Question

Downloading “apps” to your cell phone. According to an August 2011 survey by the Pew Internet & American Life Project life project, nearly 40% of adult cell phone owners have downloaded an application (“app”) to their cell phone. The next table next table gives the probability distribution for x, the number of“apps” used at least once a week by cell phone owners .downloaded an “app” to their phone. (The probabilities in the table are based on information from the Pew Internet & American Life Project survey.)

Number of apps used x	p(x)
0	0.17
1	0.10
2	0.11
3	0.11
4	0.10
5	0.10
6	0.07
7	0.05
8	0.03
9	0.02
10	0.02
11	0.02
12	0.02
13	0.02
14	0.01
15	0.01
16	0.01
17	0.01
18	0.01
19	0.005
20	0.005

$$\begin{gathered} \mathbf{a}\mathbf{.}\mathbf{}\mathbf{Show}\mathbf{}\mathbf{that}\mathbf{}\mathbf{the}\mathbf{}\mathbf{properties}\mathbf{}\mathbf{of}\mathbf{}\mathbf{a}\mathbf{}\mathbf{probability}\mathbf{}\mathbf{}\mathbf{distribution}\mathbf{}\mathbf{for}\mathbf{}\mathbf{a}\mathbf{}\mathbf{discrete}\mathbf{}\mathbf{random}\mathbf{}\mathbf{variable}\mathbf{}\mathbf{are}\mathbf{}\mathbf{satisfied}\mathbf{.} \\ \mathbf{}\mathbf{b}\mathbf{.}\mathbf{}\mathbf{Find}\mathbf{}\mathbf{P}\mathbf{(}\mathbf{x}\mathbf{>}\mathbf{10}\mathbf{)}\mathbf{}\mathbf{is}\mathbf{}\mathbf{0}\mathbf{.}\mathbf{14}\mathbf{} \\ \mathbf{c}\mathbf{.}\mathbf{}\mathbf{Find}\mathbf{}\mathbf{the}\mathbf{}\mathbf{mean}\mathbf{}\mathbf{and}\mathbf{}\mathbf{variance}\mathbf{}\mathbf{of}\mathbf{}\mathbf{x}\mathbf{.} \\ \mathbf{}\mathbf{d}\mathbf{.}\mathbf{}\mathbf{Give}\mathbf{}\mathbf{an}\mathbf{}\mathbf{interval}\mathbf{}\mathbf{that}\mathbf{}\mathbf{will}\mathbf{}\mathbf{contain}\mathbf{}\mathbf{the}\mathbf{}\mathbf{value}\mathbf{}\mathbf{of}\mathbf{}\mathbf{x}\mathbf{}\mathbf{with}\mathbf{}\mathbf{a}\mathbf{}\mathbf{probabilityof}\mathbf{}\mathbf{at}\mathbf{}\mathbf{least}\mathbf{}\mathbf{.}\mathbf{75}\mathbf{.} \end{gathered}$$

Short answer

$$\begin{gathered} \mathrm{a}.\mathrm{the}\mathrm{properties}\mathrm{of}\mathrm{a}\mathrm{probability}\mathrm{distribution}\mathrm{for}\mathrm{a}\mathrm{discrete}\mathrm{random}\mathrm{variable}\mathrm{are}\mathrm{satisfied}. \\ \mathrm{b}.\mathrm{P}\left(\mathrm{X}\ge 10\right)\mathrm{is}0.14 \\ \mathrm{c}.\mathrm{mean}\mathrm{of}\mathrm{x}.\mathrm{is}4.655\mathrm{and}\mathrm{variance}\mathrm{is}\underline{19.855} \\ \mathrm{d}.\mathrm{an}\mathrm{interval}\mathrm{that}\mathrm{will}\mathrm{contain}\mathrm{the}\mathrm{value}\mathrm{of}\mathrm{x}\mathrm{with}\mathrm{a}\mathrm{probability}\mathrm{of}\mathrm{at}\mathrm{least}.75.\mathrm{is}\left(0,13.56\right) \end{gathered}$$

1. the properties of a probability distribution for a discrete random variable are satisfied.
2. is
3. mean of x. is 4.655 and variance is 19.855
4. an interval that will contain the value of x with a probability of at least .75.is

Step-by-step solution

Given information

According to august 2011 survey, 40% of the adult cell phone owners downloaded an application to their cell phone.

