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How do I generate distribution of positive numbers only with min, max and mean?

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1

0

$begingroup$

I am trying to generate a sample of 2000 rows. I have the following values
min = 80
max = 12000
mean = 500

I want to generate only positive numbers. I tried using triangular distribution and range rule(sd = max-min/4). The values were negative.

Is there anyway I can generate only positive numbers?

Thanks,
Anusha

distributions simulation

share|cite|improve this question

asked 8 hours ago

user3437212user3437212

3244 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  What kind of distribution of values do you want?
  $endgroup$
  – Dave
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  as long as they are positive and has mean = 500, min = 80 and max = 12000
  $endgroup$
  – user3437212
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  min and max is fine but the mean is way too off in triangular distribution, do you have any other suggestion?
  $endgroup$
  – user3437212
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Strongly related: stats.stackexchange.com/q/236449/35989
  $endgroup$
  – Tim♦
  7 hours ago
  

  
  
  
  

add a comment | 

1

0

$begingroup$

I am trying to generate a sample of 2000 rows. I have the following values
min = 80
max = 12000
mean = 500

I want to generate only positive numbers. I tried using triangular distribution and range rule(sd = max-min/4). The values were negative.

Is there anyway I can generate only positive numbers?

Thanks,
Anusha

distributions simulation

share|cite|improve this question

asked 8 hours ago

user3437212user3437212

3244 bronze badges

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  What kind of distribution of values do you want?
  $endgroup$
  – Dave
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  as long as they are positive and has mean = 500, min = 80 and max = 12000
  $endgroup$
  – user3437212
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  min and max is fine but the mean is way too off in triangular distribution, do you have any other suggestion?
  $endgroup$
  – user3437212
  8 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Strongly related: stats.stackexchange.com/q/236449/35989
  $endgroup$
  – Tim♦
  7 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

I am trying to generate a sample of 2000 rows. I have the following values
min = 80
max = 12000
mean = 500

I want to generate only positive numbers. I tried using triangular distribution and range rule(sd = max-min/4). The values were negative.

Is there anyway I can generate only positive numbers?

Thanks,
Anusha

distributions simulation

share|cite|improve this question

asked 8 hours ago

user3437212user3437212

3244 bronze badges

$endgroup$

share|cite|improve this question

asked 8 hours ago

user3437212user3437212

3244 bronze badges

share|cite|improve this question

asked 8 hours ago

user3437212user3437212

3244 bronze badges

asked 8 hours ago

user3437212user3437212

3244 bronze badges

asked 8 hours ago

user3437212user3437212

3244 bronze badges

2 Answers
2

active

oldest

votes

6

$begingroup$

While the problem is very much ill-posed, since there is an infinite range of distributions satisfying these constraints, a possible solution is to find the maximum entropy distribution under the constraint of a support of $(80,12000)$ [thus using the uniform measure on that interval as the reference measure] and a mean of $mathbb E[X]=500$ is of the form
$$p(x)=exp{alpha+beta x},mathbb I_{(80,1200)}(x)$$
with
$$int_{80}^{12000} exp{alpha+beta x},text dx=1qquadtext{and}qquad
int_{80}^{12000} xexp{alpha+beta x},text dx=500$$
which leads to
$$exp{-alpha}=beta^{-1}[exp{12000beta}-exp{80beta}]$$
and$$beta^{-1}exp{alpha}[12000exp{12000beta}-80exp{80beta}]-beta^{-1}=500$$which can be solved numerically in $beta$. Leading to
$$beta^*=-.00238quadtext{and}quadalpha^*=-5.850$$which can be easily simulated as a truncated exponential distribution, by inversion of the cdf, e.g., using qexp() in R. For instance,

function(n=1)

  return(qexp(pexp(80,.00238)+runif(n)*

   (pexp(12000,.00238)-pexp(80,.00238)),.00238))

If the question is instead about simulating a sample $X_{1:2000}$ such that $$min(X_{1:2000})=80,quadmax(X_{1:2000})=1200,quadbar X_{1:2000}=500$$
there is again an infinite range of solutions, the simplest being a multinomial distribution constrained by its minimum being 80 and its maximum being 12000 since
$$underbrace{X_{(1)}}_{80}+cdots+underbrace{X_{(2000)}}_{12000} = 80 + 1998times 987920 + 12000= 2000times 500$$

share|cite|improve this answer

edited 6 hours ago

answered 7 hours ago

Xi'anXi'an

61.7k88 gold badges9999 silver badges376376 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Is your comment about $X_{1:2000}$ about ensuring that the sample (not the population) has a mean of 500? Also, two clarifications would be helpful. 1) Why do something in the form of $exp{alpha + beta X}$? 2) Once we solve for $alpha$ and $beta$, how do to simulate draws from that PDF?
  $endgroup$
  – Dave
  7 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  As indicated in the answer, the second interpretation of the question is to see all the constraints as constraints on the sample, not on the distribution. And as indicated in the answer, the (truncated) exponential form of the distribution follows from maximum entropy principles.
  $endgroup$
  – Xi'an
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Xi'an Thank you this helps!
  $endgroup$
  – user3437212
  6 hours ago
  

  
  
  
  

add a comment | 

4

$begingroup$

If you don't care about the distribution aside from min, max, and mean, then there is a simple answer.

