Chapter 5, Problem 62SE

A bookstore at the Hartsfield-Jackson Airport in Atlanta sells reading materials (paperback books, newspapers, magazines) as well as snacks (peanuts, pretzels, candy, etc.). A point-of-sale terminal collects a variety of information about customer purchases. Shown below is a table showing the number of snack items and the number of items of reading material purchased by the most recent 600 customers.

		0	Reading Material 1	2
	0	0	60	18
Snacks	1	240	90	30
	2	120	30	12

1. a. Using the data in the table construct an empirical discrete bivariate probability distribution for x = number of snack items and y = number of reading materials in a randomly selected customer purchase. what is the probability of a customer purchase consisting of one item of reading materials and two snack items? what is the probability of a customer purchasing one snack item only? why is the probability f(x = 0, y = 0) = 0?
2. b. Show the marginal probability distribution for the number of snack items purchased. Compute the expected value and variance.
3. c. What is the expected value and variance for the number of reading materials purchased by a customer?
4. d. Show the probability distribution for t = total number of items in a customer purchase. Compute its expected value and variance.

Compute the covariance and correlation coefficient between x and y. What is the relationship, if any, between the number of reading materials and number of snacks purchased on a customer visit?

To determine

Develop an empirical bivariate probability distribution for number of snack items and the number of reading materials in a randomly selected customer purchase.

Find the probability of a customer purchase consisting of one item of reading materials and two snack items, probability of a customer purchasing one snack item and to explain why the probability, $f\left(x=0,y=0\right)=0$.

Answer to Problem 62SE

The empirical bivariate probability distribution for number of snack items and the number of reading materials in a randomly selected customer purchase is as follows:

		Reading material (y)
		0	1	2	Total
Snacks	0	0.00	0.10	0.03	0.13
	1	0.40	0.15	0.05	0.60
	2	0.20	0.05	0.02	0.27
	Total	0.60	0.30	0.10	1.00

The probability of a customer purchase consisting of one item of reading materials and two snack items is 0.05.

The probability of a customer purchasing one snack item is 0.40.

Explanation of Solution

Calculation:

The table represents the number of snack items and the number of items of reading material purchased by 600 customers. The random variable x represents the number of snack items and the random variable y represents the number of reading materials in a randomly selected customer purchase.

The bivariate distribution is obtained by dividing each frequency in the table by the total number of customers. The probability of a customer purchasing 1 item of reading material and 1 item of snacks is,

$\begin{array}{c}f\left(1,1\right)=\frac{\text{Frequency}}{\text{Total number of Customers}}\\ =\frac{90}{600}\\ =0.15\end{array}$

Similarly other bivariate probabilities are obtained and given in the following table:

		Reading material (y)
		0	1	2	Total
Snacks(x)	0	$\frac{0}{600}=0.00$	$\frac{60}{600}=0.10$	$\frac{18}{600}=0.03$	$\left(\begin{array}{l}0.00+\\ 0.10+\\ 0.03\end{array}\right)=0.13$
	1	$\frac{240}{600}=0.40$	$\frac{90}{600}=0.15$	$\frac{30}{600}=0.05$	$\left(\begin{array}{l}0.40+\\ 0.15+\\ 0.05\end{array}\right)=0.60$
	2	$\frac{120}{600}=0.20$	$\frac{30}{600}=0.05$	$\frac{12}{600}=0.02$	$\left(\begin{array}{l}0.20+\\ 0.05+\\ 0.02\end{array}\right)=0.27$
	Total	$\left(\begin{array}{l}0.00+\\ 0.40+\\ 0.20\end{array}\right)=0.60$	$\left(\begin{array}{l}0.10+\\ 0.15+\\ 0.05\end{array}\right)=0.30$	$\left(\begin{array}{l}0.03+\\ 0.05+\\ 0.02\end{array}\right)=0.10$	1.00

The marginal probability distribution for the random variable x is in the rightmost column and the marginal probability distribution for the random variable y is in the bottom row.

Thus, the bivariate probability distribution for number of snack items and the number of reading materials in a randomly selected customer purchase is obtained.

From the bivariate probability distribution table, the probability of a customer purchase consisting of one item of reading materials and two snack items is given by,

$f\left(x=1,y=2\right)=0.05$

Thus, the probability of a customer purchase consisting of one item of reading materials and two snack items is 0.05.

From the bivariate probability distribution table, the probability of a customer purchase one snack item is given by,

$f\left(x=1,y=0\right)=0.40$

Thus, the probability of a customer purchasing one snack item is 0.40.

The point sale terminal is only used when someone makes a purchase. Thus, the probability value, $f\left(x=0,y=0\right)=0$.

To determine

Show the marginal probability distribution for the number of snack items purchased and compute its expected value and variance.

Answer to Problem 62SE

The marginal probability distribution for the number of snack items purchased is:

x	f(x)
0	0.13
1	0.60
2	0.27

The expected value for the number of snack items purchased is 1.14.

The variance for the number of snack items purchased is 0.3804.

Explanation of Solution

Calculation:

The marginal distribution for the number of snack items purchased is the values of the random variable x and the corresponding probabilities. This values are obtained from the rightmost column of the bivariate probability distribution for number of snack items and the number of reading materials in a randomly selected customer purchase obtained in part (a).

The marginal distribution of parts cost is given in the following table:

x	f(x)
0	0.13
1	0.60
2	0.27

Thus, the marginal distribution of parts cost is obtained.

The expected value of the discrete random variable:

The formula for the expected value of a discrete random variable is,

$\begin{array}{c}E\left(x\right)=\mu \\ ={\displaystyle \sum x\cdot f\left(x\right)}\end{array}$

The expected value for the random variable x is obtained in the following table:

x	f(x)	$x\cdot f\left(x\right)$
0	0.13	0.00
1	0.60	0.60
2	0.27	0.54
Total	1.00	1.14

Thus, the expected value for the random variable x is 1.14.

