Combining Two Distributions

It is sometimes necessary to add or subtract variables from two or more distributions to create a new variable. In this page I want to make sure you understand how to work with these combined variables. I begin with an example:

Suppose that your company cuts the threads in lug nuts for tractor wheels. The process takes two machine operations. Machine A is a boring machine that drills a hole in the hexagonal blank of steel. The drilling operation is normally distributed with a mean time of 18 seconds and a standard deviation of 3 seconds. The thread cutting machine takes 28 seconds on average, with a standard deviation of 4 seconds. If we combine these two operations into a single process by having a robot instantaneously transfer parts from Machine A to Machine B, we now have a single operation. What is the mean and standard deviation of the manufacturing time for the complete part.

To answer this question we need to know how means of distributions are combined, and how the standard deviations combine.

Means are easy. If we add the results of two distributions, the mean of the result is the sum of the means of the original distributions. In the language of statistics, if we add distributions X and Y to get Z then m_z= m_x +m_y . On the other hand, if we were to subtract the variables to get the new variable, we would apply that same operation to the mean.

Standard deviations combine much like the Pythagorean theorem (more exactly, they combine like the law of Cosines, but since we are only using independent variable for awhile, we can ignore that for now). If the standard deviations of X and Y are given by s_x and s_y then the standard deviation of the combined distribution, Z, is given by (s_z)²= (s_x)² + (s_y)². Notice that another way to say this is that for independent variables, the VARIANCE of the combined result is the sum of the VARIANCES of the two distributions.

In our example above we had m_A= 18 and m_B= 28 so the combined process will have a mean of 18+28 = 46 seconds. For the standard deviations we get (s_z)²= (s_x)² + (s_y)² and so (s_z)²= (3)² + (4)² . Adding and taking the square root gives us a standard deviation of 5 seconds for the final process.

Sample Problems: Try these and make sure you understand the ideas at work. The answers are after the last problem.

1) A six-sided and a ten-sided die are both rolled. The mean and std dev for the six-sided die are 3.5 and 1.707. For the ten-sided die the mean and std dev are 5.5 and 2.872.

a) Find the mean and variance if the results of the two dice are added.

b) Find the result if the results of the six-sided die is subtracted from the result of the ten-sided die.

2) Nutty Chocolate candy has real nuts covered with chocolate. The typical weight of the nuts in a package is normally distributed with a mean of 12 grams and a standard deviation of .8 grams. The thick chocolate coating is also normally distributed with a mean weight of 5 grams and a standard deviation of .65 grams.

a) What is the mean weight of a Nutty Chocolate candy?

b) what is the standard deviation of the weight of a nutty chocolate candy?

3) Nutty Chocolate candies (as described in prob 2) come in a package of 20 candies to a bag. What is the mean and standard deviation of the weight of a bag of Nutty Chocolate candies?

4) What weight could be printed on the package so that 99% of the packages would contain more than this minimum weight?

Answers…………………

1a) The mean sum will be 3.5+5.5 = 9. The std dev will be = 3.341

1b) The mean difference would be 5.5-3.5 = 2. The std dev would be the same as in part a, whether you added or subtracted.

2a) The weight of the candies would average the mean of the nut plus the mean of the chocolate, 12 + 5 = 17 grams

2b) The standard deviation of the candies will be = 1.031 grams.

2c) The mean of 20 candies will be 20 times the mean of one candy, so the average package weight would be 20 * 17 grams = 340 grams (that’s about 12 ounces if I remember correctly). Since all of the candies have the same standard deviation we have to square the standard deviation and add 20 of these squared values together… with 20 of the squares under that radical, but that is the same as which is about 4.610 grams per package.

2d) Since InvNorm(.01) = -2.326 we need to find a weight that has a Z-score of –2.326. Using the formula for Z-scores, and using the values we know for mu and sigma of the package, we get x= 340 – 2.326(4.610) or about 329 grams.