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Given a range $[a,b] subset mathbb R$ and we have to choose a Real number from this range
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Is it possible to find the probability that the number would be less than ((a+b)/2) and greater than (a+b)/5
probability
edited Jul 14 at 21:01
amWhy
189k25219431
asked Jul 14 at 20:54
tupug bhuj
12
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up vote
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Is it possible to find the probability that the number would be less than ((a+b)/2) and greater than (a+b)/5
probability
edited Jul 14 at 21:01
amWhy
189k25219431
asked Jul 14 at 20:54
tupug bhuj
12
3
Clearly this depends on the choice of distribution. This was the same problem with your prior question
– lulu
Jul 14 at 20:59
1
Can you add your own workings on this? Where did you get stuck, during the process?
– amWhy
Jul 14 at 20:59
@scentofthetrees: $(a+b)/5$ may be $<a$
– gammatester
Jul 14 at 21:05
1
For the uniform distribution, the probability would be the length of the interval $[max(a,(a+b)/5), (a+b)/2]$ divided by the length of the interval $[a,b]$. So, $[(a+b)/2-max(a,(a+b)/5)]/(b-a)$.
– user574380
Jul 14 at 21:12
@gammatester It may also be $>b$.
– Arthur
Jul 14 at 22:08
add a comment |Â
up vote
-1
down vote
favorite
up vote
-1
down vote
favorite
Is it possible to find the probability that the number would be less than ((a+b)/2) and greater than (a+b)/5
probability
edited Jul 14 at 21:01
amWhy
189k25219431
asked Jul 14 at 20:54
tupug bhuj
12
Is it possible to find the probability that the number would be less than ((a+b)/2) and greater than (a+b)/5
probability
edited Jul 14 at 21:01
amWhy
189k25219431
asked Jul 14 at 20:54
tupug bhuj
12
edited Jul 14 at 21:01
amWhy
189k25219431
edited Jul 14 at 21:01
amWhy
189k25219431
edited Jul 14 at 21:01
amWhy
189k25219431
189k25219431
asked Jul 14 at 20:54
tupug bhuj
12
asked Jul 14 at 20:54
tupug bhuj
12
asked Jul 14 at 20:54
tupug bhuj
12
12
3
Clearly this depends on the choice of distribution. This was the same problem with your prior question
– lulu
Jul 14 at 20:59
1
Can you add your own workings on this? Where did you get stuck, during the process?
– amWhy
Jul 14 at 20:59
@scentofthetrees: $(a+b)/5$ may be $<a$
– gammatester
Jul 14 at 21:05
1
For the uniform distribution, the probability would be the length of the interval $[max(a,(a+b)/5), (a+b)/2]$ divided by the length of the interval $[a,b]$. So, $[(a+b)/2-max(a,(a+b)/5)]/(b-a)$.
– user574380
Jul 14 at 21:12
@gammatester It may also be $>b$.
– Arthur
Jul 14 at 22:08
add a comment |Â
3
Clearly this depends on the choice of distribution. This was the same problem with your prior question
– lulu
Jul 14 at 20:59
1
Can you add your own workings on this? Where did you get stuck, during the process?
– amWhy
Jul 14 at 20:59
@scentofthetrees: $(a+b)/5$ may be $<a$
– gammatester
Jul 14 at 21:05
1
For the uniform distribution, the probability would be the length of the interval $[max(a,(a+b)/5), (a+b)/2]$ divided by the length of the interval $[a,b]$. So, $[(a+b)/2-max(a,(a+b)/5)]/(b-a)$.
– user574380
Jul 14 at 21:12
@gammatester It may also be $>b$.
– Arthur
Jul 14 at 22:08
3
3
Clearly this depends on the choice of distribution. This was the same problem with your prior question
– lulu
Jul 14 at 20:59
Clearly this depends on the choice of distribution. This was the same problem with your prior question
– lulu
Jul 14 at 20:59
1
1
Can you add your own workings on this? Where did you get stuck, during the process?
– amWhy
Jul 14 at 20:59
Can you add your own workings on this? Where did you get stuck, during the process?
– amWhy
Jul 14 at 20:59
@scentofthetrees: $(a+b)/5$ may be $<a$
– gammatester
Jul 14 at 21:05
@scentofthetrees: $(a+b)/5$ may be $<a$
– gammatester
Jul 14 at 21:05
1
1
For the uniform distribution, the probability would be the length of the interval $[max(a,(a+b)/5), (a+b)/2]$ divided by the length of the interval $[a,b]$. So, $[(a+b)/2-max(a,(a+b)/5)]/(b-a)$.
– user574380
Jul 14 at 21:12
For the uniform distribution, the probability would be the length of the interval $[max(a,(a+b)/5), (a+b)/2]$ divided by the length of the interval $[a,b]$. So, $[(a+b)/2-max(a,(a+b)/5)]/(b-a)$.
– user574380
Jul 14 at 21:12
@gammatester It may also be $>b$.
– Arthur
Jul 14 at 22:08
@gammatester It may also be $>b$.
– Arthur
Jul 14 at 22:08
add a comment |Â
1 Answer
1
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Comment (with hints): As @lulu has commented, you need to specify a particular distribution.
Often, in elementary courses, when one speaks of picking a real number $X$ from
$[a,b],$ it is understood that the distribution is $mathsfUnif(a,b),$
which has density function $f_X(x) = frac1b-a,$ for $a le x le b$ and
$f_X(x) = 0,$ for all other values of $x$.
I hope you can figure out how to solve the problem for a uniform distribution with general parameter values
$a < b,$ and I will leave that to you. (Some of the other Comments should be helpful as you do that.)
For an example, suppose $a = 5$ and $b = 20.$ Then you seek $P(5 le X le 12.5) = 7.5/15 = 0.5.$ [Other choices of $a$ and $b$ may not be so tidy; perhaps try $a = -1$ and $b = 1.$]
For $a = 5,, b=20,$ the figure below shows the density 'curve' (here, part of a 'rectangle') of the random variable $X.$ The
probability $P(5 le X le 12.5)$ corresponds to the area beneath the density
and between the vertical dotted lines.
However, if you choose a different distribution for the random variable $X,$
then you may get a different answer.
Note: I hope you will find this site helpful. In order to give you the most helpful answers, we need to see your approach to the problem, and to know
why you are having trouble with it.
answered Jul 14 at 21:26
BruceET
33.3k61440
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
0
down vote
Comment (with hints): As @lulu has commented, you need to specify a particular distribution.
Often, in elementary courses, when one speaks of picking a real number $X$ from
$[a,b],$ it is understood that the distribution is $mathsfUnif(a,b),$
which has density function $f_X(x) = frac1b-a,$ for $a le x le b$ and
$f_X(x) = 0,$ for all other values of $x$.
I hope you can figure out how to solve the problem for a uniform distribution with general parameter values
$a < b,$ and I will leave that to you. (Some of the other Comments should be helpful as you do that.)
For an example, suppose $a = 5$ and $b = 20.$ Then you seek $P(5 le X le 12.5) = 7.5/15 = 0.5.$ [Other choices of $a$ and $b$ may not be so tidy; perhaps try $a = -1$ and $b = 1.$]
For $a = 5,, b=20,$ the figure below shows the density 'curve' (here, part of a 'rectangle') of the random variable $X.$ The
probability $P(5 le X le 12.5)$ corresponds to the area beneath the density
and between the vertical dotted lines.
However, if you choose a different distribution for the random variable $X,$
then you may get a different answer.
Note: I hope you will find this site helpful. In order to give you the most helpful answers, we need to see your approach to the problem, and to know
why you are having trouble with it.
answered Jul 14 at 21:26
BruceET
33.3k61440
add a comment |Â
up vote
0
down vote
Comment (with hints): As @lulu has commented, you need to specify a particular distribution.
Often, in elementary courses, when one speaks of picking a real number $X$ from
$[a,b],$ it is understood that the distribution is $mathsfUnif(a,b),$
which has density function $f_X(x) = frac1b-a,$ for $a le x le b$ and
$f_X(x) = 0,$ for all other values of $x$.
I hope you can figure out how to solve the problem for a uniform distribution with general parameter values
$a < b,$ and I will leave that to you. (Some of the other Comments should be helpful as you do that.)
For an example, suppose $a = 5$ and $b = 20.$ Then you seek $P(5 le X le 12.5) = 7.5/15 = 0.5.$ [Other choices of $a$ and $b$ may not be so tidy; perhaps try $a = -1$ and $b = 1.$]
For $a = 5,, b=20,$ the figure below shows the density 'curve' (here, part of a 'rectangle') of the random variable $X.$ The
probability $P(5 le X le 12.5)$ corresponds to the area beneath the density
and between the vertical dotted lines.
However, if you choose a different distribution for the random variable $X,$
then you may get a different answer.
Note: I hope you will find this site helpful. In order to give you the most helpful answers, we need to see your approach to the problem, and to know
why you are having trouble with it.
answered Jul 14 at 21:26
BruceET
33.3k61440
add a comment |Â
up vote
0
down vote
up vote
0
down vote
Comment (with hints): As @lulu has commented, you need to specify a particular distribution.
Often, in elementary courses, when one speaks of picking a real number $X$ from
$[a,b],$ it is understood that the distribution is $mathsfUnif(a,b),$
which has density function $f_X(x) = frac1b-a,$ for $a le x le b$ and
$f_X(x) = 0,$ for all other values of $x$.
I hope you can figure out how to solve the problem for a uniform distribution with general parameter values
$a < b,$ and I will leave that to you. (Some of the other Comments should be helpful as you do that.)
For an example, suppose $a = 5$ and $b = 20.$ Then you seek $P(5 le X le 12.5) = 7.5/15 = 0.5.$ [Other choices of $a$ and $b$ may not be so tidy; perhaps try $a = -1$ and $b = 1.$]
For $a = 5,, b=20,$ the figure below shows the density 'curve' (here, part of a 'rectangle') of the random variable $X.$ The
probability $P(5 le X le 12.5)$ corresponds to the area beneath the density
and between the vertical dotted lines.
However, if you choose a different distribution for the random variable $X,$
then you may get a different answer.
Note: I hope you will find this site helpful. In order to give you the most helpful answers, we need to see your approach to the problem, and to know
why you are having trouble with it.
answered Jul 14 at 21:26
BruceET
33.3k61440
Comment (with hints): As @lulu has commented, you need to specify a particular distribution.
Often, in elementary courses, when one speaks of picking a real number $X$ from
$[a,b],$ it is understood that the distribution is $mathsfUnif(a,b),$
which has density function $f_X(x) = frac1b-a,$ for $a le x le b$ and
$f_X(x) = 0,$ for all other values of $x$.
I hope you can figure out how to solve the problem for a uniform distribution with general parameter values
$a < b,$ and I will leave that to you. (Some of the other Comments should be helpful as you do that.)
For an example, suppose $a = 5$ and $b = 20.$ Then you seek $P(5 le X le 12.5) = 7.5/15 = 0.5.$ [Other choices of $a$ and $b$ may not be so tidy; perhaps try $a = -1$ and $b = 1.$]
For $a = 5,, b=20,$ the figure below shows the density 'curve' (here, part of a 'rectangle') of the random variable $X.$ The
probability $P(5 le X le 12.5)$ corresponds to the area beneath the density
and between the vertical dotted lines.
However, if you choose a different distribution for the random variable $X,$
then you may get a different answer.
Note: I hope you will find this site helpful. In order to give you the most helpful answers, we need to see your approach to the problem, and to know
why you are having trouble with it.
answered Jul 14 at 21:26
BruceET
33.3k61440
edited Jul 14 at 22:06
answered Jul 14 at 21:26
BruceET
33.3k61440
answered Jul 14 at 21:26
BruceET
33.3k61440
answered Jul 14 at 21:26
BruceET
33.3k61440
33.3k61440
add a comment |Â
add a comment |Â
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3
Clearly this depends on the choice of distribution. This was the same problem with your prior question
– lulu
Jul 14 at 20:59
1
Can you add your own workings on this? Where did you get stuck, during the process?
– amWhy
Jul 14 at 20:59
@scentofthetrees: $(a+b)/5$ may be $<a$
– gammatester
Jul 14 at 21:05
1
For the uniform distribution, the probability would be the length of the interval $[max(a,(a+b)/5), (a+b)/2]$ divided by the length of the interval $[a,b]$. So, $[(a+b)/2-max(a,(a+b)/5)]/(b-a)$.
– user574380
Jul 14 at 21:12
@gammatester It may also be $>b$.
– Arthur
Jul 14 at 22:08