Expected consultant bill given distribution of time taken
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Expected consultant bill given distribution of time taken
$begingroup$
The cdf of the number of hours it takes a consultant to complete a project is given by $F(x)= dfracx^216$ for o to 4. The consultant bills $300 per hour, rounded up to the nearest half hour, for the project. What is the expected amount of the total bill?
(a)900
(b)800
(c)872
(d)950
(e)1100
My work:
$f(x)= dfracdF(x)dx$
$f(x)=dfracx8$
so
integral of $x^2/8$ from 0 to 4 = $x^3/24$ from 0 to 4 = 64/24 = 2.6667
round 2.6667 to nearest half hour is 2.5
so
2.5*300 = 800
But that's wrong; the answer to the question is 872.
Can I please have help understanding why my method is incorrect, so I can try another method while understanding why my last attempt was incorrect.
probability probability-distributions
edited Jun 2 at 12:08
user21820
41k546166
asked Jun 1 at 18:02
Jordan GreenhutJordan Greenhut
577
$endgroup$
|
show 5 more comments
$begingroup$
The cdf of the number of hours it takes a consultant to complete a project is given by $F(x)= dfracx^216$ for o to 4. The consultant bills $300 per hour, rounded up to the nearest half hour, for the project. What is the expected amount of the total bill?
(a)900
(b)800
(c)872
(d)950
(e)1100
My work:
$f(x)= dfracdF(x)dx$
$f(x)=dfracx8$
so
integral of $x^2/8$ from 0 to 4 = $x^3/24$ from 0 to 4 = 64/24 = 2.6667
round 2.6667 to nearest half hour is 2.5
so
2.5*300 = 800
But that's wrong; the answer to the question is 872.
Can I please have help understanding why my method is incorrect, so I can try another method while understanding why my last attempt was incorrect.
probability probability-distributions
edited Jun 2 at 12:08
user21820
41k546166
asked Jun 1 at 18:02
Jordan GreenhutJordan Greenhut
577
$endgroup$
7
$begingroup$
It seems to me that you have just ignored the part about rounding up to the nearest half hour. I would say that a discrete approach is necessary here.
$endgroup$
– TonyK
Jun 1 at 18:32
3
$begingroup$
The probability distribution is continuous, but the cost is a step function.
$endgroup$
– eyeballfrog
Jun 1 at 18:34
1
$begingroup$
How can you think that charging up to $150 extra is subtle?
$endgroup$
– TonyK
Jun 1 at 18:35
2
$begingroup$
Just for intuition: If you earn $10$% interest on $$1000$, compounding annually, after $1$ year you'll have $$1100$. If you compound interest continuously, you'll have $$1000e^.1 approx $1105.17$. There's a significant difference.
$endgroup$
– Ted Shifrin
Jun 1 at 18:59
2
$begingroup$
Wow, that's the answer I wanted Ted. That would get gold, if I have such powers. I forgot this part of accounting. I always wondered why we had to take accounting... I will remember the difference from compounding interest over time increments versus compounding continuously can have large implications (it should have been obvious and that's the thought I need to pass this f***er)
$endgroup$
– Jordan Greenhut
Jun 1 at 19:01
|
show 5 more comments
$begingroup$
The cdf of the number of hours it takes a consultant to complete a project is given by $F(x)= dfracx^216$ for o to 4. The consultant bills $300 per hour, rounded up to the nearest half hour, for the project. What is the expected amount of the total bill?
(a)900
(b)800
(c)872
(d)950
(e)1100
My work:
$f(x)= dfracdF(x)dx$
$f(x)=dfracx8$
so
integral of $x^2/8$ from 0 to 4 = $x^3/24$ from 0 to 4 = 64/24 = 2.6667
round 2.6667 to nearest half hour is 2.5
so
2.5*300 = 800
But that's wrong; the answer to the question is 872.
Can I please have help understanding why my method is incorrect, so I can try another method while understanding why my last attempt was incorrect.
probability probability-distributions
edited Jun 2 at 12:08
user21820
41k546166
asked Jun 1 at 18:02
Jordan GreenhutJordan Greenhut
577
$endgroup$
The cdf of the number of hours it takes a consultant to complete a project is given by $F(x)= dfracx^216$ for o to 4. The consultant bills $300 per hour, rounded up to the nearest half hour, for the project. What is the expected amount of the total bill?
(a)900
(b)800
(c)872
(d)950
(e)1100
My work:
$f(x)= dfracdF(x)dx$
$f(x)=dfracx8$
so
integral of $x^2/8$ from 0 to 4 = $x^3/24$ from 0 to 4 = 64/24 = 2.6667
round 2.6667 to nearest half hour is 2.5
so
2.5*300 = 800
But that's wrong; the answer to the question is 872.
