X_2, \dots, X_{50}$, the amounts of snowfall on each individual day, are all independent. See here for a proof of the fact that the variance of a sum of independent random variables is equal to the sum of the individual variances.