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Finding mean and variance - Discrete Probability

Variance in a Discrete Probability Distribution (Homework)Conditional mean and variancevariance and generating function - probability hat problemFinding the Mean, Variance, and Probability of a Poisson ModelFinding moment-generating function and variance of discrete RVFinding the mean and variance of an exponential probability distributionPoison distribution variance,probability. and mean.Finding expectation and varianceMoment generating function for finding mean and varianceProbability distribution mean and variance

0

$begingroup$

So I think I know how to get the variance for these, first we need to find $E(X)$ right? So $E(X)$ for $a$ would be:

$$-1 * frac14 + 0 * frac12 + 1 * frac14 = 0$$

So to find the variance, we need $E((x - E(x))^2)$

So is it $(-1 - 0)^2 + (0-0)^2 + (1-0)^2$ which is just $2$?

probability discrete-mathematics

share|cite|improve this question

edited 2 days ago

Dr. Mathva

2,296526

asked 2 days ago

StuyStuy

4351412

$endgroup$

This question has an open bounty worth +50
reputation from Stuy ending ending at 2019-03-18 21:34:47Z">in 7 days.

This question has not received enough attention.

I dont know how to find the mean from this info

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Remember to multiply by the probabilities! So $mathbbEleft[(X-mathbbE[X])^2right] = (-1-0)^2 colorbluetimesfrac14 + cdots$.
  $endgroup$
  – Minus One-Twelfth
  2 days ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  A more explicit formula for a is $Var(X)=sumlimits_x=-1^1 p(x)cdot (x-E(x))^2$
  $endgroup$
  – callculus
  2 days ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How do I find the mean? @callculus
  $endgroup$
  – Stuy
  4 hours ago
  

  
  
  
  

add a comment | 

0

$begingroup$

So I think I know how to get the variance for these, first we need to find $E(X)$ right? So $E(X)$ for $a$ would be:

$$-1 * frac14 + 0 * frac12 + 1 * frac14 = 0$$

So to find the variance, we need $E((x - E(x))^2)$

So is it $(-1 - 0)^2 + (0-0)^2 + (1-0)^2$ which is just $2$?

probability discrete-mathematics

share|cite|improve this question

edited 2 days ago

Dr. Mathva

2,296526

asked 2 days ago

StuyStuy

4351412

$endgroup$

This question has an open bounty worth +50
reputation from Stuy ending ending at 2019-03-18 21:34:47Z">in 7 days.

This question has not received enough attention.

I dont know how to find the mean from this info

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Remember to multiply by the probabilities! So $mathbbEleft[(X-mathbbE[X])^2right] = (-1-0)^2 colorbluetimesfrac14 + cdots$.
  $endgroup$
  – Minus One-Twelfth
  2 days ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  A more explicit formula for a is $Var(X)=sumlimits_x=-1^1 p(x)cdot (x-E(x))^2$
  $endgroup$
  – callculus
  2 days ago
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  How do I find the mean? @callculus
  $endgroup$
  – Stuy
  4 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

So I think I know how to get the variance for these, first we need to find $E(X)$ right? So $E(X)$ for $a$ would be:

$$-1 * frac14 + 0 * frac12 + 1 * frac14 = 0$$

So to find the variance, we need $E((x - E(x))^2)$

So is it $(-1 - 0)^2 + (0-0)^2 + (1-0)^2$ which is just $2$?

probability discrete-mathematics

share|cite|improve this question

edited 2 days ago

Dr. Mathva

2,296526

asked 2 days ago

StuyStuy

4351412

$endgroup$

share|cite|improve this question

edited 2 days ago

Dr. Mathva

2,296526

asked 2 days ago

StuyStuy

4351412

share|cite|improve this question

edited 2 days ago

Dr. Mathva

2,296526

asked 2 days ago

StuyStuy

4351412

edited 2 days ago

Dr. Mathva

2,296526

edited 2 days ago

Dr. Mathva

2,296526

asked 2 days ago

StuyStuy

4351412

asked 2 days ago

StuyStuy

4351412