The uniform probability distribution of a specific value

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I am having this problem: The rainfall in April in Famagusta follows a uniform distribution between $0.5$ inches and $3.0$ inches. I am confused in finding the probability that the amount of rainfall will be exactly 1 inch of rain for the month.

Based on the formula of a uniform probability distribution we have: $P(X=x)=frac12.5$ for every $x$ on the closed interval $[0.5; 3]$ and consequently $P(X=1)=frac12.5=0.4$.

On the other hand, based on the rectangular shape of the uniform probability distribution we have: $P(X=1)=$(height)(base)$=(frac12.5) (1-1)=0.$

But, $0$ is distinct from $0.4$. So what is the probability of exactly $1$ inch of rain? Which of the methods is wrong?

Please give me a hand.

probability-distributions uniform-distribution

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edited Jul 18 at 19:29

RayDansh

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asked Jul 18 at 19:18

User1999

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I am having this problem: The rainfall in April in Famagusta follows a uniform distribution between $0.5$ inches and $3.0$ inches. I am confused in finding the probability that the amount of rainfall will be exactly 1 inch of rain for the month.

Based on the formula of a uniform probability distribution we have: $P(X=x)=frac12.5$ for every $x$ on the closed interval $[0.5; 3]$ and consequently $P(X=1)=frac12.5=0.4$.

On the other hand, based on the rectangular shape of the uniform probability distribution we have: $P(X=1)=$(height)(base)$=(frac12.5) (1-1)=0.$

But, $0$ is distinct from $0.4$. So what is the probability of exactly $1$ inch of rain? Which of the methods is wrong?

Please give me a hand.

probability-distributions uniform-distribution

share|cite|improve this question

edited Jul 18 at 19:29

RayDansh

882214

asked Jul 18 at 19:18

User1999

33618

add a comment | 

up vote
1
down vote

favorite

up vote
1
down vote

favorite

I am having this problem: The rainfall in April in Famagusta follows a uniform distribution between $0.5$ inches and $3.0$ inches. I am confused in finding the probability that the amount of rainfall will be exactly 1 inch of rain for the month.

Based on the formula of a uniform probability distribution we have: $P(X=x)=frac12.5$ for every $x$ on the closed interval $[0.5; 3]$ and consequently $P(X=1)=frac12.5=0.4$.

On the other hand, based on the rectangular shape of the uniform probability distribution we have: $P(X=1)=$(height)(base)$=(frac12.5) (1-1)=0.$

But, $0$ is distinct from $0.4$. So what is the probability of exactly $1$ inch of rain? Which of the methods is wrong?

Please give me a hand.

probability-distributions uniform-distribution

share|cite|improve this question

edited Jul 18 at 19:29

RayDansh

882214

asked Jul 18 at 19:18

User1999

33618

I am having this problem: The rainfall in April in Famagusta follows a uniform distribution between $0.5$ inches and $3.0$ inches. I am confused in finding the probability that the amount of rainfall will be exactly 1 inch of rain for the month.

Based on the formula of a uniform probability distribution we have: $P(X=x)=frac12.5$ for every $x$ on the closed interval $[0.5; 3]$ and consequently $P(X=1)=frac12.5=0.4$.

On the other hand, based on the rectangular shape of the uniform probability distribution we have: $P(X=1)=$(height)(base)$=(frac12.5) (1-1)=0.$

But, $0$ is distinct from $0.4$. So what is the probability of exactly $1$ inch of rain? Which of the methods is wrong?

Please give me a hand.

probability-distributions uniform-distribution

share|cite|improve this question

edited Jul 18 at 19:29

RayDansh

882214

asked Jul 18 at 19:18

User1999

33618

share|cite|improve this question

share|cite|improve this question

edited Jul 18 at 19:29

RayDansh

882214

edited Jul 18 at 19:29

RayDansh

882214

edited Jul 18 at 19:29

RayDansh

882214

882214

asked Jul 18 at 19:18

User1999

33618

asked Jul 18 at 19:18

User1999

33618

asked Jul 18 at 19:18

User1999

33618

33618

add a comment | 

add a comment | 

2 Answers
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oldest

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up vote
2
down vote

Your problem is here:

Based on the formula of a uniform probability distribution we have:
$P(X=x)=1/2.5$ for every $x$ on the closed interval $[0.5; 3]$ .