Can I please have help understanding why my method is incorrect, so I can try another method while understanding why my last attempt was incorrect.
probability probability-distributions
probability probability-distributions
edited Jun 2 at 12:08
user21820
41k546166
asked Jun 1 at 18:02
Jordan GreenhutJordan Greenhut
577
edited Jun 2 at 12:08
user21820
41k546166
asked Jun 1 at 18:02
Jordan GreenhutJordan Greenhut
577
edited Jun 2 at 12:08
user21820
41k546166
edited Jun 2 at 12:08
user21820
41k546166
edited Jun 2 at 12:08
user21820
41k546166
41k546166
asked Jun 1 at 18:02
Jordan GreenhutJordan Greenhut
577
asked Jun 1 at 18:02
Jordan GreenhutJordan Greenhut
577
asked Jun 1 at 18:02
Jordan GreenhutJordan Greenhut
577
577
7
$begingroup$
It seems to me that you have just ignored the part about rounding up to the nearest half hour. I would say that a discrete approach is necessary here.
$endgroup$
– TonyK
Jun 1 at 18:32
3
$begingroup$
The probability distribution is continuous, but the cost is a step function.
$endgroup$
– eyeballfrog
Jun 1 at 18:34
1
$begingroup$
How can you think that charging up to $150 extra is subtle?
$endgroup$
– TonyK
Jun 1 at 18:35
2
$begingroup$
Just for intuition: If you earn $10$% interest on $$1000$, compounding annually, after $1$ year you'll have $$1100$. If you compound interest continuously, you'll have $$1000e^.1 approx $1105.17$. There's a significant difference.
$endgroup$
– Ted Shifrin
Jun 1 at 18:59
2
$begingroup$
Wow, that's the answer I wanted Ted. That would get gold, if I have such powers. I forgot this part of accounting. I always wondered why we had to take accounting... I will remember the difference from compounding interest over time increments versus compounding continuously can have large implications (it should have been obvious and that's the thought I need to pass this f***er)
$endgroup$
– Jordan Greenhut
Jun 1 at 19:01
|
show 5 more comments
7
$begingroup$
It seems to me that you have just ignored the part about rounding up to the nearest half hour. I would say that a discrete approach is necessary here.
$endgroup$
– TonyK
Jun 1 at 18:32
3
$begingroup$
The probability distribution is continuous, but the cost is a step function.
$endgroup$
– eyeballfrog
Jun 1 at 18:34
1
$begingroup$
How can you think that charging up to $150 extra is subtle?
$endgroup$
– TonyK
Jun 1 at 18:35
2
$begingroup$
Just for intuition: If you earn $10$% interest on $$1000$, compounding annually, after $1$ year you'll have $$1100$. If you compound interest continuously, you'll have $$1000e^.1 approx $1105.17$. There's a significant difference.
$endgroup$
– Ted Shifrin
Jun 1 at 18:59
2
$begingroup$
Wow, that's the answer I wanted Ted. That would get gold, if I have such powers. I forgot this part of accounting. I always wondered why we had to take accounting... I will remember the difference from compounding interest over time increments versus compounding continuously can have large implications (it should have been obvious and that's the thought I need to pass this f***er)
$endgroup$
– Jordan Greenhut
Jun 1 at 19:01
7
7
$begingroup$
It seems to me that you have just ignored the part about rounding up to the nearest half hour. I would say that a discrete approach is necessary here.
$endgroup$
– TonyK
Jun 1 at 18:32
$begingroup$
It seems to me that you have just ignored the part about rounding up to the nearest half hour. I would say that a discrete approach is necessary here.
$endgroup$
– TonyK
Jun 1 at 18:32
3
3
$begingroup$
The probability distribution is continuous, but the cost is a step function.
$endgroup$
– eyeballfrog
Jun 1 at 18:34
$begingroup$
The probability distribution is continuous, but the cost is a step function.
$endgroup$
– eyeballfrog
Jun 1 at 18:34
1
1
$begingroup$
How can you think that charging up to $150 extra is subtle?
$endgroup$
– TonyK
Jun 1 at 18:35
$begingroup$
How can you think that charging up to $150 extra is subtle?
$endgroup$
– TonyK
Jun 1 at 18:35
2
2
$begingroup$
Just for intuition: If you earn $10$% interest on $$1000$, compounding annually, after $1$ year you'll have $$1100$. If you compound interest continuously, you'll have $$1000e^.1 approx $1105.17$. There's a significant difference.
$endgroup$
– Ted Shifrin
Jun 1 at 18:59
$begingroup$
Just for intuition: If you earn $10$% interest on $$1000$, compounding annually, after $1$ year you'll have $$1100$. If you compound interest continuously, you'll have $$1000e^.1 approx $1105.17$. There's a significant difference.
