we are given a set of data and its corresponding probability. Since the data is given in a discrete way, that is single values as opposed to continuous values this is not a "normal distribution" since the normal distribution is used for continuous values. This is also not a binomial distribution since its values do follow that pattern as we can see in the following diagram:
Therefore, this is a discrete distribution.
The expected value of a discrete distribution of a random variable is given by the sum of the product of each variable by its corresponding probability, therefore, the expected value is:
![E=(0)(0.15)+(5)(0.17)+(10)(0.07)+(15)(0.05)+(20)(0.56)](https://img.qammunity.org/2023/formulas/mathematics/college/he7fmfma2ctfvzefcm8b646jp1wsmgtzxl.png)
Solving the operations:
![E=13.5](https://img.qammunity.org/2023/formulas/mathematics/college/cub9zwmpbbbssknzmawu9cmyqezu0tjtrx.png)