| Line 6: | Line 6: | ||
---- | ---- | ||
==Solution:== | ==Solution:== | ||
| − | If X is taken to be the number of correctly produced widgets made each day, then the expected value of X is | + | If X is taken to be the number of correctly produced widgets made each day, then the expected value of X is |
| − | <math>E(X)= \sum_{k=0}^n kr^k</math> | + | <math>E(X)= \sum_{k=0}^n kr^k</math> |
| − | Since | + | Since |
| − | <math>\frac{1-r^(n+1)}{1-r}= \sum_{k=0}^n r^k</math> | + | <math>\frac{1-r^(n+1)}{1-r}= \sum_{k=0}^n r^k</math> |
Taking the derivative <math>\frac{d}{dr}</math> of both sides will yield | Taking the derivative <math>\frac{d}{dr}</math> of both sides will yield | ||
<math>\frac{-(n+1)r^n(1-r)+(1-r^(n+1))}{(1-r)^2}= \sum_{k=0}^n kr^(k-1)</math> | <math>\frac{-(n+1)r^n(1-r)+(1-r^(n+1))}{(1-r)^2}= \sum_{k=0}^n kr^(k-1)</math> | ||
| Line 24: | Line 24: | ||
<math>r\frac{-(n+1)(r^n)(1-r)+(1-r^(n+1))}{(1-r)^2}= \sum_{k=0}^n kr^(k)</math> | <math>r\frac{-(n+1)(r^n)(1-r)+(1-r^(n+1))}{(1-r)^2}= \sum_{k=0}^n kr^(k)</math> | ||
can be used to derive the formula for <math>E(x^2)</math>. To do this, take the derivative <math>\frac{d}{dr}</math> of both sides to find that | can be used to derive the formula for <math>E(x^2)</math>. To do this, take the derivative <math>\frac{d}{dr}</math> of both sides to find that | ||
| − | <math>\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)</math> | + | <math>\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)</math> |
Multiplying both sides by r yields the expression for <math>E(x^2)</math> to be | Multiplying both sides by r yields the expression for <math>E(x^2)</math> to be | ||
| − | <math>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^2)(r^k)</math> | + | <math>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^2)(r^k)</math> |
Therefore, the formula for the variance is given by | Therefore, the formula for the variance is given by | ||
| − | <math>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</math> | + | <math>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</math> |
When 100 is inserted in for n and .9 is inserted in for r, the variance can be found to equal . | When 100 is inserted in for n and .9 is inserted in for r, the variance can be found to equal . | ||
Revision as of 14:59, 21 February 2013
Contents
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. What is the mean and the variance of the process Bob uses?
Solution:
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) $ This value is the mean, which, when 100 widgets is inserted in for n and .9 is inserted in for r, can be found to equal 90 widgets, as expected.
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^2)(r^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^2)(r^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 .
Questions/Comments/Fixes
Ask your questions/post comments/submit changes to the solution below.
- question/comment here.
- question/comment here.
- etc.