Above table gives the probability distribution of X

Verifying the properties of  probability distribution for a discrete random variable.

From the table given above, we can calculate following values, that is:

$$\begin{gathered} \stackrel{}{\sum \mathrm{P}}\left(\mathrm{x}\right)=1 \\ \stackrel{}{\sum \mathrm{X}\times \mathrm{P}}\left(\mathrm{X}\right)=4.655 \\ \stackrel{}{\sum {\mathrm{X}}^2\times \mathrm{P}}\left(\mathrm{X}\right)=41.525 \end{gathered}$$

We know that the requirements for a discrete probability distribution are

1. $0\le \mathrm{P}\left(\mathrm{x}\right)\le 1$

2.$\sum _{all\mathit{}y}\mathrm{P}\left(\mathrm{x}\right)=1$

Therefore, from the given data,

We can see that,

$0\le \mathrm{P}\left(\mathrm{x}\right)\le 1$

And also,$\stackrel{}{\sum \mathrm{P}}\left(\mathrm{x}\right)=1$

Hence the probability distribution is satisfying the given conditions for a discrete random variable.

Calculating the P(x≥10)

b.

b.

$$\begin{gathered} \mathrm{P}\left(\mathrm{X}\ge 10\right)=\mathrm{P}\left(\mathrm{X}=10\right)+\mathrm{P}\left(\mathrm{X}=11\right)+...+\mathrm{P}\left(\mathrm{X}=20\right) \\ =0.02+0.02+0.02+0.02+0.01+0.01+0.01+0.01+0.01 \\ 0.005+0.005 \\ =0.14 \end{gathered}$$

Therefore$\mathrm{P}(\mathrm{X}\ge 10)=0.14$

Calculating the mean and variance of X

c.

c.

Let the mean of X is given by:

$$\begin{gathered} \mathrm{E}\left(\mathrm{X}\right)=\sum _{}\mathrm{x}\times \mathrm{P}\left(\mathrm{x}\right) \\ =4.655 \end{gathered}$$

Therefore,

$E\left(X\right)=4.655$

The variance is given by:

$\mathit{V}\mathit{a}\mathit{r}\mathbf{\left(}\mathit{X}\mathbf{\right)}\mathbf{=}\mathit{E}\mathbf{\left(}{\mathit{X}}^{\mathbf{2}}\mathbf{\right)}\mathbf{-}{\left[E\left(X\right)\right]}^{\mathbf{2}}$

$$\begin{gathered} \mathrm{Var}\left(\mathrm{X}\right)=\sum _{}{\mathrm{x}}^2\times \mathrm{P}\left(\mathrm{x}\right)-{\left[\sum _{}\mathrm{x}\times \mathrm{P}\left(\mathrm{x}\right)\right]}^2 \\ =41.525-21.669025 \\ =19.855975 \\ \mathrm{Var}\left(\mathrm{X}\right)=19.855975 \end{gathered}$$

Calculating the interval that will contain the value of Y with a probability of at least 0.75

d.

d.

To calculate an interval that will contain the value of x with a probability of at least .75, we specify

$\begin{array}{l}1\frac{1}{k^2}=0.75\\ \frac{1}{k^2}=1-0.75\\ \frac{1}{k^2}=0.25\\ k^2=\frac{1}{0.25}\\ =4\\ =2\end{array}$

Thus, the interval$\mu -2\sigma$ to$\mu +2\sigma$ will contain at least 75% of the probability.

This interval is given by

$$\begin{gathered} \mu -2\sigma =4.655-2\left(4.56004\right) \\ =4.655-8.912008752 \\ =-4.25700875 \end{gathered}$$

Hence,$\mathrm{\mu }-2\mathrm{\sigma }=-4.26$

$$\begin{gathered} \mu -2\sigma =4.655+2\left(4.56004\right) \\ =4.655-8.91200 \\ =13.5670 \end{gathered}$$

Hence,$\mu -2\sigma =13.56$ .