People visiting video rental stores often rent more than one DVD at a time. The probability distribution for DVD rentals per customer at Video To Go is given in the following table. There is a five-video limit per customer at this store, so nobody ever rents more than five DVDs.

$\begin{array}{|ll|}\hline x& P\left(x\right)\\ 0& 0.03\\ 1& 0.50\\ 2& 0.24\\ 3& \\ 4& 0.07\\ 5& 0.04\\ \hline\end{array}$

1. Describe the random variable X in words.
2. Find the probability that a customer rents three DVDs.
3. Find the probability that a customer rents at least four DVDs.
4. Find the probability that a customer rents at most two DVDs.

Another shop, Entertainment Headquarters, rents DVDs and video games. The probability distribution for DVD rentals per customer at this shop is given as follows. They also have a five-DVD limit per customer.

$\begin{array}{|ll|}\hline \mathit{x}& \mathit{P}\left(\mathit{x}\right)\\ 0& 0.35\\ 1& 0.25\\ 2& 0.20\\ 3& 0.10\\ 4& 0.05\\ 5& 0.05\\ \hline\end{array}$

e. At which store is the expected number of DVDs rented per customer higher?

f. If Video to Go estimates that they will have 300 customers next week, how many DVDs do they expect to rent next week? Answer in sentence form.

g. If Video to Go expects 300 customers next week, and Entertainment HQ projects that they will have 420 customers, for which store is the expected number of DVD rentals for next week higher? Explain.

h. Which of the two video stores experiences more variation in the number of DVD rentals per customer? How do you know that?

Given in the question that, the probability distribution for DVD rentals per customer at Video To Go is given in the following table. There is a five-video limit per customer at this store, so nobody ever rents more than five DVDs.

We have to describe the random variable $X$in words.

A random variable is a numerical representation of a statistical experiment's outcome. The random variable $X$can be determined as follows using the problem's provided information:

$X=ThequantityofDVDsrentedbya\sin gleconsumer.$

Given in the question that,

$\begin{array}{|ll|}\hline x& P\left(x\right)\\ 0& 0.03\\ 1& 0.50\\ 2& 0.24\\ 3& \\ 4& 0.07\\ 5& 0.04\\ \hline\end{array}$

We have to find the probability that a customer rents three DVDs.

The sum of probability is $1$, as we know.

As a result, we have:

The probability of a customer renting three DVDs can be estimated as follows:

$$\begin{gathered} P(x=3)=1-(0.03+0.50+0.24+0.07+0.04) \\ =1-0.88 \\ =0.12 \end{gathered}$$

Given in the question that,

$\begin{array}{|ll|}\hline \mathit{x}& \mathit{P}\left(\mathit{x}\right)\\ 0& 0.35\\ 1& 0.25\\ 2& 0.20\\ 3& 0.10\\ 4& 0.05\\ 5& 0.05\\ \hline\end{array}$

We have to find the probability that a customer rents at least four DVDs.

The following formula can be used to calculate the probability that a consumer will rent at least four DVDs:

$$\begin{gathered} P(x\ge 4)=P(x=4)+P(x=5) \\ =0.07+0.04 \\ =0.11 \end{gathered}$$

Given in the question that,

$\begin{array}{|ll|}\hline \mathit{x}& \mathit{P}\left(\mathit{x}\right)\\ 0& 0.35\\ 1& 0.25\\ 2& 0.20\\ 3& 0.10\\ 4& 0.05\\ 5& 0.05\\ \hline\end{array}$

We have to find the probability that a customer rents at most two DVDs.

The following formula can be used to calculate the probability that a consumer will rent at least two DVDs:

$$\begin{gathered} P(x\le 2)=P(x=0)+P(x=1)+P(x=2) \\ =0.03+0.50+0.24 \\ =0.77 \end{gathered}$$

The probability of a customer renting no more than two DVDs is 0.77.

Given in the question that,

$\begin{array}{|ll|}\hline \mathit{x}& \mathit{P}\left(\mathit{x}\right)\\ 0& 0.35\\ 1& 0.25\\ 2& 0.20\\ 3& 0.10\\ 4& 0.05\\ 5& 0.05\\ \hline\end{array}$

We need to determine which retailer has a larger predicted quantity of DVDs rented per customer.