$endgroup$
– Ted Shifrin
Jun 1 at 18:59
2
2
$begingroup$
Wow, that's the answer I wanted Ted. That would get gold, if I have such powers. I forgot this part of accounting. I always wondered why we had to take accounting... I will remember the difference from compounding interest over time increments versus compounding continuously can have large implications (it should have been obvious and that's the thought I need to pass this f***er)
$endgroup$
– Jordan Greenhut
Jun 1 at 19:01
$begingroup$
Wow, that's the answer I wanted Ted. That would get gold, if I have such powers. I forgot this part of accounting. I always wondered why we had to take accounting... I will remember the difference from compounding interest over time increments versus compounding continuously can have large implications (it should have been obvious and that's the thought I need to pass this f***er)
$endgroup$
– Jordan Greenhut
Jun 1 at 19:01
|
show 5 more comments
4 Answers
4
active
oldest
votes
$begingroup$
Because it gets rounded up to the nearest half hour, the expected fee is simply
$$$150left(frac0.5^216-frac0^216right)+$300left(frac1^216-frac0.5^216right)+cdots+$1200left(frac4^216-frac3.5^216right)$$
(the probability it lands in each half hour multiplied by the amount received in that half hour)
which is just $$$1200-$150left(frac3.5^216+cdots+frac0.5^216+frac0^216right)=$871.875approx$872$$
answered Jun 1 at 18:48
auscryptauscrypt
7,416614
$endgroup$
1
$begingroup$
Why can't it be continuous? It's rounded... that's not doing it for me
$endgroup$
– Jordan Greenhut
Jun 1 at 18:49
1
$begingroup$
@JordanGreenhut If the amount of time it takes is continuous, but the billing is done based on a rounded form of this time, then the amount billed is discrete, since the time associated to the bill is always some multiple of half an hour.
$endgroup$
– Ian
Jun 1 at 18:51
$begingroup$
I can kinda see that. So time must be viewed as half hour increments, because that's the window the consultant offers their time. "Why wont continuous work?" is a better question
$endgroup$
– Jordan Greenhut
Jun 1 at 18:53
$begingroup$
A continuous treatment of the problem without the rounding would work, but it would most likely give a different answer, unless all the rounding errors somehow cancel out.
$endgroup$
– Ian
Jun 1 at 19:28
add a comment |
$begingroup$
The problem here doesn't really have to do with discrete vs continuous. What you have done is to correctly calculate the expected time the project will take, not the expected bill. It is not correct to multiply the expected time by $300$ to get the expected bill, even if you round up that time.
Why is this? Let's say $X$ is the random variable representing the time taken, and $Y$ is the random variable representing the bill.
If the consultant did not round to the nearest half hour, then we would just have $Y=300X$.
In that case, you could have used the following (correct) reasoning:
$$textSince Y=300X,quad E[Y]=300E[X].$$
However, we actually have $Y=300,mathttrnd(X)$, where $mathttrnd$ is the function that rounds up to the nearest half hour. Your method used the following (incorrect) reasoning:
$$textSince Y=300,mathttrnd(X),quad E[Y]=300,mathttrnd(E[X])quadcolorredtextFalse!$$
While expectation is linear, it does not commute with arbitrary functions (in this case the rounding function).
You have to compute the expectation of the $Y$ variable (the cost) directly, there is no way to get it just from the expectation of the $X$ variable (the time) - because the relationship between the two variables is more complicated than just a linear function.
P.S. The problem also says the time is rounded up to the nearest half hour, so your 2.66667 should have been rounded up to 3, not down to 2.5. That still wouldn't give the right answer for the reasons I explained above, but keep it in mind if you want to try the problem again.
answered Jun 2 at 5:32
CarmeisterCarmeister
2,9692925
$endgroup$
add a comment |
$begingroup$
Let $X$ be the amount of the total bill, in dollars. Then $X$ is certainly a discrete random variable. Its possible values are $0, 150, 300, 450, ldots, 1200$. To treat $X$ as a continuous random variable is an error, because it is simply not so.
answered Jun 1 at 18:54
littleOlittleO
31.8k651114
$endgroup$
$begingroup$
Thank you, that helped a little
$endgroup$
– Jordan Greenhut
Jun 1 at 18:55
add a comment |
$begingroup$
The probability for the duration to be between $frack-12$ and $frac k2$ is $frac2k-164$ and the charge would be $150k$. Thus, the expected charge would be
$$
beginalign
frac7532sum_k=1^8left(2k^2-kright)
&=frac7532sum_k=1^8left[4binomk2+binomk1right]\
&=frac7532left[4binomk+13+binomk+12right]_k=0^k=8\
&=frac7532left[4binom93+binom92right]\
&=frac7532cdot372\[9pt]
&=871tfrac78
endalign
$$
answered Jun 1 at 19:16
robjohn♦robjohn
274k28320655
$endgroup$
$begingroup$
Fancy. Can you explain it further? How do you arrive at the left hand side equation?
