If $U$ is uniformly distributed with mean $5$ and variance $3$, what is $P(U<4)$
-1
$begingroup$
I'm stuck on this question, can someone help me, many thanks.
If $U$ is uniformly distributed with mean $5$ and variance $3$, what is $P(U<4)$?
update(this problem has been solved):
I made a mistake when calculating, the result should be under the condition : x ranges from a to b. The final result is 1/3.
probability statistics uniform-distribution
asked Jan 17 at 12:42
wawawawawawa
33
$endgroup$
add a comment |
-1
$begingroup$
I'm stuck on this question, can someone help me, many thanks.
If $U$ is uniformly distributed with mean $5$ and variance $3$, what is $P(U<4)$?
update(this problem has been solved):
I made a mistake when calculating, the result should be under the condition : x ranges from a to b. The final result is 1/3.
probability statistics uniform-distribution
asked Jan 17 at 12:42
wawawawawawa
33
$endgroup$
2
$begingroup$
uniformly distributed over what set?
$endgroup$
– mathworker21
Jan 17 at 12:44
1
$begingroup$
Where did you get stuck with solving this? What have you tried already? Questions without context and visible effort often get downvoted and closed.
$endgroup$
– postmortes
Jan 17 at 12:58
add a comment |
-1
-1
-1
$begingroup$
I'm stuck on this question, can someone help me, many thanks.
If $U$ is uniformly distributed with mean $5$ and variance $3$, what is $P(U<4)$?
update(this problem has been solved):
I made a mistake when calculating, the result should be under the condition : x ranges from a to b. The final result is 1/3.
probability statistics uniform-distribution
asked Jan 17 at 12:42
wawawawawawa
33
$endgroup$
I'm stuck on this question, can someone help me, many thanks.
If $U$ is uniformly distributed with mean $5$ and variance $3$, what is $P(U<4)$?
update(this problem has been solved):
I made a mistake when calculating, the result should be under the condition : x ranges from a to b. The final result is 1/3.
probability statistics uniform-distribution
probability statistics uniform-distribution
asked Jan 17 at 12:42
wawawawawawa
33
asked Jan 17 at 12:42
wawawawawawa
33
edited Jan 19 at 12:50
wawawa
asked Jan 17 at 12:42
wawawawawawa
33
asked Jan 17 at 12:42
wawawawawawa
33
asked Jan 17 at 12:42
wawawawawawa
33
33
2
$begingroup$
uniformly distributed over what set?
$endgroup$
– mathworker21
Jan 17 at 12:44
1
$begingroup$
Where did you get stuck with solving this? What have you tried already? Questions without context and visible effort often get downvoted and closed.
$endgroup$
– postmortes
Jan 17 at 12:58
add a comment |
2
$begingroup$
uniformly distributed over what set?
$endgroup$
– mathworker21
Jan 17 at 12:44
1
$begingroup$
Where did you get stuck with solving this? What have you tried already? Questions without context and visible effort often get downvoted and closed.
$endgroup$
– postmortes
Jan 17 at 12:58
2
2
$begingroup$
uniformly distributed over what set?
$endgroup$
– mathworker21
Jan 17 at 12:44
$begingroup$
uniformly distributed over what set?
$endgroup$
– mathworker21
Jan 17 at 12:44
1
1
$begingroup$
Where did you get stuck with solving this? What have you tried already? Questions without context and visible effort often get downvoted and closed.
$endgroup$
– postmortes
Jan 17 at 12:58
$begingroup$
Where did you get stuck with solving this? What have you tried already? Questions without context and visible effort often get downvoted and closed.
$endgroup$
– postmortes
Jan 17 at 12:58
add a comment |
2 Answers
2
active
oldest
votes
2
$begingroup$
Hint:- Let us suppose $U$ follows Uniform distribution with parameter $a$ and $b$.
Mean=$E(U)=frac{b+a}{2}=5$ and Variance $=V(U)=frac{(b-a)^2}{12}=3$.
$endgroup$
1
$begingroup$
You are assuming that $U$ is continuous Uniform. All that is stated is that it is Uniform (which might be continuous or discrete)
$endgroup$
– wolfies
Jan 17 at 12:47
$begingroup$
Yes I know this formula, the result is 1/3 right? I just realized I made a mistake, x should be in the range from a to b
$endgroup$
– wawawa
Jan 17 at 12:48
$begingroup$
@ Cecilia Yes, you are correct. I forgot to mention one thing that you solve the two equations under the restriction that $a<b$.
