Mixing
Uniform Distribution - Is my solution correct?
$begingroup$
I'm preparing for an exam in Probability and Statistics and since I am self-studying, I don't have answers to refer to. So would be really great if someone would please take the time and check if my solution for the following problem is correct. Thank you in advance!
X corresponds to the duration of Mary's commute to work, and Y to the duration of John's. X is uniformly distributed in [15,25] and Y in [15,30]. X and Y are assumed to be independent.
What is the probability that Mary and John both take more than 20 minutes to get to work?
So we are looking for: P(X>20,Y>20)=P(X>20)xP(Y>20). In other words we are looking for: P[1-(P(X≦20),1-(P(Y≦20).
As per the rule P with Uniform Distribution:
P([c,d])= $int_c^dfrac{1}{b-a}dx=frac{d-c}{b-a}$
So in our cases:
$P(X≦20)= P([15,20])= int_{15}^{20}frac{1}{25-15}dx=frac{20-15}{25-15}=0.5$ from here
follows that $1-P(X≦20)= 1-0.5=0.5$
So $P(X>20)=0.5$
$P(Y≦20)= P([15,20])= int_{15}^{20}frac{1}{30-15}dx=frac{20-15}{30-15}=0.33$ from here
follows that $1-P(Y≦20)= 1-0.33=0.67$
So $P(Y>20)=0.57$
Going back to
P(X>20,Y>20)=P(X>20)xP(Y>20)=0.5x0.67=0.335
In conclusion, that probability that both Mary and John take more than 20 minutes to get to work is 0.335.
probability statistics uniform-distribution
asked Feb 2 at 16:05
VRTVRT
957
$endgroup$
add a comment |
$begingroup$
I'm preparing for an exam in Probability and Statistics and since I am self-studying, I don't have answers to refer to. So would be really great if someone would please take the time and check if my solution for the following problem is correct. Thank you in advance!
X corresponds to the duration of Mary's commute to work, and Y to the duration of John's. X is uniformly distributed in [15,25] and Y in [15,30]. X and Y are assumed to be independent.
What is the probability that Mary and John both take more than 20 minutes to get to work?
So we are looking for: P(X>20,Y>20)=P(X>20)xP(Y>20). In other words we are looking for: P[1-(P(X≦20),1-(P(Y≦20).
As per the rule P with Uniform Distribution:
P([c,d])= $int_c^dfrac{1}{b-a}dx=frac{d-c}{b-a}$
So in our cases:
$P(X≦20)= P([15,20])= int_{15}^{20}frac{1}{25-15}dx=frac{20-15}{25-15}=0.5$ from here
follows that $1-P(X≦20)= 1-0.5=0.5$
So $P(X>20)=0.5$
$P(Y≦20)= P([15,20])= int_{15}^{20}frac{1}{30-15}dx=frac{20-15}{30-15}=0.33$ from here
follows that $1-P(Y≦20)= 1-0.33=0.67$
So $P(Y>20)=0.57$
Going back to
P(X>20,Y>20)=P(X>20)xP(Y>20)=0.5x0.67=0.335
In conclusion, that probability that both Mary and John take more than 20 minutes to get to work is 0.335.
probability statistics uniform-distribution
asked Feb 2 at 16:05
VRTVRT
957
$endgroup$
$begingroup$
In your situation, it can behoove you to use some programming language to double check with a few lines of code if your answer is at least plausible. For instance, in R: X = runif(n = 1e6, min = 15, max = 25) > 20; Y = runif(n = 1e6, min = 15, max = 30) > 20; XY = cbind(X,Y); mean(rowSums(XY) == 2) [1] 0.332743
$endgroup$
– Antoni Parellada
Feb 2 at 16:25
add a comment |
$begingroup$
I'm preparing for an exam in Probability and Statistics and since I am self-studying, I don't have answers to refer to. So would be really great if someone would please take the time and check if my solution for the following problem is correct. Thank you in advance!