Take 96.476510067114100 percent of draws as 80 and 3.523489932885910 percent of draws as 12000. On average, you get 500, and you have your min and max. I calculated the percentages by solving a system of equations

$$a + b =1$$ $$80a + 12000b = 500$$

The first equation establishes the the values must sum to one, making sure that we are dealing with probabilities. The second equation get us our average of 500.

D <- rep(NA,2000) # define a vector of NAs to hold your sampled values

for (i in 1:2000){

    X <- rbinom(1,1,0.96476510067114100) # determine which value you'll take, 80 or 12000

    if (X==0){D[i] <- 12000} # declare observation i as 12000

    if (X==1){D[i] <- 80} # declare observation i as 80

}

share|cite|improve this answer

edited 4 hours ago

answered 7 hours ago

DaveDave

7021010 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks Dave! But I don't want only 80s and 12000s in my distribution. I would like a range of values but all positive
  $endgroup$
  – user3437212
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'm sorry I was not clear earlier
  $endgroup$
  – user3437212
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  We can construct a different distribution, but what requirements do you have?
  $endgroup$
  – Dave
  7 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  normal or a poisson but no negative values
  $endgroup$
  – user3437212
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  $begingroup$
  Normal is out, since it can take all real values. More helpful, though, would be to know what you're considering either of those distributions. So...why normal or Poisson?
  $endgroup$
  – Dave
  7 hours ago
  

  
  
  
  

 | 
show 1 more comment

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6

$begingroup$

While the problem is very much ill-posed, since there is an infinite range of distributions satisfying these constraints, a possible solution is to find the maximum entropy distribution under the constraint of a support of $(80,12000)$ [thus using the uniform measure on that interval as the reference measure] and a mean of $mathbb E[X]=500$ is of the form
$$p(x)=exp{alpha+beta x},mathbb I_{(80,1200)}(x)$$
with
$$int_{80}^{12000} exp{alpha+beta x},text dx=1qquadtext{and}qquad
int_{80}^{12000} xexp{alpha+beta x},text dx=500$$
which leads to
$$exp{-alpha}=beta^{-1}[exp{12000beta}-exp{80beta}]$$
and$$beta^{-1}exp{alpha}[12000exp{12000beta}-80exp{80beta}]-beta^{-1}=500$$which can be solved numerically in $beta$. Leading to
$$beta^*=-.00238quadtext{and}quadalpha^*=-5.850$$which can be easily simulated as a truncated exponential distribution, by inversion of the cdf, e.g., using qexp() in R. For instance,

function(n=1)

  return(qexp(pexp(80,.00238)+runif(n)*

   (pexp(12000,.00238)-pexp(80,.00238)),.00238))

If the question is instead about simulating a sample $X_{1:2000}$ such that $$min(X_{1:2000})=80,quadmax(X_{1:2000})=1200,quadbar X_{1:2000}=500$$
there is again an infinite range of solutions, the simplest being a multinomial distribution constrained by its minimum being 80 and its maximum being 12000 since
$$underbrace{X_{(1)}}_{80}+cdots+underbrace{X_{(2000)}}_{12000} = 80 + 1998times 987920 + 12000= 2000times 500$$

share|cite|improve this answer

edited 6 hours ago

answered 7 hours ago

Xi'anXi'an

61.7k88 gold badges9999 silver badges376376 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Is your comment about $X_{1:2000}$ about ensuring that the sample (not the population) has a mean of 500? Also, two clarifications would be helpful. 1) Why do something in the form of $exp{alpha + beta X}$? 2) Once we solve for $alpha$ and $beta$, how do to simulate draws from that PDF?
  $endgroup$
  – Dave
  7 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  As indicated in the answer, the second interpretation of the question is to see all the constraints as constraints on the sample, not on the distribution. And as indicated in the answer, the (truncated) exponential form of the distribution follows from maximum entropy principles.
  $endgroup$
  – Xi'an
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Xi'an Thank you this helps!
  $endgroup$
  – user3437212
  6 hours ago
  

  
  
