I'm struggling with the following problem:
(ed: Don't be lazy. Just type it out. )
A certain small freight elevator has a max. capacity $C$, which is Normally distributed, with mean $400\text{lbs.}$, and standard deviation $4\text{lbs.}$
The weight of the boxes being loaded into the elevator is a R.V., with mean $30\text{lbs.}$, and standard deviation $0.3\text{lbs.}$.
How many boxes may be loaded into this elevator before the probability of disaster exceeds $0.2$?
Consider the random variable, $C$, to be independent of the weight of the boxes.
I know that with a normal distribution, I'm going to want to standardize at least one of them.
I thought of using a Z table to look up Probability( $E \geq Z$ ), where:
$E$ is the sum of the weights of n boxes (so expected value of $E = (n)(30)$ )
$Z$ is the normalized random variable $C$ (so $Z = (C - 400)/16$ )
But I'm not sure how to find a numerical solution to this. If $E$ is not fixed, how can I lookup a probability in a table.
I really appreciate any help on this.
Hint: The sum of independent normally distributed random variables is a normally distributed random variable with (1) its mean being the sum of the means, (2) its variance being the sum of the variances.
You want to find $n$ such that $\mathsf P((\sum_{k=1}^n B_k) - C \geq 0)\gt 0.2$, where $\{B_k\}$ are the weights of boxes and C is the capacity of the elevator; which are each normally distributed R.V., with the given means and variances. (Note: do you remember what is the relation between variance and standard deviation?)
Then so too is $X \mathop{:=} (\sum_{k=1}^n B_k) - C$ a normally distributed random variable. Find its mean and variance, then the probability it exceeds zero. (Or rather, find what value of $n$ makes the probability $0.2$.)
$\Box$