Chapter 5, Problem 8SR

Consumers were surveyed on the relative number of visits to a Sears store (often, occasional, and never) and if the store was located in an enclosed mall (yes and no). When variables are measured nominally, such as these data, the results are usually summarized in a contingency table.

What is the probability of selecting a shopper who:

1. (a) Visited a Sears store often?
2. (b) Visited a Sears store in an enclosed mall?
3. (c) Visited a Sears store in an enclosed mall or visited a Sears store often?
4. (d) Visited a Sears store often, given that the shopper went to a Sears store in an enclosed mall?

In addition:

1. (e) Are the number of visits and the enclosed mall variables independent?
2. (f) What is the probability of selecting a shopper who visited a Sears store often and it was in an enclosed mall?
3. (g) Draw a tree diagram and determine the various joint probabilities.

To determine

Obtain the probability of selecting a shopper who visits a Sears store often.

Answer to Problem 8SR

The probability of selecting a shopper who visits a Sears store often is 0.4103.

Explanation of Solution

Here, number of shopper who visited Sears often is 80. Total number of shoppers is 195.

The required probability is obtained as given below:

$\begin{array}{c}P\left(\text{Shoppers visited often}\right)=\frac{\left(\text{Number of shoppers visited often}\right)}{\text{Total number of shoppers}}\\ =\frac{80}{195}\\ =\text{0}\text{.4103}\end{array}$

Thus, the probability of selecting a shopper who visits a Sears store often is 0.4103.

To determine

Obtain the probability of selecting a shopper in an enclosed mall.

Answer to Problem 8SR

The probability of selecting a shopper in an enclosed mall is 0.4615.

Explanation of Solution

Here, number of shopper who visited an enclosed mall is 90.

The required probability is obtained as given below:

$\begin{array}{c}P\left(\text{Shoppers visited enclosed mall}\right)=\frac{\left(\begin{array}{l}\text{Number of shoppers }\\ \text{visited enclosed mall}\end{array}\right)}{\text{Total number of shoppers}}\\ =\frac{90}{195}\\ =\text{0}\text{.4615}\end{array}$

Thus, the probability of selecting a shopper in an enclosed mall is 0.4615.

To determine

Obtain the probability of selecting a shopper in an enclosed mall or visited Sears store often.

Answer to Problem 8SR

The probability of selecting a shopper in an enclosed mall or visited Sears store often is 0.5641.

Explanation of Solution

Here, number of shopper who visited an enclosed mall and often is 60.

$\begin{array}{c}P\left(\begin{array}{l}\text{Visited in an }\\ \text{enclosed mall}\\ \text{or often}\end{array}\right)=P\left(\begin{array}{l}\text{Visited in an}\\ \text{enclosed mall}\end{array}\right)+P\left(\text{Visited often}\right)-P\left(\begin{array}{l}\text{Visited in an }\\ \text{enclosed mall}\\ \text{and often}\end{array}\right)\\ =\frac{90}{195}+\frac{80}{195}-\frac{60}{195}\\ =\frac{\text{110}}{195}\\ =\text{0}\text{.5641}\end{array}$

Thus, the probability of selecting a shopper in an enclosed mall or visited Sears store often is 0.5641.

To determine

Obtain the probability of selecting a shopper who visited Sears store often given that shopper went to a Sears store in an enclosed mall.

Answer to Problem 8SR

The probability of selecting a shopper who visited Sears store often given that shopper went to a Sears store in an enclosed mall is 0.6667.

Explanation of Solution

The required probability is obtained as given below:

$\begin{array}{c}P\left(\begin{array}{l}\text{Visited often | Visited}\\ \text{in an enclosed mall}\end{array}\right)=\frac{P\left(\begin{array}{l}\text{Visited often and Visited}\\ \text{in an enclosed mall}\end{array}\right)}{P\left(\text{Visited in an enclosed mall}\right)}\\ =\frac{\frac{60}{195}}{\frac{90}{195}}\\ =\frac{60}{90}\\ =\text{0}\text{.6667}\end{array}$

Thus, the probability of selecting a shopper who visited Sears store often given that shopper went to a Sears store in an enclosed mall is 0.6667.

To determine

Check whether number of visits and the enclosed mall variables independent.

Answer to Problem 8SR

Number of visits and the enclosed mall variables are not independent.

Explanation of Solution

Two events A and B are independent if $P\left(A|B\right)=P\left(A\right)$

From part (d), $P\left(\begin{array}{l}\text{Visited often | Visited}\\ \text{in an enclosed mall}\end{array}\right)=\text{0}\text{.6667}$

From part (a), $P\left(\text{Shoppers visited often}\right)=\text{0}\text{.4103}$

Thus, $P\left(A|B\right)\ne P\left(A\right)$. Hence, number of visits and the enclosed mall variables are not independent.

To determine

Obtain the probability of selecting a shopper in an enclosed mall and visited Sears store often.

Answer to Problem 8SR

The probability of selecting a shopper in an enclosed mall and visited Sears store often is 0.3077.

Explanation of Solution

Here, number of shopper who visited an enclosed mall and often is 60.

$\begin{array}{c}P\left(\begin{array}{l}\text{Visited in an }\\ \text{enclosed mall}\\ \text{and often}\end{array}\right)=\frac{\left(\begin{array}{l}\text{Number of shoppers }\\ \text{visited enclosed mall and often}\end{array}\right)}{\text{Total number of shoppers}}\\ =\frac{60}{195}\\ =\text{0}\text{.3077}\end{array}$

Thus, the probability of selecting a shopper in an enclosed mall and visited Sears store often is 0.3077.

To determine

Construct a tree diagram that shows probabilities, conditional probabilities and joint probabilities.

Explanation of Solution

The first branch represents the shoppers visited enclosed mall which is divided into two categories. In the second branch, each category is subdivided into three different visits. The probabilities under the second branch represents conditional probabilities. The product of the probabilities in two branches represent joint probabilities.

The tree diagram is as given below: