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==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<br> |
| − | <math>E(X)= \sum_{k=0}^n kr^k</math> | + | <math>E(X)= \sum_{k=0}^n kr^k</math><br> |
| − | Since | + | Since<br> |
| − | <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><br> |
| − | 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<br> |
| − | <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><br> |
| − | Multiply both sides by r to see the form of the expected value in the problem: | + | Multiply both sides by r to see the form of the expected value in the problem:<br> |
| − | <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><br> |
| − | 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. | + | 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.<p> |
| − | To find the variance, the formula | + | To find the variance, the formula<br> |
| − | <math>VAR=E(x^2)-(E(x))^2</math> | + | <math>VAR=E(x^2)-(E(x))^2</math><br> |
| − | can be used. | + | can be used.<br> |
| − | <math>E(x^2)</math> can be expanded to find that | + | <math>E(x^2)</math> can be expanded to find that<br> |
| − | <math>VAR=(\sum_{k=0}^n (k^2)(r^k))-(E(x))^2</math> | + | <math>VAR=(\sum_{k=0}^n (k^2)(r^k))-(E(x))^2</math><br> |
| − | The formula | + | The formula<br> |
| − | <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><br> |
| − | 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<br> |
| − | <math>\frac{-(n+1)r^n(1-r)+(1-r^(n+1))}{(1-r)^2}\ | + | <math>\frac{-(n+1)r^n(1-r)+(1-r^(n+1))}{(1-r)^2}<\math><br><math>+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><br> |
| − | 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<br> |
| − | <math>r\frac{-(n+1)r^n(1-r)+(1-r^(n+1))}{(1-r)^2}\ | + | <math>r\frac{-(n+1)r^n(1-r)+(1-r^(n+1))}{(1-r)^2}<\math><br><math>+(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><br> |
| − | Therefore, the formula for the variance is given by | + | Therefore, the formula for the variance is given by <br> |
| − | <math>VAR=r\frac{-(n+1)r^n(1-r)+(1-r^(n+1))}{(1-r)^2}\ | + | <math>VAR=r\frac{-(n+1)r^n(1-r)+(1-r^(n+1))}{(1-r)^2}<\math><br><math>+(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><br> |
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 16:05, 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) $
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}<\math><br><math>+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}<\math><br><math>+(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}<\math><br><math>+(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 .
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