X corresponds to the duration of Mary's commute to work, and Y to the duration of John's. X is uniformly distributed in [15,25] and Y in [15,30]. X and Y are assumed to be independent.
What is the probability that Mary and John both take more than 20 minutes to get to work?
So we are looking for: P(X>20,Y>20)=P(X>20)xP(Y>20). In other words we are looking for: P[1-(P(X≦20),1-(P(Y≦20).
As per the rule P with Uniform Distribution:
P([c,d])= $int_c^dfrac{1}{b-a}dx=frac{d-c}{b-a}$
So in our cases:
$P(X≦20)= P([15,20])= int_{15}^{20}frac{1}{25-15}dx=frac{20-15}{25-15}=0.5$ from here
follows that $1-P(X≦20)= 1-0.5=0.5$
So $P(X>20)=0.5$
$P(Y≦20)= P([15,20])= int_{15}^{20}frac{1}{30-15}dx=frac{20-15}{30-15}=0.33$ from here
follows that $1-P(Y≦20)= 1-0.33=0.67$
So $P(Y>20)=0.57$
Going back to
P(X>20,Y>20)=P(X>20)xP(Y>20)=0.5x0.67=0.335
In conclusion, that probability that both Mary and John take more than 20 minutes to get to work is 0.335.
probability statistics uniform-distribution
asked Feb 2 at 16:05
VRTVRT
957
$endgroup$
I'm preparing for an exam in Probability and Statistics and since I am self-studying, I don't have answers to refer to. So would be really great if someone would please take the time and check if my solution for the following problem is correct. Thank you in advance!
X corresponds to the duration of Mary's commute to work, and Y to the duration of John's. X is uniformly distributed in [15,25] and Y in [15,30]. X and Y are assumed to be independent.
What is the probability that Mary and John both take more than 20 minutes to get to work?
So we are looking for: P(X>20,Y>20)=P(X>20)xP(Y>20). In other words we are looking for: P[1-(P(X≦20),1-(P(Y≦20).
As per the rule P with Uniform Distribution:
P([c,d])= $int_c^dfrac{1}{b-a}dx=frac{d-c}{b-a}$
So in our cases:
$P(X≦20)= P([15,20])= int_{15}^{20}frac{1}{25-15}dx=frac{20-15}{25-15}=0.5$ from here
follows that $1-P(X≦20)= 1-0.5=0.5$
So $P(X>20)=0.5$
$P(Y≦20)= P([15,20])= int_{15}^{20}frac{1}{30-15}dx=frac{20-15}{30-15}=0.33$ from here
follows that $1-P(Y≦20)= 1-0.33=0.67$
So $P(Y>20)=0.57$
Going back to
P(X>20,Y>20)=P(X>20)xP(Y>20)=0.5x0.67=0.335
In conclusion, that probability that both Mary and John take more than 20 minutes to get to work is 0.335.