The following is the expected value of Video To Go:

$Expectedvalue=\sum x·P\left(x\right)$

$$\begin{gathered} =0\times 0.03+1\times 0.5+2\times 0.24+3\times 0.12+4\times 0.07+5\times 0.04 \\ =0+0.5+0.48+0.36+0.28+0.2 \\ =1.82 \end{gathered}$$

Entertainment Headquarters is anticipated to be worth:

$$\begin{gathered} \text{Expected value}=\sum x·P\left(x\right) \\ =0\times 0.35+1\times 0.25+2\times 0.20+3\times 0.10+4\times 0.05+5\times 0.05 \\ =0+0.25+0.4+0.3+0.2+0.25 \\ =1.4 \end{gathered}$$

As a result, the estimated quantity of DVDs rented per customer at Entertainment Headquarters is higher.

Given in the question that,

$\begin{array}{|ll|}\hline \mathit{x}& \mathit{P}\left(\mathit{x}\right)\\ 0& 0.35\\ 1& 0.25\\ 2& 0.20\\ 3& 0.10\\ 4& 0.05\\ 5& 0.05\\ \hline\end{array}$

The following is the predicted value of video To Go:

$$\begin{gathered} \text{Expected value}=\sum x·P\left(x\right) \\ =0\times 0.03+1\times 0.5+2\times 0.24+3\times 0.12+4\times 0.07+5\times 0.04 \\ =0+0.5+0.48+0.36+0.28+0.2 \\ =1.82 \end{gathered}$$

As a result, the quantity of videos leased by 300 Video To Go consumers is estimated to be:

$300\times 1.82=546.$

Given in the question that,

$\begin{array}{|ll|}\hline \mathit{x}& \mathit{P}\left(\mathit{x}\right)\\ 0& 0.35\\ 1& 0.25\\ 2& 0.20\\ 3& 0.10\\ 4& 0.05\\ 5& 0.05\\ \hline\end{array}$

The following is the expected value of Video To Go:

$$\begin{gathered} \text{Expected value}=\sum x·P\left(x\right) \\ =0\times 0.03+1\times 0.5+2\times 0.24+3\times 0.12+4\times 0.07+5\times 0.04 \\ =0+0.5+0.48+0.36+0.28+0.2 \\ =1.82 \end{gathered}$$

Entertainment HQ is anticipated to be worth:

$$\begin{gathered} \text{Expected value}=\sum x·P\left(x\right) \\ =0\times 0.35+1\times 0.25+2\times 0.20+3\times 0.10+4\times 0.05+5\times 0.05 \\ =0+0.25+0.4+0.3+0.2+0.25 \\ =1.4 \end{gathered}$$

The following is the estimated amount of videos leased by 300 Video To Go customers:

$300\times 1.82=546$

Customers of 420 Entertainment Headquarters can anticipate to rent the following amount of videos:

$420\times 1.4=588\text{.}$

As a result, Entertainment Headquarters will continue to rent out additional videos.

Given in the question that,

$\begin{array}{|ll|}\hline \mathit{x}& \mathit{P}\left(\mathit{x}\right)\\ 0& 0.35\\ 1& 0.25\\ 2& 0.20\\ 3& 0.10\\ 4& 0.05\\ 5& 0.05\\ \hline\end{array}$

Let's compute the standard deviation of Video to Go is:

$$\begin{gathered} \text{Standard deviation}=\sqrt{\sum (x-Mean{)}^{2}P(x)} \\ =\sqrt{\begin{array}{l}(0-1.82{)}^2\times 0.03+(1-1.82{)}^2\times 0.50+(2-1.82{)}^2\times 0.24+(3-1.82{)}^2\\ \times 0.12+(4-1.82{)}^2\times 0.07+(5-1.82{)}^2\times 0.04\end{array}} \\ =1.1609 \end{gathered}$$

The entertainment headquarters standard deviation is:

$$\begin{gathered} \text{Standard deviation}=\sqrt{\sum (x-Mean{)}^{2}P(x)} \\ =\sqrt{\begin{array}{l}(0-1.4{)}^2\times 0.35+(1-1.4{)}^2\times 0.25+(2-1.4{)}^2\times 0.20+(3-1.4{)}^2\times 0.10\\ +(4-1.4{)}^2\times 0.05+(5-1.4{)}^2\times 0.05\end{array}} \\ =1.4293 \end{gathered}$$

As a result, the entertainment headquarters has a wider range of options: