Dude91's Third Bonus Point Problem

Question:

Bob owns a company that produces n=100 widgets each day. The probability that a widget is produced without defect is r=.9.
a)What is the mean and the variance of the process Bob uses? Solve algebraically first, then solve numerically.
b)What effect does increasing r to .99 have on the variance? Please note that round-off error can be somewhat significant when performing this calculation; a high-precision calculator may be required.

Solution:

Part A

If X is taken to be the number of correctly produced widgets made each day, then the expected value of X is
$ E(X)= \sum_{k=0}^n kr^k $
Since
$ \frac{1-r^{n+1}}{1-r}= \sum_{k=0}^n r^k $
Taking the derivative $ \frac{d}{dr} $ of both sides will yield
$ \frac{-(n+1)r^n(1-r)+(1-r^{n+1})}{(1-r)^2}= \sum_{k=0}^n kr^{k-1} $
Multiply both sides by r to see the form of the expected value in the problem:
$ r\frac{-(n+1)r^n(1-r)+(1-r^{n+1})}{(1-r)^2}= \sum_{k=0}^n kr^k $

To find the variance, the formula
$ VAR=E(x^2)-(E(x))^2 $
can be used.
$ E(x^2) $ can be expanded to find that
$ VAR=(\sum_{k=0}^n k^2r^k)-(E(x))^2 $
The formula
$ r\frac{-(n+1)r^n(1-r)+(1-r^{n+1})}{(1-r)^2}= \sum_{k=0}^n kr^k $
can be used to derive the formula for $ E(x^2) $. To do this, take the derivative $ \frac{d}{dr} $ of both sides to find that
$ \frac{-(n+1)r^n(1-r)+(1-r^{n+1})}{(1-r)^2} $
$ +r\frac{((1-r)^2)(-n(n+1)r^{n-1}+n(n+1)r^n)-(-(n+1)r^n(1-r)+(1-r^{n+1}))(-2+2r)}{(1-r)^4}= \sum_{k=0}^n (k^2)r^{k-1} $
Multiplying both sides by r yields the expression for $ E(x^2) $ to be
$ r\frac{-(n+1)r^n(1-r)+(1-r^{n+1})}{(1-r)^2} $
$ +r^2\frac{((1-r)^2)(-n(n+1)r^{n-1}+n(n+1)r^n)-(-(n+1)r^n(1-r)+(1-r^{n+1}))(-2+2r)}{(1-r)^4}= \sum_{k=0}^n k^2r^k $
Therefore, the formula for the variance is given by
$ VAR=r\frac{-(n+1)r^n(1-r)+(1-r^{n+1})}{(1-r)^2} $
$ +r^2\frac{((1-r)^2)(-n(n+1)r^{n-1}+n(n+1)r^n)-(-(n+1)r^n(1-r)+(1-r^{n+1}))(-2+2r)}{(1-r)^4} $
$ -r^2(\frac{-(n+1)r^n(1-r)+(1-r^{n+1})}{(1-r)^2})^2 $
When 100 is inserted in for n and .9 is inserted in for r, the variance can be found to equal -6378 widgets$ ^2 $.

Part B

Since the formula for the variance is given by
$ VAR=r\frac{-(n+1)r^n(1-r)+(1-r^{n+1})}{(1-r)^2} $
$ +r^2\frac{((1-r)^2)(-n(n+1)r^{n-1}+n(n+1)r^n)-(-(n+1)r^n(1-r)+(1-r^{n+1}))(-2+2r)}{(1-r)^4} $
$ -r^2(\frac{-(n+1)r^n(1-r)+(1-r^{n+1})}{(1-r)^2})^2 $
Inserting 100 for n and .99 for r greatly increases the magnitude of the variance so that the variance is equal to -6874182 widgets$ ^2 $.

Questions/Comments/Fixes

Ask your questions/post comments/submit changes to the solution below.

• Can the variance be negative?
• I like that question: it is using formulas and techniques we saw in the lecture, so it is a good review of the material. But it is not an easy question because of all the computations one has to do to figure out the summations. Is there a way to shorten the explanation? Also, could one answer part b) without using a calculator? (Once again, calculators will not be allowed on the exam.) -pm
• Is the expected value
  $ E(X)= \sum_{k=0}^n kr^k $
  or
  $ E(X)= \sum_{k=0}^n kr^k(1-r)^{n-k} $
  ? -ag
• question/comment here.
• etc.