$endgroup$
– Jordan Greenhut
Jun 1 at 19:20
$begingroup$
$frac7532left(2k^2-kright)=frac2k-164cdot150k$ from the line above.
$endgroup$
– robjohn♦
Jun 1 at 19:24
$begingroup$
$sum_kbinomkn=binomk+1n+1$ follows from Pascal's recurrence: $binomk+1n+1=binomkn+1+binomkn$
$endgroup$
– robjohn♦
Jun 1 at 19:26
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
Because it gets rounded up to the nearest half hour, the expected fee is simply
$$$150left(frac0.5^216-frac0^216right)+$300left(frac1^216-frac0.5^216right)+cdots+$1200left(frac4^216-frac3.5^216right)$$
(the probability it lands in each half hour multiplied by the amount received in that half hour)
which is just $$$1200-$150left(frac3.5^216+cdots+frac0.5^216+frac0^216right)=$871.875approx$872$$
answered Jun 1 at 18:48
auscryptauscrypt
7,416614
$endgroup$
1
$begingroup$
Why can't it be continuous? It's rounded... that's not doing it for me
$endgroup$
– Jordan Greenhut
Jun 1 at 18:49
1
$begingroup$
@JordanGreenhut If the amount of time it takes is continuous, but the billing is done based on a rounded form of this time, then the amount billed is discrete, since the time associated to the bill is always some multiple of half an hour.
$endgroup$
– Ian
Jun 1 at 18:51
$begingroup$
I can kinda see that. So time must be viewed as half hour increments, because that's the window the consultant offers their time. "Why wont continuous work?" is a better question
$endgroup$
– Jordan Greenhut
Jun 1 at 18:53
$begingroup$
A continuous treatment of the problem without the rounding would work, but it would most likely give a different answer, unless all the rounding errors somehow cancel out.
$endgroup$
– Ian
Jun 1 at 19:28
add a comment |
$begingroup$
Because it gets rounded up to the nearest half hour, the expected fee is simply
$$$150left(frac0.5^216-frac0^216right)+$300left(frac1^216-frac0.5^216right)+cdots+$1200left(frac4^216-frac3.5^216right)$$
(the probability it lands in each half hour multiplied by the amount received in that half hour)
which is just $$$1200-$150left(frac3.5^216+cdots+frac0.5^216+frac0^216right)=$871.875approx$872$$
answered Jun 1 at 18:48
auscryptauscrypt
7,416614
$endgroup$
1
$begingroup$
Why can't it be continuous? It's rounded... that's not doing it for me
$endgroup$
– Jordan Greenhut
Jun 1 at 18:49
1
$begingroup$
@JordanGreenhut If the amount of time it takes is continuous, but the billing is done based on a rounded form of this time, then the amount billed is discrete, since the time associated to the bill is always some multiple of half an hour.
$endgroup$
– Ian
Jun 1 at 18:51
$begingroup$
I can kinda see that. So time must be viewed as half hour increments, because that's the window the consultant offers their time. "Why wont continuous work?" is a better question
$endgroup$
– Jordan Greenhut
Jun 1 at 18:53
$begingroup$
A continuous treatment of the problem without the rounding would work, but it would most likely give a different answer, unless all the rounding errors somehow cancel out.
$endgroup$
– Ian
Jun 1 at 19:28
add a comment |
$begingroup$
Because it gets rounded up to the nearest half hour, the expected fee is simply
$$$150left(frac0.5^216-frac0^216right)+$300left(frac1^216-frac0.5^216right)+cdots+$1200left(frac4^216-frac3.5^216right)$$
(the probability it lands in each half hour multiplied by the amount received in that half hour)
which is just $$$1200-$150left(frac3.5^216+cdots+frac0.5^216+frac0^216right)=$871.875approx$872$$
answered Jun 1 at 18:48
auscryptauscrypt
7,416614
$endgroup$
Because it gets rounded up to the nearest half hour, the expected fee is simply
$$$150left(frac0.5^216-frac0^216right)+$300left(frac1^216-frac0.5^216right)+cdots+$1200left(frac4^216-frac3.5^216right)$$
(the probability it lands in each half hour multiplied by the amount received in that half hour)
which is just $$$1200-$150left(frac3.5^216+cdots+frac0.5^216+frac0^216right)=$871.875approx$872$$
answered Jun 1 at 18:48
auscryptauscrypt
7,416614
answered Jun 1 at 18:48
auscryptauscrypt
7,416614
answered Jun 1 at 18:48
auscryptauscrypt
7,416614
answered Jun 1 at 18:48
auscryptauscrypt
7,416614
7,416614
1
$begingroup$
Why can't it be continuous? It's rounded... that's not doing it for me
$endgroup$
– Jordan Greenhut
Jun 1 at 18:49
1
$begingroup$
@JordanGreenhut If the amount of time it takes is continuous, but the billing is done based on a rounded form of this time, then the amount billed is discrete, since the time associated to the bill is always some multiple of half an hour.
$endgroup$
– Ian
Jun 1 at 18:51
$begingroup$
I can kinda see that. So time must be viewed as half hour increments, because that's the window the consultant offers their time. "Why wont continuous work?" is a better question
$endgroup$
– Jordan Greenhut
Jun 1 at 18:53
$begingroup$
A continuous treatment of the problem without the rounding would work, but it would most likely give a different answer, unless all the rounding errors somehow cancel out.
$endgroup$
– Ian
Jun 1 at 19:28
add a comment |
1
$begingroup$
Why can't it be continuous? It's rounded... that's not doing it for me
$endgroup$
– Jordan Greenhut
Jun 1 at 18:49
1
$begingroup$
@JordanGreenhut If the amount of time it takes is continuous, but the billing is done based on a rounded form of this time, then the amount billed is discrete, since the time associated to the bill is always some multiple of half an hour.
$endgroup$
– Ian
Jun 1 at 18:51
$begingroup$
I can kinda see that. So time must be viewed as half hour increments, because that's the window the consultant offers their time. "Why wont continuous work?" is a better question
$endgroup$
– Jordan Greenhut
Jun 1 at 18:53
$begingroup$
A continuous treatment of the problem without the rounding would work, but it would most likely give a different answer, unless all the rounding errors somehow cancel out.
$endgroup$
– Ian
Jun 1 at 19:28
1
1
$begingroup$
Why can't it be continuous? It's rounded... that's not doing it for me
$endgroup$
– Jordan Greenhut
Jun 1 at 18:49
$begingroup$
Why can't it be continuous? It's rounded... that's not doing it for me
$endgroup$
– Jordan Greenhut
Jun 1 at 18:49
1
1
$begingroup$
@JordanGreenhut If the amount of time it takes is continuous, but the billing is done based on a rounded form of this time, then the amount billed is discrete, since the time associated to the bill is always some multiple of half an hour.
$endgroup$
– Ian
Jun 1 at 18:51
$begingroup$
@JordanGreenhut If the amount of time it takes is continuous, but the billing is done based on a rounded form of this time, then the amount billed is discrete, since the time associated to the bill is always some multiple of half an hour.
$endgroup$
– Ian
Jun 1 at 18:51
$begingroup$
I can kinda see that. So time must be viewed as half hour increments, because that's the window the consultant offers their time. "Why wont continuous work?" is a better question
$endgroup$
– Jordan Greenhut
Jun 1 at 18:53
$begingroup$
I can kinda see that. So time must be viewed as half hour increments, because that's the window the consultant offers their time. "Why wont continuous work?" is a better question
$endgroup$
– Jordan Greenhut
Jun 1 at 18:53
$begingroup$
A continuous treatment of the problem without the rounding would work, but it would most likely give a different answer, unless all the rounding errors somehow cancel out.
$endgroup$
– Ian
Jun 1 at 19:28
$begingroup$
A continuous treatment of the problem without the rounding would work, but it would most likely give a different answer, unless all the rounding errors somehow cancel out.
$endgroup$
– Ian
Jun 1 at 19:28
add a comment |
$begingroup$
The problem here doesn't really have to do with discrete vs continuous. What you have done is to correctly calculate the expected time the project will take, not the expected bill. It is not correct to multiply the expected time by $300$ to get the expected bill, even if you round up that time.
Why is this? Let's say $X$ is the random variable representing the time taken, and $Y$ is the random variable representing the bill.
If the consultant did not round to the nearest half hour, then we would just have $Y=300X$.
In that case, you could have used the following (correct) reasoning:
$$textSince Y=300X,quad E[Y]=300E[X].$$
However, we actually have $Y=300,mathttrnd(X)$, where $mathttrnd$ is the function that rounds up to the nearest half hour. Your method used the following (incorrect) reasoning:
$$textSince Y=300,mathttrnd(X),quad E[Y]=300,mathttrnd(E[X])quadcolorredtextFalse!$$
While expectation is linear, it does not commute with arbitrary functions (in this case the rounding function).
You have to compute the expectation of the $Y$ variable (the cost) directly, there is no way to get it just from the expectation of the $X$ variable (the time) - because the relationship between the two variables is more complicated than just a linear function.
P.S. The problem also says the time is rounded up to the nearest half hour, so your 2.66667 should have been rounded up to 3, not down to 2.5. That still wouldn't give the right answer for the reasons I explained above, but keep it in mind if you want to try the problem again.
answered Jun 2 at 5:32
CarmeisterCarmeister
2,9692925
$endgroup$
add a comment |
$begingroup$
The problem here doesn't really have to do with discrete vs continuous. What you have done is to correctly calculate the expected time the project will take, not the expected bill. It is not correct to multiply the expected time by $300$ to get the expected bill, even if you round up that time.
