Answer:
[tex]f(x)=\frac{e^{-\lambda} \lambda^x}{x!} , x=0,1,2,3,4,...[/tex]
The possible values for the random variable would be:
X=0,1,2,3,4,.....,
All the positive natural integers.
Step-by-step explanation:
Previous concepts
Let X the random variable that represent the number of phone calls an author recieves in a day. We know that [tex]X \sim Poisson(\lambda=7.6)[/tex]
The probability mass function for the random variable is given by:
[tex]f(x)=\frac{e^{-\lambda} \lambda^x}{x!} , x=0,1,2,3,4,...[/tex]
And f(x)=0 for other case.
For this distribution the expected value is the same parameter [tex]\lambda[/tex]
[tex]E(X)=\mu =\lambda[/tex]
Solution to the problem
The posible values for the random variable would be:
X=0,1,2,3,4,.....,
All the positive natural integers.
