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Given CDF of random variable X, Find P(X≤2), and P(1

-1

$begingroup$

cumulative distribution function of the random X is given by

$$ F(x) = begin{cases}
1-(1+x)e^{-x} & text{for } xgt 0 \
0 & text{elsewhere}
end{cases} $$

1. Find $P(X leq2 )$
   


2. Find $P(1lt Xlt 3)$

Hi, i am not sure of how to solve this problem. Do i just have to integrate the values in given cdf equation?

probability statistics

share|cite|improve this question

edited Mar 24 at 20:19

Infiaria

48111

asked Mar 24 at 19:55

Sara RafiqSara Rafiq

275

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  What was a cumulative distribution function again...?
  $endgroup$
  – Saucy O'Path
  Mar 24 at 19:59
  
  
  

  
  
  
  

add a comment | 

-1

$begingroup$

cumulative distribution function of the random X is given by

$$ F(x) = begin{cases}
1-(1+x)e^{-x} & text{for } xgt 0 \
0 & text{elsewhere}
end{cases} $$

1. Find $P(X leq2 )$
   


2. Find $P(1lt Xlt 3)$

Hi, i am not sure of how to solve this problem. Do i just have to integrate the values in given cdf equation?

probability statistics

share|cite|improve this question

edited Mar 24 at 20:19

Infiaria

48111

asked Mar 24 at 19:55

Sara RafiqSara Rafiq

275

$endgroup$

• 
  
  
  

  
  2
  

  
  
  
  
  
  

  
  $begingroup$
  What was a cumulative distribution function again...?
  $endgroup$
  – Saucy O'Path
  Mar 24 at 19:59
  
  
  

  
  
  
  

add a comment | 

$begingroup$

cumulative distribution function of the random X is given by

$$ F(x) = begin{cases}
1-(1+x)e^{-x} & text{for } xgt 0 \
0 & text{elsewhere}
end{cases} $$

1. Find $P(X leq2 )$
   


2. Find $P(1lt Xlt 3)$

Hi, i am not sure of how to solve this problem. Do i just have to integrate the values in given cdf equation?

probability statistics

share|cite|improve this question

edited Mar 24 at 20:19

Infiaria

48111

asked Mar 24 at 19:55

Sara RafiqSara Rafiq

275

$endgroup$

share|cite|improve this question

edited Mar 24 at 20:19

Infiaria

48111

asked Mar 24 at 19:55

Sara RafiqSara Rafiq

275

share|cite|improve this question

edited Mar 24 at 20:19

Infiaria

48111

asked Mar 24 at 19:55

Sara RafiqSara Rafiq

275

edited Mar 24 at 20:19

Infiaria

48111

edited Mar 24 at 20:19

Infiaria

48111

asked Mar 24 at 19:55

Sara RafiqSara Rafiq

275

asked Mar 24 at 19:55

Sara RafiqSara Rafiq

275

2 Answers
2

active

oldest

votes

0

$begingroup$

Hint

By definition of CDF for a continuous random variable we have:

$$F_X(x)triangleqPr{X<x}=Pr{Xle x}$$and $$Pr{a<x<b}=Pr{x<b}-Pr{x<a}$$

share|cite|improve this answer

answered Mar 24 at 20:26

Mostafa AyazMostafa Ayaz

18.1k31040

$endgroup$

add a comment | 

0

$begingroup$

A cumulative distribution function shows the probability of $X$ being less than a given value $x$, ie $P(X<x)$. Therefore $P(X<2) = 1-(1+2)e^{-2} = 1-3e^{-2}$ and $P(1<X<3)$ = $P(X<3) - P(X<1) = 1-(1+3)e^{-3}-(1-(1+1)e^{-1}) = 2e^{-1}-4e^{-3}$

share|cite|improve this answer

answered Mar 24 at 20:29

automaticallyGeneratedautomaticallyGenerated

1358

$endgroup$

add a comment | 

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0

$begingroup$

Hint

By definition of CDF for a continuous random variable we have:

$$F_X(x)triangleqPr{X<x}=Pr{Xle x}$$and $$Pr{a<x<b}=Pr{x<b}-Pr{x<a}$$

share|cite|improve this answer

answered Mar 24 at 20:26

Mostafa AyazMostafa Ayaz

18.1k31040

$endgroup$

add a comment | 

0

$begingroup$

Hint

By definition of CDF for a continuous random variable we have:

$$F_X(x)triangleqPr{X<x}=Pr{Xle x}$$and $$Pr{a<x<b}=Pr{x<b}-Pr{x<a}$$

share|cite|improve this answer

answered Mar 24 at 20:26

Mostafa AyazMostafa Ayaz

18.1k31040

$endgroup$

add a comment | 

$begingroup$

Hint

By definition of CDF for a continuous random variable we have:

$$F_X(x)triangleqPr{X<x}=Pr{Xle x}$$and $$Pr{a<x<b}=Pr{x<b}-Pr{x<a}$$

share|cite|improve this answer

answered Mar 24 at 20:26

Mostafa AyazMostafa Ayaz

18.1k31040

$endgroup$

share|cite|improve this answer

answered Mar 24 at 20:26

Mostafa AyazMostafa Ayaz

18.1k31040

answered Mar 24 at 20:26

Mostafa AyazMostafa Ayaz

18.1k31040

answered Mar 24 at 20:26

Mostafa AyazMostafa Ayaz

18.1k31040

0

$begingroup$

A cumulative distribution function shows the probability of $X$ being less than a given value $x$, ie $P(X<x)$. Therefore $P(X<2) = 1-(1+2)e^{-2} = 1-3e^{-2}$ and $P(1<X<3)$ = $P(X<3) - P(X<1) = 1-(1+3)e^{-3}-(1-(1+1)e^{-1}) = 2e^{-1}-4e^{-3}$

share|cite|improve this answer

answered Mar 24 at 20:29

automaticallyGeneratedautomaticallyGenerated

1358

$endgroup$

add a comment | 

0

$begingroup$

A cumulative distribution function shows the probability of $X$ being less than a given value $x$, ie $P(X<x)$. Therefore $P(X<2) = 1-(1+2)e^{-2} = 1-3e^{-2}$ and $P(1<X<3)$ = $P(X<3) - P(X<1) = 1-(1+3)e^{-3}-(1-(1+1)e^{-1}) = 2e^{-1}-4e^{-3}$

share|cite|improve this answer

answered Mar 24 at 20:29

automaticallyGeneratedautomaticallyGenerated

1358

$endgroup$

add a comment | 

$begingroup$

A cumulative distribution function shows the probability of $X$ being less than a given value $x$, ie $P(X<x)$. Therefore $P(X<2) = 1-(1+2)e^{-2} = 1-3e^{-2}$ and $P(1<X<3)$ = $P(X<3) - P(X<1) = 1-(1+3)e^{-3}-(1-(1+1)e^{-1}) = 2e^{-1}-4e^{-3}$

share|cite|improve this answer

answered Mar 24 at 20:29

automaticallyGeneratedautomaticallyGenerated

1358

$endgroup$

share|cite|improve this answer

answered Mar 24 at 20:29

automaticallyGeneratedautomaticallyGenerated

1358

answered Mar 24 at 20:29

automaticallyGeneratedautomaticallyGenerated

1358

answered Mar 24 at 20:29

automaticallyGeneratedautomaticallyGenerated

1358