Biology Forums - Study Force

Suppose that the demand for a company’s product in weeks 1, 2, and 3 are each normally distributed and the mean demand during each of these three weeks is 50, 45, and 65, respectively. Suppose the standard deviation of the demand during each of these three weeks is known to be 10, 5, and 15, respectively. It turns out that if we can assume that these three demands are probabilistically independent then the total demand for the three week period is also normally distributed. And, the mean demand for the entire three week period is the sum of the individual means. Likewise, the variance of the demand for the entire three week period is the sum of the individual weekly variances. But be careful! The standard deviation of the demand for the entire 3 week period is not the sum of the individual standard deviations. Square roots don’t work that way!

Now, suppose that the company currently has 180 units in stock, and it will not be receiving any further shipments from its supplier for at least 3 weeks. What is the probability that the company will run out of units?
 

m(3wks)= 50+45+65 = 165

V(3wks) = 100+ 25 + 225 = 350, sqrt(350) = 18.708

z = (180-165)/18.7 = 15/18.7 = .8021

P(z > .8021) = 21.12%

Let X: demand for 3 weeks
 E(X) = mean of X= 45+50+65 =160
Var(X) = ( 102+52+152) = 350
Stdev(X) = √350 = 18.708
X is normally distributed with ( 160,350)
Now we want P( X >180) = P( (180-160)/18.7087) = P( z >1.069) =0.14253
There is a 14.253% chance that  company will run out of units

Thank you for your help! :)

Who gets best answer?

 I appreciate both of you helping, but Vanbu's answer I went with because he/she had 45+50+65 =160 where the other had these numbers equaling 165.  An honest mistake; but I still went with Vanbu. Thanks again! ;D

So who is really right?

Go with Vanbu's.