Why is this? Let's say $X$ is the random variable representing the time taken, and $Y$ is the random variable representing the bill.
If the consultant did not round to the nearest half hour, then we would just have $Y=300X$.
In that case, you could have used the following (correct) reasoning:
$$textSince Y=300X,quad E[Y]=300E[X].$$
However, we actually have $Y=300,mathttrnd(X)$, where $mathttrnd$ is the function that rounds up to the nearest half hour. Your method used the following (incorrect) reasoning:
$$textSince Y=300,mathttrnd(X),quad E[Y]=300,mathttrnd(E[X])quadcolorredtextFalse!$$
While expectation is linear, it does not commute with arbitrary functions (in this case the rounding function).
You have to compute the expectation of the $Y$ variable (the cost) directly, there is no way to get it just from the expectation of the $X$ variable (the time) - because the relationship between the two variables is more complicated than just a linear function.
P.S. The problem also says the time is rounded up to the nearest half hour, so your 2.66667 should have been rounded up to 3, not down to 2.5. That still wouldn't give the right answer for the reasons I explained above, but keep it in mind if you want to try the problem again.
answered Jun 2 at 5:32
CarmeisterCarmeister
2,9692925
$endgroup$
add a comment |
$begingroup$
The problem here doesn't really have to do with discrete vs continuous. What you have done is to correctly calculate the expected time the project will take, not the expected bill. It is not correct to multiply the expected time by $300$ to get the expected bill, even if you round up that time.
Why is this? Let's say $X$ is the random variable representing the time taken, and $Y$ is the random variable representing the bill.
If the consultant did not round to the nearest half hour, then we would just have $Y=300X$.
In that case, you could have used the following (correct) reasoning:
$$textSince Y=300X,quad E[Y]=300E[X].$$
However, we actually have $Y=300,mathttrnd(X)$, where $mathttrnd$ is the function that rounds up to the nearest half hour. Your method used the following (incorrect) reasoning:
$$textSince Y=300,mathttrnd(X),quad E[Y]=300,mathttrnd(E[X])quadcolorredtextFalse!$$
While expectation is linear, it does not commute with arbitrary functions (in this case the rounding function).
You have to compute the expectation of the $Y$ variable (the cost) directly, there is no way to get it just from the expectation of the $X$ variable (the time) - because the relationship between the two variables is more complicated than just a linear function.
P.S. The problem also says the time is rounded up to the nearest half hour, so your 2.66667 should have been rounded up to 3, not down to 2.5. That still wouldn't give the right answer for the reasons I explained above, but keep it in mind if you want to try the problem again.
answered Jun 2 at 5:32
CarmeisterCarmeister
2,9692925
$endgroup$
The problem here doesn't really have to do with discrete vs continuous. What you have done is to correctly calculate the expected time the project will take, not the expected bill. It is not correct to multiply the expected time by $300$ to get the expected bill, even if you round up that time.
Why is this? Let's say $X$ is the random variable representing the time taken, and $Y$ is the random variable representing the bill.
If the consultant did not round to the nearest half hour, then we would just have $Y=300X$.
In that case, you could have used the following (correct) reasoning:
$$textSince Y=300X,quad E[Y]=300E[X].$$
However, we actually have $Y=300,mathttrnd(X)$, where $mathttrnd$ is the function that rounds up to the nearest half hour. Your method used the following (incorrect) reasoning:
$$textSince Y=300,mathttrnd(X),quad E[Y]=300,mathttrnd(E[X])quadcolorredtextFalse!$$
While expectation is linear, it does not commute with arbitrary functions (in this case the rounding function).
You have to compute the expectation of the $Y$ variable (the cost) directly, there is no way to get it just from the expectation of the $X$ variable (the time) - because the relationship between the two variables is more complicated than just a linear function.
P.S. The problem also says the time is rounded up to the nearest half hour, so your 2.66667 should have been rounded up to 3, not down to 2.5. That still wouldn't give the right answer for the reasons I explained above, but keep it in mind if you want to try the problem again.
answered Jun 2 at 5:32
CarmeisterCarmeister
2,9692925
answered Jun 2 at 5:32
CarmeisterCarmeister
2,9692925
answered Jun 2 at 5:32
CarmeisterCarmeister
2,9692925
answered Jun 2 at 5:32
CarmeisterCarmeister
2,9692925
2,9692925
add a comment |
add a comment |
$begingroup$
Let $X$ be the amount of the total bill, in dollars. Then $X$ is certainly a discrete random variable. Its possible values are $0, 150, 300, 450, ldots, 1200$. To treat $X$ as a continuous random variable is an error, because it is simply not so.
answered Jun 1 at 18:54
littleOlittleO
31.8k651114
$endgroup$
$begingroup$
Thank you, that helped a little
$endgroup$
– Jordan Greenhut
Jun 1 at 18:55
add a comment |
$begingroup$
Let $X$ be the amount of the total bill, in dollars. Then $X$ is certainly a discrete random variable. Its possible values are $0, 150, 300, 450, ldots, 1200$. To treat $X$ as a continuous random variable is an error, because it is simply not so.
answered Jun 1 at 18:54
littleOlittleO
31.8k651114
$endgroup$
$begingroup$
Thank you, that helped a little
$endgroup$
– Jordan Greenhut
Jun 1 at 18:55
add a comment |
$begingroup$
Let $X$ be the amount of the total bill, in dollars. Then $X$ is certainly a discrete random variable. Its possible values are $0, 150, 300, 450, ldots, 1200$. To treat $X$ as a continuous random variable is an error, because it is simply not so.
answered Jun 1 at 18:54
littleOlittleO
31.8k651114
$endgroup$
Let $X$ be the amount of the total bill, in dollars. Then $X$ is certainly a discrete random variable. Its possible values are $0, 150, 300, 450, ldots, 1200$. To treat $X$ as a continuous random variable is an error, because it is simply not so.
answered Jun 1 at 18:54
littleOlittleO
31.8k651114
edited Jun 1 at 18:58
answered Jun 1 at 18:54
littleOlittleO
31.8k651114
answered Jun 1 at 18:54
littleOlittleO
31.8k651114
answered Jun 1 at 18:54
littleOlittleO
31.8k651114
31.8k651114
$begingroup$
Thank you, that helped a little
$endgroup$
– Jordan Greenhut
Jun 1 at 18:55
add a comment |
$begingroup$
Thank you, that helped a little
$endgroup$
– Jordan Greenhut
Jun 1 at 18:55
$begingroup$
Thank you, that helped a little
$endgroup$
– Jordan Greenhut
Jun 1 at 18:55
$begingroup$
Thank you, that helped a little
$endgroup$
– Jordan Greenhut
Jun 1 at 18:55
add a comment |
$begingroup$
The probability for the duration to be between $frack-12$ and $frac k2$ is $frac2k-164$ and the charge would be $150k$. Thus, the expected charge would be
$$
beginalign
frac7532sum_k=1^8left(2k^2-kright)
&=frac7532sum_k=1^8left[4binomk2+binomk1right]\
&=frac7532left[4binomk+13+binomk+12right]_k=0^k=8\
&=frac7532left[4binom93+binom92right]\
&=frac7532cdot372\[9pt]
&=871tfrac78
endalign
$$
answered Jun 1 at 19:16
robjohn♦robjohn
274k28320655
$endgroup$
$begingroup$
Fancy. Can you explain it further? How do you arrive at the left hand side equation?
$endgroup$
– Jordan Greenhut
Jun 1 at 19:20
$begingroup$
$frac7532left(2k^2-kright)=frac2k-164cdot150k$ from the line above.
$endgroup$
– robjohn♦
Jun 1 at 19:24
$begingroup$
$sum_kbinomkn=binomk+1n+1$ follows from Pascal's recurrence: $binomk+1n+1=binomkn+1+binomkn$
$endgroup$
– robjohn♦
Jun 1 at 19:26
add a comment |
$begingroup$
The probability for the duration to be between $frack-12$ and $frac k2$ is $frac2k-164$ and the charge would be $150k$. Thus, the expected charge would be
$$
beginalign
frac7532sum_k=1^8left(2k^2-kright)
&=frac7532sum_k=1^8left[4binomk2+binomk1right]\
&=frac7532left[4binomk+13+binomk+12right]_k=0^k=8\
&=frac7532left[4binom93+binom92right]\
&=frac7532cdot372\[9pt]
&=871tfrac78
endalign
$$
answered Jun 1 at 19:16
robjohn♦robjohn
274k28320655
$endgroup$
$begingroup$
Fancy. Can you explain it further? How do you arrive at the left hand side equation?
$endgroup$
– Jordan Greenhut
Jun 1 at 19:20
$begingroup$
$frac7532left(2k^2-kright)=frac2k-164cdot150k$ from the line above.
