Question

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.

a. Describe the random variable $X$in words.

b. Find the probability that a customer rents three DVDs.

c. Find the probability that a customer rents at least four DVDs.

d. 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.

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?

Short answer

(a) X is the quantity of DVDs on rent per customer.

(b)$0.12$.

(c) $0.11$.

(d)$0.77$.

(e) At the Entertainment Headquarters, the expected number of DVDs rented per customer is higher

(f)$546$

(g)$588$

(h) The entertainment headquarters has more variation

Step-by-step solution

Given Information 

Given a table of the probability distribution for DVD rentals per customer at the video rental store.

Explanation (Part a)

X is the quantity of DVDs on rent per customer

Explanation (Part b)

The likelihood of just someone leasing three DVDs is

$$\begin{gathered} P(X=3)=1-0.03-0.5-0.24-0.07-0.04 \\ =0.12 \end{gathered}$$

Explanation (Part c)

The probability that the customer takes at least$4$DVDs is

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

Explanation (Part d)

The probability that the customer takes two or more DVDs is

$$\begin{gathered} P(X\le 2)=P(X=0)+P(X=1)+P(X=2) \\ =0.03+0.5+0.24 \\ =0.77 \end{gathered}$$

Explanation (Part e) 

In the first store, the value obtained from supplied DVDs is

$$\begin{gathered} {\mu }_1=\sum _{i=0}^5 i.P(X=i) \\ =0\left(0.03\right)+1\left(0.5\right)+2\left(0.24\right)+3\left(0.12\right)+4\left(0.07\right)+5 \\ =0+0.5+0.48+0.36+0.28+0.2 \\ =1.82 \end{gathered}$$

The value obtained from Entertainment headquarters:

$$\begin{gathered} {\mu }_1=\sum _{i=0}^5 iP(X=i) \\ =0\left(0.35\right)+1\left(0.25\right)+2\left(0.2\right)+3\left(0.1\right)+4 \\ =0+0.25+0.4+0.3+0.2+0.25 \\ =1.4 \end{gathered}$$

Hence the predicted amount of DVDs acquired by a visitor is higher.

Explanation (Part f)

For Video To Go The expected value is,

Expected value is$\sum x.P\left(x\right)$

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

Hence, the number of videos expectedly rented to$300$video To Go customers are

$300\times 1.82=546$

Explanation (Part g)

For Video To Go the expected value is

Expected value

=$\sum x.P\left(x\right)$

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

The value of Entertainment Headquarters is expected as

Expected value =$\sum _{}x.P\left(x\right)$

$$\begin{gathered} =0\times 0.35+1\times 0.25+2\times 0.20+3\times 0.10+4\times 0.05+5\times 0.05 \\ =1.4 \end{gathered}$$

The expected number of videos rented is$300\times 1.82=546$

For the videos rented to the expected number is

$420\times 1.4=588$

Explanation (part h)

The expected value of Video To Go is:

Expected value$=\underset{}{\sum x.P\left(x\right)}$

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

The expected value of Entertainment Headquarters is:

Expected value$=\underset{}{\sum x.P\left(x\right)}$

$$\begin{gathered} =0\times 0.35+1\times 0.25+2\times 0.20+3\times 0.10+4\times 0.05+5\times 0.05 \\ =0+0.25+0.4+0.3+0.2+0.25 \\ =1.4 \end{gathered}$$

Calculating the quality deviations from both sets

$$\begin{gathered} {\sigma }_1=\sqrt{\sum _{i=0}^5 {\left(xi-{\mu }_1\right)}^{2}P(X=i)} \\ =\sqrt{{\left(0-1.82\right)}^2\times 0.03+{\left(1-1.82\right)}^2\times 0.50+{\left(2-1.82\right)}^2\times 0.24+{\left(3-1.82\right)}^2\times 0.12+{\left(4-1.82\right)}^2\times 0.07+{\left(5-1.82\right)}^2\times 0.04} \\ =1.161 \\ {\sigma }_2=\sqrt{\sum _{i=0}^5 {\left(xi-{\mu }_2\right)}^{2}P(X=i)} \\ =\sqrt{{\left(0-1.4\right)}^2\times 0.35+{\left(1-1.4\right)}^2\times 0.25+{\left(2-1.4\right)}^2\times 0.20+{\left(3-1.4\right)}^2\times 0.10+{\left(4-1.4\right)}^2\times 0.05+{\left(5-1.4\right)}^2\times 0.05} \\ =1.429 \end{gathered}$$