probability statistics uniform-distribution
probability statistics uniform-distribution
asked Feb 2 at 16:05
VRTVRT
957
asked Feb 2 at 16:05
VRTVRT
957
asked Feb 2 at 16:05
VRTVRT
957
asked Feb 2 at 16:05
VRTVRT
957
asked Feb 2 at 16:05
VRTVRT
957
957
$begingroup$
In your situation, it can behoove you to use some programming language to double check with a few lines of code if your answer is at least plausible. For instance, in R: X = runif(n = 1e6, min = 15, max = 25) > 20; Y = runif(n = 1e6, min = 15, max = 30) > 20; XY = cbind(X,Y); mean(rowSums(XY) == 2) [1] 0.332743
$endgroup$
– Antoni Parellada
Feb 2 at 16:25
add a comment |
$begingroup$
In your situation, it can behoove you to use some programming language to double check with a few lines of code if your answer is at least plausible. For instance, in R: X = runif(n = 1e6, min = 15, max = 25) > 20; Y = runif(n = 1e6, min = 15, max = 30) > 20; XY = cbind(X,Y); mean(rowSums(XY) == 2) [1] 0.332743
$endgroup$
– Antoni Parellada
Feb 2 at 16:25
$begingroup$
In your situation, it can behoove you to use some programming language to double check with a few lines of code if your answer is at least plausible. For instance, in R: X = runif(n = 1e6, min = 15, max = 25) > 20; Y = runif(n = 1e6, min = 15, max = 30) > 20; XY = cbind(X,Y); mean(rowSums(XY) == 2) [1] 0.332743
$endgroup$
– Antoni Parellada
Feb 2 at 16:25
$begingroup$
In your situation, it can behoove you to use some programming language to double check with a few lines of code if your answer is at least plausible. For instance, in R: X = runif(n = 1e6, min = 15, max = 25) > 20; Y = runif(n = 1e6, min = 15, max = 30) > 20; XY = cbind(X,Y); mean(rowSums(XY) == 2) [1] 0.332743
$endgroup$
– Antoni Parellada
Feb 2 at 16:25
add a comment |
2 Answers
2
active
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votes
$begingroup$
Generally looks good. I wouldn’t introduce decimals and start rounding—the exact answer is $1/3$, which is a bit different from $0.335$.
Also I’m not sure whether the notation $P([a,b])$ is standard. I typically write $P(aleq{X}leq{b})$, but that might be my background.
Also I’m not sure what you meant here:
P[1-(P(X≦20),1-(P(Y≦20)
I think you meant $$[1-P(Xleq20)]cdot[1-P(Yleq20)],$$ so be careful with notation.
answered Feb 2 at 16:11
David M.David M.
2,232421
$endgroup$
$begingroup$
Thank you so much! :)
$endgroup$
– VRT
Feb 2 at 16:25
add a comment |
$begingroup$
To check this intuitively, You could as well calculated it easily $P(X>20)$ and $P(Y>20)$.
Uniform distribution means every event is equally likely
So for X commute time is uniformly distributed between 15 and 25.
So the probability of reaching exactly at a particular minute between 15 and 25 is $1/10$. Similarly for Y it is $1/15$. Now how many minutes above 20 are there for X and Y in their respective ranges?
5 for X and 10 for Y. So, $P(X> 20)=5/10=1/2$ and $P(Y>20)=10/15=2/3$. Since the events are independent,
$P(X>20, Y>20)= 1/2 *2/3 =1/3$. If you look at the Probablity graph of the distribution, the integral that you posted is nothing but the rectangular area between $c=21$ and $d=25$ for X and $c=21$ and $d=30$ for Y. The height is the probablity which is $1/10$ and $1/15$ Respectively. Since it’s the same for every event you form a nice $y=f(x) = 1/(b-a)$ line.Now if the distribution is not Uniform, then each such event would have differing probability and so the integral area under the curve will vary.
answered Feb 2 at 17:05
Gopal AnantharamanGopal Anantharaman
826
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add a comment |
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2 Answers
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2 Answers
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$begingroup$
Generally looks good. I wouldn’t introduce decimals and start rounding—the exact answer is $1/3$, which is a bit different from $0.335$.
Also I’m not sure whether the notation $P([a,b])$ is standard. I typically write $P(aleq{X}leq{b})$, but that might be my background.
Also I’m not sure what you meant here:
P[1-(P(X≦20),1-(P(Y≦20)
I think you meant $$[1-P(Xleq20)]cdot[1-P(Yleq20)],$$ so be careful with notation.
answered Feb 2 at 16:11
David M.David M.
2,232421
$endgroup$
$begingroup$
Thank you so much! :)
$endgroup$
– VRT
Feb 2 at 16:25
add a comment |
$begingroup$
Generally looks good. I wouldn’t introduce decimals and start rounding—the exact answer is $1/3$, which is a bit different from $0.335$.
