# Using Central Limit Theorem to find a probability of disaster

A small elevator has a maximum capacity $C$, which is normally distributed, with mean $400$ kg., and standard deviation $4$ kg. The weight of the boxes being loaded into the elevator is a random variable with mean $30$ kg., and standard deviation $0.3$ kg.

Assume that the weights of the boxes and maximum capacity are independent random variables. How many boxes may be loaded into the elevator before the probability of disaster exceeds $20\%$?

I've found so far that:

$$P\left(∑X−C≥0\right)>0.2$$

With $X$ being the weight of the boxes and $C$ being the weight of the elevator.

But I can't seem to figure out where to go from here.

According to the CLT, the sum $S_n$ of weights of $n$ boxes is approximately normal with mean $30n$ and variance $0.3^2 n$. So $X_n = C - S_n$ is approximately normal with mean $400 - 30 n$ and variance $4^2 + 0.3^2 n$. So $P(X < 0) \approx .\ldots$

• Could you care to explain how you found P( X < 0)? – X1XX Mar 23 '16 at 3:46

$E[\Sigma X - C] = 30n-400$ the variance of $\Sigma X - C = n\sigma_x^2+\sigma_c^2 = n(0.09)+16$ And, the standard deviation of, $\Sigma X - C = \sqrt{n(0.09)+16}$

If $30n-400 = 0$ there is a 50% chance that the elevator will be overloaded. To back into a 20% chance we look it up on our tables and we get that $P^{-1}(0.20) = -0.84$. That is if the expected weight is inside of 0.84 standard deviations we are above our risk tolerance.

$30n-400 < -0.84\sqrt{n(0.09)+16}$

13 boxes is the largest integer that satisfies the equation.

• Would you mind explaining which table you used to find P(0.2) = -0.84? – X1XX Mar 23 '16 at 4:04
• I didn't actually use a table. I typed +normsinv(0.2) into an excel spreadsheet. – Doug M Mar 23 '16 at 4:37

The distribution of the difference between two normal distributions is $N(\mu_1-\mu_2,\sqrt{\sigma^2_1+\sigma^2_2})$

Using the Central Limit Theorem, we can say that the distribution of the weight of the boxes is approximately $N(30b,.3\sqrt b)$ where b is the number of boxes.

At this point, I would just use trial and error. If there are 13 boxes, the distribution of the difference between the capacity and the weight of the boxes is $N(10,4.144)$ and P(X<0)=.008 where x equals the capacity minus the weight. So 13 boxes would still be okay. If there are 14 boxes, the mean weight would be 420 so there is for sure more than a .5 chance that the weight exceeds the capacity.