$endgroup$
– robjohn♦
Jun 1 at 19:24
$begingroup$
$sum_kbinomkn=binomk+1n+1$ follows from Pascal's recurrence: $binomk+1n+1=binomkn+1+binomkn$
$endgroup$
– robjohn♦
Jun 1 at 19:26
add a comment |
$begingroup$
The probability for the duration to be between $frack-12$ and $frac k2$ is $frac2k-164$ and the charge would be $150k$. Thus, the expected charge would be
$$
beginalign
frac7532sum_k=1^8left(2k^2-kright)
&=frac7532sum_k=1^8left[4binomk2+binomk1right]\
&=frac7532left[4binomk+13+binomk+12right]_k=0^k=8\
&=frac7532left[4binom93+binom92right]\
&=frac7532cdot372\[9pt]
&=871tfrac78
endalign
$$
answered Jun 1 at 19:16
robjohn♦robjohn
274k28320655
$endgroup$
The probability for the duration to be between $frack-12$ and $frac k2$ is $frac2k-164$ and the charge would be $150k$. Thus, the expected charge would be
$$
beginalign
frac7532sum_k=1^8left(2k^2-kright)
&=frac7532sum_k=1^8left[4binomk2+binomk1right]\
&=frac7532left[4binomk+13+binomk+12right]_k=0^k=8\
&=frac7532left[4binom93+binom92right]\
&=frac7532cdot372\[9pt]
&=871tfrac78
endalign
$$
answered Jun 1 at 19:16
robjohn♦robjohn
274k28320655
answered Jun 1 at 19:16
robjohn♦robjohn
274k28320655
answered Jun 1 at 19:16
robjohn♦robjohn
274k28320655
answered Jun 1 at 19:16
robjohn♦robjohn
274k28320655
274k28320655
$begingroup$
Fancy. Can you explain it further? How do you arrive at the left hand side equation?
$endgroup$
– Jordan Greenhut
Jun 1 at 19:20
$begingroup$
$frac7532left(2k^2-kright)=frac2k-164cdot150k$ from the line above.
$endgroup$
– robjohn♦
Jun 1 at 19:24
$begingroup$
$sum_kbinomkn=binomk+1n+1$ follows from Pascal's recurrence: $binomk+1n+1=binomkn+1+binomkn$
$endgroup$
– robjohn♦
Jun 1 at 19:26
add a comment |
$begingroup$
Fancy. Can you explain it further? How do you arrive at the left hand side equation?
$endgroup$
– Jordan Greenhut
Jun 1 at 19:20
$begingroup$
$frac7532left(2k^2-kright)=frac2k-164cdot150k$ from the line above.
$endgroup$
– robjohn♦
Jun 1 at 19:24
$begingroup$
$sum_kbinomkn=binomk+1n+1$ follows from Pascal's recurrence: $binomk+1n+1=binomkn+1+binomkn$
$endgroup$
– robjohn♦
Jun 1 at 19:26
$begingroup$
Fancy. Can you explain it further? How do you arrive at the left hand side equation?
$endgroup$
– Jordan Greenhut
Jun 1 at 19:20
$begingroup$
Fancy. Can you explain it further? How do you arrive at the left hand side equation?
$endgroup$
– Jordan Greenhut
Jun 1 at 19:20
$begingroup$
$frac7532left(2k^2-kright)=frac2k-164cdot150k$ from the line above.
$endgroup$
– robjohn♦
Jun 1 at 19:24
$begingroup$
$frac7532left(2k^2-kright)=frac2k-164cdot150k$ from the line above.
$endgroup$
– robjohn♦
Jun 1 at 19:24
$begingroup$
$sum_kbinomkn=binomk+1n+1$ follows from Pascal's recurrence: $binomk+1n+1=binomkn+1+binomkn$
$endgroup$
– robjohn♦
Jun 1 at 19:26
$begingroup$
$sum_kbinomkn=binomk+1n+1$ follows from Pascal's recurrence: $binomk+1n+1=binomkn+1+binomkn$
$endgroup$
– robjohn♦
Jun 1 at 19:26
add a comment |
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7
$begingroup$
It seems to me that you have just ignored the part about rounding up to the nearest half hour. I would say that a discrete approach is necessary here.
$endgroup$
– TonyK
Jun 1 at 18:32
3
$begingroup$
The probability distribution is continuous, but the cost is a step function.
$endgroup$
– eyeballfrog
Jun 1 at 18:34
1
$begingroup$
How can you think that charging up to $150 extra is subtle?
$endgroup$
– TonyK
Jun 1 at 18:35
2
$begingroup$
Just for intuition: If you earn $10$% interest on $$1000$, compounding annually, after $1$ year you'll have $$1100$. If you compound interest continuously, you'll have $$1000e^.1 approx $1105.17$. There's a significant difference.
$endgroup$
– Ted Shifrin
Jun 1 at 18:59
2
$begingroup$
Wow, that's the answer I wanted Ted. That would get gold, if I have such powers. I forgot this part of accounting. I always wondered why we had to take accounting... I will remember the difference from compounding interest over time increments versus compounding continuously can have large implications (it should have been obvious and that's the thought I need to pass this f***er)
$endgroup$
– Jordan Greenhut
Jun 1 at 19:01