Also I’m not sure whether the notation $P([a,b])$ is standard. I typically write $P(aleq{X}leq{b})$, but that might be my background.
Also I’m not sure what you meant here:
P[1-(P(X≦20),1-(P(Y≦20)
I think you meant $$[1-P(Xleq20)]cdot[1-P(Yleq20)],$$ so be careful with notation.
answered Feb 2 at 16:11
David M.David M.
2,232421
$endgroup$
$begingroup$
Thank you so much! :)
$endgroup$
– VRT
Feb 2 at 16:25
add a comment |
$begingroup$
Generally looks good. I wouldn’t introduce decimals and start rounding—the exact answer is $1/3$, which is a bit different from $0.335$.
Also I’m not sure whether the notation $P([a,b])$ is standard. I typically write $P(aleq{X}leq{b})$, but that might be my background.
Also I’m not sure what you meant here:
P[1-(P(X≦20),1-(P(Y≦20)
I think you meant $$[1-P(Xleq20)]cdot[1-P(Yleq20)],$$ so be careful with notation.
answered Feb 2 at 16:11
David M.David M.
2,232421
$endgroup$
Generally looks good. I wouldn’t introduce decimals and start rounding—the exact answer is $1/3$, which is a bit different from $0.335$.
Also I’m not sure whether the notation $P([a,b])$ is standard. I typically write $P(aleq{X}leq{b})$, but that might be my background.
Also I’m not sure what you meant here:
P[1-(P(X≦20),1-(P(Y≦20)
I think you meant $$[1-P(Xleq20)]cdot[1-P(Yleq20)],$$ so be careful with notation.
answered Feb 2 at 16:11
David M.David M.
2,232421
answered Feb 2 at 16:11
David M.David M.
2,232421
answered Feb 2 at 16:11
David M.David M.
2,232421
answered Feb 2 at 16:11
David M.David M.
2,232421
2,232421
$begingroup$
Thank you so much! :)
$endgroup$
– VRT
Feb 2 at 16:25
add a comment |
$begingroup$
Thank you so much! :)
$endgroup$
– VRT
Feb 2 at 16:25
$begingroup$
Thank you so much! :)
$endgroup$
– VRT
Feb 2 at 16:25
$begingroup$
Thank you so much! :)
$endgroup$
– VRT
Feb 2 at 16:25
add a comment |
$begingroup$
To check this intuitively, You could as well calculated it easily $P(X>20)$ and $P(Y>20)$.
Uniform distribution means every event is equally likely
So for X commute time is uniformly distributed between 15 and 25.
So the probability of reaching exactly at a particular minute between 15 and 25 is $1/10$. Similarly for Y it is $1/15$. Now how many minutes above 20 are there for X and Y in their respective ranges?
5 for X and 10 for Y. So, $P(X> 20)=5/10=1/2$ and $P(Y>20)=10/15=2/3$. Since the events are independent,
$P(X>20, Y>20)= 1/2 *2/3 =1/3$. If you look at the Probablity graph of the distribution, the integral that you posted is nothing but the rectangular area between $c=21$ and $d=25$ for X and $c=21$ and $d=30$ for Y. The height is the probablity which is $1/10$ and $1/15$ Respectively. Since it’s the same for every event you form a nice $y=f(x) = 1/(b-a)$ line.Now if the distribution is not Uniform, then each such event would have differing probability and so the integral area under the curve will vary.
answered Feb 2 at 17:05
Gopal AnantharamanGopal Anantharaman
826
$endgroup$
add a comment |
$begingroup$
To check this intuitively, You could as well calculated it easily $P(X>20)$ and $P(Y>20)$.
Uniform distribution means every event is equally likely
So for X commute time is uniformly distributed between 15 and 25.
So the probability of reaching exactly at a particular minute between 15 and 25 is $1/10$. Similarly for Y it is $1/15$. Now how many minutes above 20 are there for X and Y in their respective ranges?
