Probability Mass Function given by f(x) = c(x^2 + 4), where

abc4616

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The P.M.F. (probability mass function) of a discrete random variable X is given by f(x) = c(x^2 + 4), where X = (0, 1, 2, 3), and c is a constant.

a) Determine value of c.
b) Find the mean and variance of X.

Does anyone know how to answer this?
 
Solve the equation k=03f(k)=1\displaystyle \sum\limits_{k = 0}^3 {f(k)} = 1 for c.

The expected value (the mean) is E(X)=k=03k  f(k).\displaystyle E(X) = \sum\limits_{k = 0}^3 {k\;f(k)} .

The variance is V(X)=k=03k2f(k)(E(X))2.\displaystyle V(X) = \sum\limits_{k = 0}^3 {k^2 f(k) - \left( {E(X)} \right)^2 }.
 
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