$endgroup$
– user440191
Jan 17 at 12:50
$begingroup$
@ wolfies you are absolutely right. I should have mentioned that I am assuming continuous uniform distribution. However, if one assume that it is the case of discrete uniform, I think thee will be a issue regarding the consideration of the set over which it is uniformly distributed ( as mentioned by @mathworker above)
$endgroup$
– user440191
Jan 17 at 12:58
$begingroup$
I urge @Cecilia to consider the discrete case when the set is ${1,2,dots N}$ and see what happens.
$endgroup$
– user440191
Jan 17 at 12:59
add a comment |
0
$begingroup$
As the mean is directly in the middle between $a$ and $b$ you can set
$a = 5-x$ and $b=5+x$ for $x>0$
Now solve
$$sigma^2 = frac{(b-a)^2}{12}= frac{(2x)^2}{12}=frac{x^2}{3} = 3stackrel{x>0}{Rightarrow} x= 3$$
So, $U$ "lives" on $[a,b]=[2,8]$. It follows
$$P(U<4) = frac{4-2}{8-2}=frac{2}{6}=frac{1}{3}$$
answered Jan 17 at 12:58
trancelocationtrancelocation
14.1k1829
$endgroup$
$begingroup$
Thank you so much for the answer, however @Bhargob answers the question before you haha and I've already chosen his answer, thanks for your time and I really appreciated it.
$endgroup$
– wawawa
Jan 17 at 14:59
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
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active
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2
$begingroup$
Hint:- Let us suppose $U$ follows Uniform distribution with parameter $a$ and $b$.
Mean=$E(U)=frac{b+a}{2}=5$ and Variance $=V(U)=frac{(b-a)^2}{12}=3$.
$endgroup$
1
$begingroup$
You are assuming that $U$ is continuous Uniform. All that is stated is that it is Uniform (which might be continuous or discrete)
$endgroup$
– wolfies
Jan 17 at 12:47
$begingroup$
Yes I know this formula, the result is 1/3 right? I just realized I made a mistake, x should be in the range from a to b
$endgroup$
– wawawa
Jan 17 at 12:48
$begingroup$
@ Cecilia Yes, you are correct. I forgot to mention one thing that you solve the two equations under the restriction that $a<b$.
$endgroup$
– user440191
Jan 17 at 12:50
$begingroup$
@ wolfies you are absolutely right. I should have mentioned that I am assuming continuous uniform distribution. However, if one assume that it is the case of discrete uniform, I think thee will be a issue regarding the consideration of the set over which it is uniformly distributed ( as mentioned by @mathworker above)
$endgroup$
– user440191
Jan 17 at 12:58
$begingroup$
I urge @Cecilia to consider the discrete case when the set is ${1,2,dots N}$ and see what happens.
$endgroup$
– user440191
Jan 17 at 12:59
add a comment |
2
$begingroup$
Hint:- Let us suppose $U$ follows Uniform distribution with parameter $a$ and $b$.
Mean=$E(U)=frac{b+a}{2}=5$ and Variance $=V(U)=frac{(b-a)^2}{12}=3$.
$endgroup$
1
$begingroup$
You are assuming that $U$ is continuous Uniform. All that is stated is that it is Uniform (which might be continuous or discrete)
$endgroup$
– wolfies
Jan 17 at 12:47
$begingroup$
Yes I know this formula, the result is 1/3 right? I just realized I made a mistake, x should be in the range from a to b
$endgroup$
– wawawa
Jan 17 at 12:48
$begingroup$
@ Cecilia Yes, you are correct. I forgot to mention one thing that you solve the two equations under the restriction that $a<b$.
$endgroup$
– user440191
Jan 17 at 12:50
$begingroup$
@ wolfies you are absolutely right. I should have mentioned that I am assuming continuous uniform distribution. However, if one assume that it is the case of discrete uniform, I think thee will be a issue regarding the consideration of the set over which it is uniformly distributed ( as mentioned by @mathworker above)
$endgroup$
– user440191
Jan 17 at 12:58
$begingroup$
I urge @Cecilia to consider the discrete case when the set is ${1,2,dots N}$ and see what happens.
$endgroup$
– user440191
Jan 17 at 12:59
add a comment |
2
2
2
$begingroup$
Hint:- Let us suppose $U$ follows Uniform distribution with parameter $a$ and $b$.
Mean=$E(U)=frac{b+a}{2}=5$ and Variance $=V(U)=frac{(b-a)^2}{12}=3$.
$endgroup$
Hint:- Let us suppose $U$ follows Uniform distribution with parameter $a$ and $b$.