  
  

add a comment | 

6

$begingroup$

While the problem is very much ill-posed, since there is an infinite range of distributions satisfying these constraints, a possible solution is to find the maximum entropy distribution under the constraint of a support of $(80,12000)$ [thus using the uniform measure on that interval as the reference measure] and a mean of $mathbb E[X]=500$ is of the form
$$p(x)=exp{alpha+beta x},mathbb I_{(80,1200)}(x)$$
with
$$int_{80}^{12000} exp{alpha+beta x},text dx=1qquadtext{and}qquad
int_{80}^{12000} xexp{alpha+beta x},text dx=500$$
which leads to
$$exp{-alpha}=beta^{-1}[exp{12000beta}-exp{80beta}]$$
and$$beta^{-1}exp{alpha}[12000exp{12000beta}-80exp{80beta}]-beta^{-1}=500$$which can be solved numerically in $beta$. Leading to
$$beta^*=-.00238quadtext{and}quadalpha^*=-5.850$$which can be easily simulated as a truncated exponential distribution, by inversion of the cdf, e.g., using qexp() in R. For instance,

function(n=1)

  return(qexp(pexp(80,.00238)+runif(n)*

   (pexp(12000,.00238)-pexp(80,.00238)),.00238))

If the question is instead about simulating a sample $X_{1:2000}$ such that $$min(X_{1:2000})=80,quadmax(X_{1:2000})=1200,quadbar X_{1:2000}=500$$
there is again an infinite range of solutions, the simplest being a multinomial distribution constrained by its minimum being 80 and its maximum being 12000 since
$$underbrace{X_{(1)}}_{80}+cdots+underbrace{X_{(2000)}}_{12000} = 80 + 1998times 987920 + 12000= 2000times 500$$

share|cite|improve this answer

edited 6 hours ago

answered 7 hours ago

Xi'anXi'an

61.7k88 gold badges9999 silver badges376376 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Is your comment about $X_{1:2000}$ about ensuring that the sample (not the population) has a mean of 500? Also, two clarifications would be helpful. 1) Why do something in the form of $exp{alpha + beta X}$? 2) Once we solve for $alpha$ and $beta$, how do to simulate draws from that PDF?
  $endgroup$
  – Dave
  7 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  As indicated in the answer, the second interpretation of the question is to see all the constraints as constraints on the sample, not on the distribution. And as indicated in the answer, the (truncated) exponential form of the distribution follows from maximum entropy principles.
  $endgroup$
  – Xi'an
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @Xi'an Thank you this helps!
  $endgroup$
  – user3437212
  6 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

While the problem is very much ill-posed, since there is an infinite range of distributions satisfying these constraints, a possible solution is to find the maximum entropy distribution under the constraint of a support of $(80,12000)$ [thus using the uniform measure on that interval as the reference measure] and a mean of $mathbb E[X]=500$ is of the form
$$p(x)=exp{alpha+beta x},mathbb I_{(80,1200)}(x)$$
with
$$int_{80}^{12000} exp{alpha+beta x},text dx=1qquadtext{and}qquad
int_{80}^{12000} xexp{alpha+beta x},text dx=500$$
which leads to
$$exp{-alpha}=beta^{-1}[exp{12000beta}-exp{80beta}]$$
and$$beta^{-1}exp{alpha}[12000exp{12000beta}-80exp{80beta}]-beta^{-1}=500$$which can be solved numerically in $beta$. Leading to
$$beta^*=-.00238quadtext{and}quadalpha^*=-5.850$$which can be easily simulated as a truncated exponential distribution, by inversion of the cdf, e.g., using qexp() in R. For instance,

function(n=1)

  return(qexp(pexp(80,.00238)+runif(n)*

   (pexp(12000,.00238)-pexp(80,.00238)),.00238))

If the question is instead about simulating a sample $X_{1:2000}$ such that $$min(X_{1:2000})=80,quadmax(X_{1:2000})=1200,quadbar X_{1:2000}=500$$
there is again an infinite range of solutions, the simplest being a multinomial distribution constrained by its minimum being 80 and its maximum being 12000 since
$$underbrace{X_{(1)}}_{80}+cdots+underbrace{X_{(2000)}}_{12000} = 80 + 1998times 987920 + 12000= 2000times 500$$

share|cite|improve this answer

edited 6 hours ago

answered 7 hours ago

Xi'anXi'an

61.7k88 gold badges9999 silver badges376376 bronze badges

$endgroup$

function(n=1)

  return(qexp(pexp(80,.00238)+runif(n)*

   (pexp(12000,.00238)-pexp(80,.00238)),.00238))

share|cite|improve this answer

edited 6 hours ago

answered 7 hours ago

Xi'anXi'an

61.7k88 gold badges9999 silver badges376376 bronze badges

answered 7 hours ago

Xi'anXi'an

61.7k88 gold badges9999 silver badges376376 bronze badges

answered 7 hours ago

Xi'anXi'an

61.7k88 gold badges9999 silver badges376376 bronze badges

4

$begingroup$

If you don't care about the distribution aside from min, max, and mean, then there is a simple answer.

