# Thread: HELP: Normal distribution probability problem

1. ## HELP: Normal distribution probability problem

I have the following problem:

A certain small freight elevator has a max. capacity C, which is Normally distributed, with mean 400 lbs., and standard deviation 4 lbs.

The weight of the boxes being loaded into the elevator is a R.V., with mean 30 lbs., and standard deviation 0.3 lbs.

How many boxes may be loaded into this elevator before the probability of disaster exceeds 0.2?

I have some idea of how to get started, but I can't seem to find the right formula. Can any one help? Thanks in advance, any help is greatly appreciated.

Mikhail

2. $P(disaster)=P\left(C<\sum_{i=1}^nB_i\right)$

C is your capacity rv
Bi is the rv associated with the weight of the ith box.
n is the fixed number of boxes.
IF you have independence between the C and the boxes, then

$C-\sum_{i=1}^nB_i$ is a normal random variable and you want

$P\left(C-\sum_{i=1}^nB_i<0\right)=.2$

YOU can find the 20th percentile of a standard normal and the distribution of

$W=C-\sum_{i=1}^nB_i$

3. Thank you!