Problem 33E

The Bureau of the Census reports that the median family income for all families in the United States during the year 2003 was $43,318. That is, half of all American families had incomes exceeding this amount, and half had incomes equal to or below this amount. Suppose that four families are surveyed and that each one reveals whether its income exceeded $43,318 in 2003.

a List the points in the sample space.

b Identify the simple events in each of the following events:

A: At least two had incomes exceeding $43,318.

B: Exactly two had incomes exceeding $43,318.

C: Exactly one had income less than or equal to $43,318.

c Make use of the given interpretation for the median to assign probabilities to the simple events and find P( A), P( B), and P(C).

Solution

Step 1 of 3

a) we have to write the sample space

The median income is 43,318$

Let E be the family has the income more than 43,318 $

Let F be the family has the income less than or equal to 43,318 $

The no.of surveyed families=4

The no.of combinations=42 =16

Hence the sample space is

S={(EEEE), (EEEF), (EEFE), (EEFF), (EFEE), (EFEF), (EFFE), (EFFF),

(FEEE), (FEEF), (FEFE), (FEFF), (FFEE), (FFEF), (FFFE), (FFFF)}