5 for X and 10 for Y. So, $P(X> 20)=5/10=1/2$ and $P(Y>20)=10/15=2/3$. Since the events are independent,
$P(X>20, Y>20)= 1/2 *2/3 =1/3$. If you look at the Probablity graph of the distribution, the integral that you posted is nothing but the rectangular area between $c=21$ and $d=25$ for X and $c=21$ and $d=30$ for Y. The height is the probablity which is $1/10$ and $1/15$ Respectively. Since it’s the same for every event you form a nice $y=f(x) = 1/(b-a)$ line.Now if the distribution is not Uniform, then each such event would have differing probability and so the integral area under the curve will vary.
answered Feb 2 at 17:05
Gopal AnantharamanGopal Anantharaman
826
$endgroup$
add a comment |
$begingroup$
To check this intuitively, You could as well calculated it easily $P(X>20)$ and $P(Y>20)$.
Uniform distribution means every event is equally likely
So for X commute time is uniformly distributed between 15 and 25.
So the probability of reaching exactly at a particular minute between 15 and 25 is $1/10$. Similarly for Y it is $1/15$. Now how many minutes above 20 are there for X and Y in their respective ranges?
5 for X and 10 for Y. So, $P(X> 20)=5/10=1/2$ and $P(Y>20)=10/15=2/3$. Since the events are independent,
$P(X>20, Y>20)= 1/2 *2/3 =1/3$. If you look at the Probablity graph of the distribution, the integral that you posted is nothing but the rectangular area between $c=21$ and $d=25$ for X and $c=21$ and $d=30$ for Y. The height is the probablity which is $1/10$ and $1/15$ Respectively. Since it’s the same for every event you form a nice $y=f(x) = 1/(b-a)$ line.Now if the distribution is not Uniform, then each such event would have differing probability and so the integral area under the curve will vary.
answered Feb 2 at 17:05
Gopal AnantharamanGopal Anantharaman
826
$endgroup$
To check this intuitively, You could as well calculated it easily $P(X>20)$ and $P(Y>20)$.
Uniform distribution means every event is equally likely
So for X commute time is uniformly distributed between 15 and 25.
So the probability of reaching exactly at a particular minute between 15 and 25 is $1/10$. Similarly for Y it is $1/15$. Now how many minutes above 20 are there for X and Y in their respective ranges?
5 for X and 10 for Y. So, $P(X> 20)=5/10=1/2$ and $P(Y>20)=10/15=2/3$. Since the events are independent,
$P(X>20, Y>20)= 1/2 *2/3 =1/3$. If you look at the Probablity graph of the distribution, the integral that you posted is nothing but the rectangular area between $c=21$ and $d=25$ for X and $c=21$ and $d=30$ for Y. The height is the probablity which is $1/10$ and $1/15$ Respectively. Since it’s the same for every event you form a nice $y=f(x) = 1/(b-a)$ line.Now if the distribution is not Uniform, then each such event would have differing probability and so the integral area under the curve will vary.
answered Feb 2 at 17:05
Gopal AnantharamanGopal Anantharaman
826
answered Feb 2 at 17:05
Gopal AnantharamanGopal Anantharaman
826
answered Feb 2 at 17:05
Gopal AnantharamanGopal Anantharaman
826
answered Feb 2 at 17:05
Gopal AnantharamanGopal Anantharaman
826
826
add a comment |
add a comment |
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$begingroup$
In your situation, it can behoove you to use some programming language to double check with a few lines of code if your answer is at least plausible. For instance, in R: X = runif(n = 1e6, min = 15, max = 25) > 20; Y = runif(n = 1e6, min = 15, max = 30) > 20; XY = cbind(X,Y); mean(rowSums(XY) == 2) [1] 0.332743
$endgroup$
– Antoni Parellada
Feb 2 at 16:25