Take 96.476510067114100 percent of draws as 80 and 3.523489932885910 percent of draws as 12000. On average, you get 500, and you have your min and max. I calculated the percentages by solving a system of equations

$$a + b =1$$ $$80a + 12000b = 500$$

The first equation establishes the the values must sum to one, making sure that we are dealing with probabilities. The second equation get us our average of 500.

D <- rep(NA,2000) # define a vector of NAs to hold your sampled values

for (i in 1:2000){

    X <- rbinom(1,1,0.96476510067114100) # determine which value you'll take, 80 or 12000

    if (X==0){D[i] <- 12000} # declare observation i as 12000

    if (X==1){D[i] <- 80} # declare observation i as 80

}

share|cite|improve this answer

edited 4 hours ago

answered 7 hours ago

DaveDave

7021010 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks Dave! But I don't want only 80s and 12000s in my distribution. I would like a range of values but all positive
  $endgroup$
  – user3437212
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'm sorry I was not clear earlier
  $endgroup$
  – user3437212
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  We can construct a different distribution, but what requirements do you have?
  $endgroup$
  – Dave
  7 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  normal or a poisson but no negative values
  $endgroup$
  – user3437212
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  $begingroup$
  Normal is out, since it can take all real values. More helpful, though, would be to know what you're considering either of those distributions. So...why normal or Poisson?
  $endgroup$
  – Dave
  7 hours ago
  

  
  
  
  

 | 
show 1 more comment

4

$begingroup$

If you don't care about the distribution aside from min, max, and mean, then there is a simple answer.

Take 96.476510067114100 percent of draws as 80 and 3.523489932885910 percent of draws as 12000. On average, you get 500, and you have your min and max. I calculated the percentages by solving a system of equations

$$a + b =1$$ $$80a + 12000b = 500$$

The first equation establishes the the values must sum to one, making sure that we are dealing with probabilities. The second equation get us our average of 500.

D <- rep(NA,2000) # define a vector of NAs to hold your sampled values

for (i in 1:2000){

    X <- rbinom(1,1,0.96476510067114100) # determine which value you'll take, 80 or 12000

    if (X==0){D[i] <- 12000} # declare observation i as 12000

    if (X==1){D[i] <- 80} # declare observation i as 80

}

share|cite|improve this answer

edited 4 hours ago

answered 7 hours ago

DaveDave

7021010 bronze badges

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thanks Dave! But I don't want only 80s and 12000s in my distribution. I would like a range of values but all positive
  $endgroup$
  – user3437212
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I'm sorry I was not clear earlier
  $endgroup$
  – user3437212
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  We can construct a different distribution, but what requirements do you have?
  $endgroup$
  – Dave
  7 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  normal or a poisson but no negative values
  $endgroup$
  – user3437212
  7 hours ago
  

  
  
  
  


• 
  
  
  

  
  5
  

  
  
  
  
  
  

  
  $begingroup$
  Normal is out, since it can take all real values. More helpful, though, would be to know what you're considering either of those distributions. So...why normal or Poisson?
  $endgroup$
  – Dave
  7 hours ago
  

  
  
  
  

 | 
show 1 more comment

$begingroup$

If you don't care about the distribution aside from min, max, and mean, then there is a simple answer.

Take 96.476510067114100 percent of draws as 80 and 3.523489932885910 percent of draws as 12000. On average, you get 500, and you have your min and max. I calculated the percentages by solving a system of equations

$$a + b =1$$ $$80a + 12000b = 500$$

The first equation establishes the the values must sum to one, making sure that we are dealing with probabilities. The second equation get us our average of 500.

D <- rep(NA,2000) # define a vector of NAs to hold your sampled values

for (i in 1:2000){

    X <- rbinom(1,1,0.96476510067114100) # determine which value you'll take, 80 or 12000

    if (X==0){D[i] <- 12000} # declare observation i as 12000

    if (X==1){D[i] <- 80} # declare observation i as 80

}

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edited 4 hours ago

answered 7 hours ago

DaveDave

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$endgroup$

D <- rep(NA,2000) # define a vector of NAs to hold your sampled values

for (i in 1:2000){

    X <- rbinom(1,1,0.96476510067114100) # determine which value you'll take, 80 or 12000

    if (X==0){D[i] <- 12000} # declare observation i as 12000

    if (X==1){D[i] <- 80} # declare observation i as 80

}

share|cite|improve this answer

edited 4 hours ago

answered 7 hours ago

DaveDave

7021010 bronze badges

answered 7 hours ago

DaveDave

7021010 bronze badges

answered 7 hours ago

DaveDave

7021010 bronze badges