Mean=$E(U)=frac{b+a}{2}=5$ and Variance $=V(U)=frac{(b-a)^2}{12}=3$.
answered Jan 17 at 12:45
user440191
1
$begingroup$
You are assuming that $U$ is continuous Uniform. All that is stated is that it is Uniform (which might be continuous or discrete)
$endgroup$
– wolfies
Jan 17 at 12:47
$begingroup$
Yes I know this formula, the result is 1/3 right? I just realized I made a mistake, x should be in the range from a to b
$endgroup$
– wawawa
Jan 17 at 12:48
$begingroup$
@ Cecilia Yes, you are correct. I forgot to mention one thing that you solve the two equations under the restriction that $a<b$.
$endgroup$
– user440191
Jan 17 at 12:50
$begingroup$
@ wolfies you are absolutely right. I should have mentioned that I am assuming continuous uniform distribution. However, if one assume that it is the case of discrete uniform, I think thee will be a issue regarding the consideration of the set over which it is uniformly distributed ( as mentioned by @mathworker above)
$endgroup$
– user440191
Jan 17 at 12:58
$begingroup$
I urge @Cecilia to consider the discrete case when the set is ${1,2,dots N}$ and see what happens.
$endgroup$
– user440191
Jan 17 at 12:59
add a comment |
1
$begingroup$
You are assuming that $U$ is continuous Uniform. All that is stated is that it is Uniform (which might be continuous or discrete)
$endgroup$
– wolfies
Jan 17 at 12:47
$begingroup$
Yes I know this formula, the result is 1/3 right? I just realized I made a mistake, x should be in the range from a to b
$endgroup$
– wawawa
Jan 17 at 12:48
$begingroup$
@ Cecilia Yes, you are correct. I forgot to mention one thing that you solve the two equations under the restriction that $a<b$.
$endgroup$
– user440191
Jan 17 at 12:50
$begingroup$
@ wolfies you are absolutely right. I should have mentioned that I am assuming continuous uniform distribution. However, if one assume that it is the case of discrete uniform, I think thee will be a issue regarding the consideration of the set over which it is uniformly distributed ( as mentioned by @mathworker above)
$endgroup$
– user440191
Jan 17 at 12:58
$begingroup$
I urge @Cecilia to consider the discrete case when the set is ${1,2,dots N}$ and see what happens.
$endgroup$
– user440191
Jan 17 at 12:59
1
1
$begingroup$
You are assuming that $U$ is continuous Uniform. All that is stated is that it is Uniform (which might be continuous or discrete)
$endgroup$
– wolfies
Jan 17 at 12:47
$begingroup$
You are assuming that $U$ is continuous Uniform. All that is stated is that it is Uniform (which might be continuous or discrete)
$endgroup$
– wolfies
Jan 17 at 12:47
$begingroup$
Yes I know this formula, the result is 1/3 right? I just realized I made a mistake, x should be in the range from a to b
$endgroup$
– wawawa
Jan 17 at 12:48
$begingroup$
Yes I know this formula, the result is 1/3 right? I just realized I made a mistake, x should be in the range from a to b
$endgroup$
– wawawa
Jan 17 at 12:48
$begingroup$
@ Cecilia Yes, you are correct. I forgot to mention one thing that you solve the two equations under the restriction that $a<b$.
$endgroup$
– user440191
Jan 17 at 12:50
$begingroup$
@ Cecilia Yes, you are correct. I forgot to mention one thing that you solve the two equations under the restriction that $a<b$.
$endgroup$
– user440191
Jan 17 at 12:50
$begingroup$
@ wolfies you are absolutely right. I should have mentioned that I am assuming continuous uniform distribution. However, if one assume that it is the case of discrete uniform, I think thee will be a issue regarding the consideration of the set over which it is uniformly distributed ( as mentioned by @mathworker above)
$endgroup$
– user440191
Jan 17 at 12:58
$begingroup$
@ wolfies you are absolutely right. I should have mentioned that I am assuming continuous uniform distribution. However, if one assume that it is the case of discrete uniform, I think thee will be a issue regarding the consideration of the set over which it is uniformly distributed ( as mentioned by @mathworker above)
$endgroup$
– user440191
Jan 17 at 12:58
$begingroup$
I urge @Cecilia to consider the discrete case when the set is ${1,2,dots N}$ and see what happens.
$endgroup$
– user440191
Jan 17 at 12:59
$begingroup$
I urge @Cecilia to consider the discrete case when the set is ${1,2,dots N}$ and see what happens.
