Question

The random variable x represents the number of phone calls an author receives in a day, and it has a Poisson distribution with a mean of 7.6 calls. What are the possible values of x

Answer 1

Answer:

$f(x)=\frac{e^{-\lambda} \lambda^x}{x!} , x=0,1,2,3,4,...$

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 $X \sim Poisson(\lambda=7.6)$

The probability mass function for the random variable is given by:

$f(x)=\frac{e^{-\lambda} \lambda^x}{x!} , x=0,1,2,3,4,...$

And f(x)=0 for other case.

For this distribution the expected value is the same parameter $\lambda$

$E(X)=\mu =\lambda$

Solution to the problem

The posible values for the random variable would be:

X=0,1,2,3,4,.....,

All the positive natural integers.