That "formula" works when your sample space is finite and you use the number of points in the denominator. But your sample space is an interval. It's a finite interval, but it contains infinitely many points. The probability that you get exactly one inch of rainfall is $0$.

share|cite|improve this answer

answered Jul 18 at 19:23

Ethan Bolker

35.7k54199

add a comment | 

up vote
1
down vote

You can show this regardless because of the definition of a continuous random variable.

A continuous random variable has a density function $f(x)$ with $f(x) geq 0$ with
$$int_-infty^infty f(x) dx = 1 $$
and
$$ P(a < X < b) =int_a^b f(x) dx $$

then the consequence that the probability a continuous random variable $X$ takes on a particular value is $0$

$$ P(X=c) = int_c^c f(c) dx = 0$$

but you can show this specifically for the uniform distribution like this. The above

$X sim U(a,b)$

specifically $ X sim U(.5,3)$ this gives us a density function of

$f(x) =beginalignbegincases frac13-.5 & textrm for frac12 leq x leq 3\ \ 0 & textrm for x <frac12 textrm or x > 3 endcases endalign$

Then we have

$$ P(X=1) = int_1^1 frac12.5 dx = 0$$

Note the density function for uniform variables is simply

$f(x) =beginalignbegincases frac1b-a & textrm for a leq x leq b \ \ 0 & textrm for x <a textrm or x > b endcases endalign$

share|cite|improve this answer

edited Jul 18 at 20:49

answered Jul 18 at 20:30

RHowe

1,000815

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2 Answers
2

active

oldest

votes

2 Answers
2

active

oldest

votes

active

oldest

votes

active

oldest

votes

up vote
2
down vote

Your problem is here:

Based on the formula of a uniform probability distribution we have:
$P(X=x)=1/2.5$ for every $x$ on the closed interval $[0.5; 3]$ .

That "formula" works when your sample space is finite and you use the number of points in the denominator. But your sample space is an interval. It's a finite interval, but it contains infinitely many points. The probability that you get exactly one inch of rainfall is $0$.

share|cite|improve this answer

answered Jul 18 at 19:23

Ethan Bolker

35.7k54199

add a comment | 

up vote
2
down vote

Your problem is here:

Based on the formula of a uniform probability distribution we have:
$P(X=x)=1/2.5$ for every $x$ on the closed interval $[0.5; 3]$ .

That "formula" works when your sample space is finite and you use the number of points in the denominator. But your sample space is an interval. It's a finite interval, but it contains infinitely many points. The probability that you get exactly one inch of rainfall is $0$.

share|cite|improve this answer

answered Jul 18 at 19:23

Ethan Bolker

35.7k54199

add a comment | 

up vote
2
down vote

up vote
2
down vote

Your problem is here:

Based on the formula of a uniform probability distribution we have:
$P(X=x)=1/2.5$ for every $x$ on the closed interval $[0.5; 3]$ .

That "formula" works when your sample space is finite and you use the number of points in the denominator. But your sample space is an interval. It's a finite interval, but it contains infinitely many points. The probability that you get exactly one inch of rainfall is $0$.

share|cite|improve this answer

answered Jul 18 at 19:23

Ethan Bolker

35.7k54199

Your problem is here:

Based on the formula of a uniform probability distribution we have:
$P(X=x)=1/2.5$ for every $x$ on the closed interval $[0.5; 3]$ .

That "formula" works when your sample space is finite and you use the number of points in the denominator. But your sample space is an interval. It's a finite interval, but it contains infinitely many points. The probability that you get exactly one inch of rainfall is $0$.

share|cite|improve this answer

answered Jul 18 at 19:23

Ethan Bolker

35.7k54199

share|cite|improve this answer

share|cite|improve this answer

answered Jul 18 at 19:23

Ethan Bolker

35.7k54199

answered Jul 18 at 19:23

Ethan Bolker

35.7k54199

answered Jul 18 at 19:23

Ethan Bolker

35.7k54199

35.7k54199

add a comment | 

add a comment | 

up vote
1
down vote

You can show this regardless because of the definition of a continuous random variable.