$endgroup$
– user440191
Jan 17 at 12:59
add a comment |
0
$begingroup$
As the mean is directly in the middle between $a$ and $b$ you can set
$a = 5-x$ and $b=5+x$ for $x>0$
Now solve
$$sigma^2 = frac{(b-a)^2}{12}= frac{(2x)^2}{12}=frac{x^2}{3} = 3stackrel{x>0}{Rightarrow} x= 3$$
So, $U$ "lives" on $[a,b]=[2,8]$. It follows
$$P(U<4) = frac{4-2}{8-2}=frac{2}{6}=frac{1}{3}$$
answered Jan 17 at 12:58
trancelocationtrancelocation
14.1k1829
$endgroup$
$begingroup$
Thank you so much for the answer, however @Bhargob answers the question before you haha and I've already chosen his answer, thanks for your time and I really appreciated it.
$endgroup$
– wawawa
Jan 17 at 14:59
add a comment |
0
$begingroup$
As the mean is directly in the middle between $a$ and $b$ you can set
$a = 5-x$ and $b=5+x$ for $x>0$
Now solve
$$sigma^2 = frac{(b-a)^2}{12}= frac{(2x)^2}{12}=frac{x^2}{3} = 3stackrel{x>0}{Rightarrow} x= 3$$
So, $U$ "lives" on $[a,b]=[2,8]$. It follows
$$P(U<4) = frac{4-2}{8-2}=frac{2}{6}=frac{1}{3}$$
answered Jan 17 at 12:58
trancelocationtrancelocation
14.1k1829
$endgroup$
$begingroup$
Thank you so much for the answer, however @Bhargob answers the question before you haha and I've already chosen his answer, thanks for your time and I really appreciated it.
$endgroup$
– wawawa
Jan 17 at 14:59
add a comment |
0
0
0
$begingroup$
As the mean is directly in the middle between $a$ and $b$ you can set
$a = 5-x$ and $b=5+x$ for $x>0$
Now solve
$$sigma^2 = frac{(b-a)^2}{12}= frac{(2x)^2}{12}=frac{x^2}{3} = 3stackrel{x>0}{Rightarrow} x= 3$$
So, $U$ "lives" on $[a,b]=[2,8]$. It follows
$$P(U<4) = frac{4-2}{8-2}=frac{2}{6}=frac{1}{3}$$
answered Jan 17 at 12:58
trancelocationtrancelocation
14.1k1829
$endgroup$
As the mean is directly in the middle between $a$ and $b$ you can set
$a = 5-x$ and $b=5+x$ for $x>0$
Now solve
$$sigma^2 = frac{(b-a)^2}{12}= frac{(2x)^2}{12}=frac{x^2}{3} = 3stackrel{x>0}{Rightarrow} x= 3$$
So, $U$ "lives" on $[a,b]=[2,8]$. It follows
$$P(U<4) = frac{4-2}{8-2}=frac{2}{6}=frac{1}{3}$$
answered Jan 17 at 12:58
trancelocationtrancelocation
14.1k1829
answered Jan 17 at 12:58
trancelocationtrancelocation
14.1k1829
answered Jan 17 at 12:58
trancelocationtrancelocation
14.1k1829
answered Jan 17 at 12:58
trancelocationtrancelocation
14.1k1829
14.1k1829
$begingroup$
Thank you so much for the answer, however @Bhargob answers the question before you haha and I've already chosen his answer, thanks for your time and I really appreciated it.
$endgroup$
– wawawa
Jan 17 at 14:59
add a comment |
$begingroup$
Thank you so much for the answer, however @Bhargob answers the question before you haha and I've already chosen his answer, thanks for your time and I really appreciated it.
$endgroup$
– wawawa
Jan 17 at 14:59
$begingroup$
Thank you so much for the answer, however @Bhargob answers the question before you haha and I've already chosen his answer, thanks for your time and I really appreciated it.
$endgroup$
– wawawa
Jan 17 at 14:59
$begingroup$
Thank you so much for the answer, however @Bhargob answers the question before you haha and I've already chosen his answer, thanks for your time and I really appreciated it.
$endgroup$
– wawawa
Jan 17 at 14:59
add a comment |
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2
$begingroup$
uniformly distributed over what set?
$endgroup$
– mathworker21
Jan 17 at 12:44
1
$begingroup$
Where did you get stuck with solving this? What have you tried already? Questions without context and visible effort often get downvoted and closed.
$endgroup$
– postmortes
Jan 17 at 12:58