Apps not working on smartphones. In a Pew Research Center survey titled U.S. Smartphone Use in 2015, more than 2,000 smartphone users were asked how often the applications (apps) they downloaded on their cell phones were not working correctly. Responses were recorded as follows: 1 = Frequently, 2 = Occasionally, 3 = Rarely, and 4 = Never. The probability distribution for the numerical response, x, is provided in the table.

1. Verify that the properties of a probability distribution for a discrete random variable are satisfied.
2. Find $\mathbf{\text{P}}\mathbf{\left(}\mathbf{\text{x>2}}\mathbf{\right)}$
3. Find $\mathbf{\text{E}}\mathbf{\left(}\mathbf{\text{x}}\mathbf{\right)}$. Interpret this value practically.

The first property which needs to be verified is whether the sum of all the properties is equivalent to 1, and this probability distribution verifies that as shown below:

$$\begin{gathered} \mathbf{\text{Summation\hspace{0.17em}of\hspace{0.17em}the\hspace{0.17em}probabilities=0.10+0.39+0.40+0.11}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{=}\mathbf{1} \end{gathered}$$

The second property is that the probabilities must be between 0 and 1, and this distribution showcases that all the probabilities are within that range.Each of the four probabilities is greater than 0 and less than 1.

The probability distribution must contain probabilities of different values of x, which must lie between 0 and 1. The probabilities of the four responses follow those criteria in this case.

The calculation $\mathbf{\text{P}}\mathbf{\left(}\mathbf{\text{x>2}}\mathbf{\right)}$is shown below:

role="math" localid="1653639865341" $$\begin{gathered} \mathbf{\text{P}}\mathbf{\left(}\mathbf{\text{x>2}}\mathbf{\right)}\mathbf{\text{=P}}\mathbf{\left(}\mathbf{\text{x=3}}\mathbf{\right)}\mathbf{\text{+P}}\mathbf{\left(}\mathbf{\text{x=4}}\mathbf{\right)} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{\text{=0.40+0.11}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{\text{=0.51}} \end{gathered}$$

Therefore, the$\mathbf{\text{P}}\mathbf{\left(}\mathbf{\text{x>2}}\mathbf{\right)}$ is 0.51

The formula for calculating $\mathbf{\text{E}}\mathbf{\left(}\mathbf{\text{x}}\mathbf{\right)}$is shown below:

$\mathbf{\text{E}}\mathbf{\left(}\mathbf{\text{x}}\mathbf{\right)}\mathbf{\text{=Summation\hspace{0.17em}of\hspace{0.17em}xp}}\mathbf{\left(}\mathbf{\text{x}}\mathbf{\right)}$

Here,$\mathbf{X}$ it represents the responses and$\mathbf{p}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$ represents the associated probabilities.

The calculation $\mathbf{\text{E}}\mathbf{\left(}\mathbf{\text{x}}\mathbf{\right)}$ is shown below:

$$\begin{gathered} \mathbf{\text{E}}\mathbf{\left(}\mathbf{\text{x}}\mathbf{\right)}\mathbf{\text{=1×0.10+2×0.39+3×0.40+4×0.11}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{\text{=0.10+0.78+1.20+0.44}} \\ \mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{}\mathbf{\text{=2.52}} \end{gathered}$$

Therefore, the $\mathbf{E}\mathbf{\left(}\mathbf{x}\mathbf{\right)}$ is 2.52, which indicates the mean of the responses by taking the associated probabilities of occurrences.