A continuous random variable has a density function $f(x)$ with $f(x) geq 0$ with
$$int_-infty^infty f(x) dx = 1 $$
and
$$ P(a < X < b) =int_a^b f(x) dx $$

then the consequence that the probability a continuous random variable $X$ takes on a particular value is $0$

$$ P(X=c) = int_c^c f(c) dx = 0$$

but you can show this specifically for the uniform distribution like this. The above

$X sim U(a,b)$

specifically $ X sim U(.5,3)$ this gives us a density function of

$f(x) =beginalignbegincases frac13-.5 & textrm for frac12 leq x leq 3\ \ 0 & textrm for x <frac12 textrm or x > 3 endcases endalign$

Then we have

$$ P(X=1) = int_1^1 frac12.5 dx = 0$$

Note the density function for uniform variables is simply

$f(x) =beginalignbegincases frac1b-a & textrm for a leq x leq b \ \ 0 & textrm for x <a textrm or x > b endcases endalign$

share|cite|improve this answer

edited Jul 18 at 20:49

answered Jul 18 at 20:30

RHowe

1,000815

add a comment | 

up vote
1
down vote

You can show this regardless because of the definition of a continuous random variable.

A continuous random variable has a density function $f(x)$ with $f(x) geq 0$ with
$$int_-infty^infty f(x) dx = 1 $$
and
$$ P(a < X < b) =int_a^b f(x) dx $$

then the consequence that the probability a continuous random variable $X$ takes on a particular value is $0$

$$ P(X=c) = int_c^c f(c) dx = 0$$

but you can show this specifically for the uniform distribution like this. The above

$X sim U(a,b)$

specifically $ X sim U(.5,3)$ this gives us a density function of

$f(x) =beginalignbegincases frac13-.5 & textrm for frac12 leq x leq 3\ \ 0 & textrm for x <frac12 textrm or x > 3 endcases endalign$

Then we have

$$ P(X=1) = int_1^1 frac12.5 dx = 0$$

Note the density function for uniform variables is simply

$f(x) =beginalignbegincases frac1b-a & textrm for a leq x leq b \ \ 0 & textrm for x <a textrm or x > b endcases endalign$

share|cite|improve this answer

edited Jul 18 at 20:49

answered Jul 18 at 20:30

RHowe

1,000815

add a comment | 

up vote
1
down vote

up vote
1
down vote

You can show this regardless because of the definition of a continuous random variable.

A continuous random variable has a density function $f(x)$ with $f(x) geq 0$ with
$$int_-infty^infty f(x) dx = 1 $$
and
$$ P(a < X < b) =int_a^b f(x) dx $$

then the consequence that the probability a continuous random variable $X$ takes on a particular value is $0$

$$ P(X=c) = int_c^c f(c) dx = 0$$

but you can show this specifically for the uniform distribution like this. The above

$X sim U(a,b)$

specifically $ X sim U(.5,3)$ this gives us a density function of

$f(x) =beginalignbegincases frac13-.5 & textrm for frac12 leq x leq 3\ \ 0 & textrm for x <frac12 textrm or x > 3 endcases endalign$

Then we have

$$ P(X=1) = int_1^1 frac12.5 dx = 0$$

Note the density function for uniform variables is simply

$f(x) =beginalignbegincases frac1b-a & textrm for a leq x leq b \ \ 0 & textrm for x <a textrm or x > b endcases endalign$

share|cite|improve this answer

edited Jul 18 at 20:49

answered Jul 18 at 20:30

RHowe

1,000815

You can show this regardless because of the definition of a continuous random variable.

A continuous random variable has a density function $f(x)$ with $f(x) geq 0$ with
$$int_-infty^infty f(x) dx = 1 $$
and
$$ P(a < X < b) =int_a^b f(x) dx $$

then the consequence that the probability a continuous random variable $X$ takes on a particular value is $0$

$$ P(X=c) = int_c^c f(c) dx = 0$$

but you can show this specifically for the uniform distribution like this. The above

$X sim U(a,b)$

specifically $ X sim U(.5,3)$ this gives us a density function of

$f(x) =beginalignbegincases frac13-.5 & textrm for frac12 leq x leq 3\ \ 0 & textrm for x <frac12 textrm or x > 3 endcases endalign$

Then we have

$$ P(X=1) = int_1^1 frac12.5 dx = 0$$

Note the density function for uniform variables is simply

$f(x) =beginalignbegincases frac1b-a & textrm for a leq x leq b \ \ 0 & textrm for x <a textrm or x > b endcases endalign$

share|cite|improve this answer

edited Jul 18 at 20:49

answered Jul 18 at 20:30

RHowe

1,000815

share|cite|improve this answer

share|cite|improve this answer

edited Jul 18 at 20:49

edited Jul 18 at 20:49

edited Jul 18 at 20:49

answered Jul 18 at 20:30

RHowe

1,000815

answered Jul 18 at 20:30

RHowe

1,000815

answered Jul 18 at 20:30

RHowe

1,000815

1,000815

add a comment | 

add a comment | 

 

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