The formula for the variance of the discrete random variable x is,

${\sigma }^2={\displaystyle \sum \left[{\left(x-\mu \right)}^2.f\left(x\right)\right]}$

The variance of the random variable x is obtained using the following table:

x	f(x )	$\left(x-\mu \right)$	${\left(x-\mu \right)}^2$	${\left(x-\mu \right)}^2.f\left(x\right)$
0	0.13	–1.14	1.2996	0.16895
1	0.60	–0.14	0.0196	0.01176
2	0.27	0.86	0.7396	0.19969
Total	1.00	–0.42	2.0588	0.3804

Therefore,

${\sigma }^2=0.3804$

Thus, the variance of the random variable x is 0.3804.

To determine

Compute the expected value and variance for the number of reading materials purchased by a customer.

Answer to Problem 62SE

The expected value for the number of reading materials purchased is 0.50.

The variance for the number of reading materials purchased is 0.45.

Explanation of Solution

Calculation:

The expected value of the discrete random variable:

The formula for the expected value of a discrete random variable is,

$\begin{array}{c}E\left(y\right)=\mu \\ ={\displaystyle \sum y\cdot f\left(y\right)}\end{array}$

The expected value for the random variable y is obtained in the following table:

y	f(y)	$y\cdot f\left(y\right)$
0	0.60	0.0
1	0.30	0.30
2	0.10	0.2
Total	1.0	0.5

Thus, the expected value for the random variable y is 0.5.

The formula for the variance of the discrete random variable y is,

${\sigma }^2={\displaystyle \sum \left[{\left(y-\mu \right)}^2.f\left(y\right)\right]}$

The variance of the random variable y is obtained using the following table:

y	f(y)	$\left(y-\mu \right)$	${\left(y-\mu \right)}^2$	${\left(y-\mu \right)}^2.f\left(y\right)$
0	0.60	–0.5	0.25	0.15
1	0.30	0.5	0.25	0.075
2	0.10	1.5	2.25	0.225
Total	1.00			0.45

Therefore,

${\sigma }^2=0.45$

Thus, the variance of the random variable y is 0.45.

That is, the expected value for the number of reading materials purchased is 0.50 and the variance is 0.45.

To determine

Show the probability distribution for total number of items in a customer purchase and to compute its expected value and variance.

Answer to Problem 62SE

The probability distribution for total number of items in a customer purchase is given below:

t	f(t)
1	0.50
2	0.38
3	0.10
4	0.02

The expected number of items purchased is 1.64.

The variance for the total number of items purchased is 0.5504.

Explanation of Solution

Calculation:

The total number of items purchased is represented using the random variable t. the random variable t take the value 1 when $x=0,y=1$ and when $x=1,y=0$. That is,

$f\left(t=1\right)=f\left(x=0,y=1\right)+f\left(x=1,y=0\right)$

Substitute the values from the bivariate probability distribution table in part (a),

$\begin{array}{c}f\left(t=1\right)=f\left(x=0,y=1\right)+f\left(x=1,y=0\right)\\ =0.40+0.10\\ =0.50\end{array}$

Similarly, other probability values can be obtained. Thus, the probability distribution for the random variable t is:

$t=x+y$	f(t)
1	0.50
2	0.38
3	0.10
4	0.02

To determine

Compute the covariance and correlation coefficient between x and y.

Identify the relationship between the number of reading materials and the number of snacks purchased on a customer visit.

Answer to Problem 62SE

The covariance of x and y is –0.14 and the correlation coefficient is –0.3384.

There is a negative correlation between the number of reading materials and the number of snacks purchased on a customer visit.

Explanation of Solution

Calculation:

Covariance for the random variables x and y is given by,

${\sigma }_{xy}=\frac{\left[\text{Var}\left(x+y\right)-\text{Var}\left(x\right)-\text{Var}\left(y\right)\right]}{2}$

Substitute the values, $\text{Var}\left(x+y\right)=0.5504,\text{Var}\left(x\right)=0.3804$ and $\text{Var}\left(y\right)=0.45$

$\begin{array}{c}{\sigma }_{xy}=\frac{\left[\text{Var}\left(x+y\right)-\text{Var}\left(x\right)-\text{Var}\left(y\right)\right]}{2}\\ =\frac{0.5504-0.3804-0.45}{2}\\ =-0.14\end{array}$

Correlation between the random variables x and y is given by,

${\rho }_{xy}=\frac{{\sigma }_{xy}}{{\sigma }_x{\sigma }_y}$

The standard deviation is obtained by taking the square root of variance.

$\begin{array}{c}{\sigma }_x=\sqrt{\text{Var}\left(x\right)}\\ =\sqrt{0.3804}\\ =0.6168\end{array}$

$\begin{array}{c}{\sigma }_y=\sqrt{\text{Var}\left(y\right)}\\ =\sqrt{0.45}\\ =0.6708\end{array}$

Substitute the values in the correlation formula,

$\begin{array}{c}{\rho }_{xy}=\frac{{\sigma }_{xy}}{{\sigma }_x{\sigma }_y}\\ =\frac{-0.14}{\left(0.6168\right)\left(0.6708\right)}\\ =-0.3384\end{array}$

The covariance of x and y is –0.14 and the correlation coefficient is –0.3384.

Thus, correlation coefficient of x and y is –0.3384 which is negative. Thus, there is a negative correlation between the number of reading materials and the number of snacks purchased on a customer visit. That is, as the purchase of reading materials increase the number of snack items purchased decreases.