Question

the table below relates the amount of time consumers spend shopping on the internet per month with their annual income. what is the probability that a randomly selected consumer either spends 0 - 2 hours per month shopping on the internet or has an annual income below $40,000?
the probability that a consumer either has an annual income below $40,000 or spends 0 - 2 hours per month shopping on the internet is
(type a decimal rounded to two decimal places.)

Explanation:

Step1: Use the formula for \(P(A\cup B)\)

The formula for the probability of the union of two events \(A\) and \(B\) is \(P(A\cup B)=P(A)+P(B)-P(A\cap B)\). Let \(A\) be the event that a consumer has an annual income below \(\$40,000\) and \(B\) be the event that a consumer spends \(0 - 2\) hours per month shopping on the Internet.

Step2: Calculate \(P(A)\), \(P(B)\) and \(P(A\cap B)\)

• \(P(A)=\frac{n(A)}{n(S)}\), where \(n(A) = 575\) (total number of consumers with income below \(\$40,000\)) and \(n(S)=1559\) (total number of consumers). So \(P(A)=\frac{575}{1559}\).
• \(P(B)=\frac{n(B)}{n(S)}\), where \(n(B) = 514\) (total number of consumers who spend \(0 - 2\) hours) and \(n(S) = 1559\). So \(P(B)=\frac{514}{1559}\).
• \(P(A\cap B)=\frac{n(A\cap B)}{n(S)}\), where \(n(A\cap B)=251\) (number of consumers with income below \(\$40,000\) and spending \(0 - 2\) hours) and \(n(S)=1559\). So \(P(A\cap B)=\frac{251}{1559}\).

Step3: Substitute into the formula

$$
LATEXBLOCK0
$$

Answer:

\(0.54\)