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=Dude91's Third Bonus Point Problem= | =Dude91's Third Bonus Point Problem= | ||
==Question:== | ==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? | + | Bob owns a company that produces n=100 widgets each day. The probability that a widget is produced without defect is r=.9.<br> a)What is the mean and the variance of the process Bob uses? Solve algebraically first, then solve numerically.<br> 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. |
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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 | + | ===Part A=== |
| − | <math>E(X)= \sum_{k=0}^n kr^k</math> | + | If X is taken to be the number of correctly produced widgets made each day, then the expected value of X is<br> |
| − | Since | + | <math>E(X)= \sum_{k=0}^n kr^k</math><br> |
| − | <math>\frac{1-r^ | + | Since<br> |
| − | Taking the derivative <math>\frac{d}{dr}</math> of both sides will yield | + | <math>\frac{1-r^{n+1}}{1-r}= \sum_{k=0}^n r^k</math><br> |
| − | <math>\frac{-(n+1)r^n(1-r)+(1-r^ | + | Taking the derivative <math>\frac{d}{dr}</math> of both sides will yield<br> |
| − | Multiply both sides by r to see the form of the expected value in the problem: | + | <math>\frac{-(n+1)r^n(1-r)+(1-r^{n+1})}{(1-r)^2}= \sum_{k=0}^n kr^{k-1}</math><br> |
| − | <math>r\frac{-(n+1)r^n(1-r)+(1-r^ | + | Multiply both sides by r to see the form of the expected value in the problem:<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. | + | <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.<p> | |
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| + | To find the variance, the formula<br> | ||
| + | <math>VAR=E(x^2)-(E(x))^2</math><br> | ||
| + | can be used.<br> | ||
| + | <math>E(x^2)</math> can be expanded to find that<br> | ||
| + | <math>VAR=(\sum_{k=0}^n k^2r^k)-(E(x))^2</math><br> | ||
| + | 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><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<br> | ||
| + | <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<br> | ||
| + | <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^2r^k</math><br> | ||
| + | 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><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}</math><br><math>-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 -6378 widgets<math>^2</math>. | ||
| + | ===Part B=== | ||
| + | Since 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><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}</math><br><math>-r^2(\frac{-(n+1)r^n(1-r)+(1-r^{n+1})}{(1-r)^2})^2</math><br> | ||
| + | Inserting 100 for n and .99 for r greatly increases the magnitude of the variance so that the variance is equal to -6874182 widgets<math>^2</math>. | ||
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== Questions/Comments/Fixes == | == Questions/Comments/Fixes == | ||
Ask your questions/post comments/submit changes to the solution below. | Ask your questions/post comments/submit changes to the solution below. | ||
| − | + | * Can the variance be negative? | |
| − | *question | + | * 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 <br><math>E(X)= \sum_{k=0}^n kr^k</math><br> or <br><math>E(X)= \sum_{k=0}^n kr^k(1-r)^{n-k}</math><br>? -ag | ||
*question/comment here. | *question/comment here. | ||
*etc. | *etc. | ||
[[Bonus_point_3_ECE302_Spring2012_Boutin|Back to third bonus point opportunity, ECE302 Spring 2013]] | [[Bonus_point_3_ECE302_Spring2012_Boutin|Back to third bonus point opportunity, ECE302 Spring 2013]] | ||
Latest revision as of 05:59, 22 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.
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.
