Chapter 13.3, Problem 65AYU

To determine

To calculate: The probability that a household has an annual income of less than $\$50,000$ using following table which is based on a survey of annual incomes in $100$ households .

$\boxed{\begin{array}{llllll}\text{Income}\hfill & \$0-24,999\hfill & \$25,000-49,999\hfill & \$50,000-74,999\hfill & \$75,000-99,999\hfill & \$100,000\text{or}\text{more}\hfill \\ \begin{array}{l}\text{Number}\text{of}\\ \text{households}\end{array}\hfill & 22\hfill & 23\hfill & 17\hfill & 12\hfill & 26\hfill \end{array}}$

Answer to Problem 65AYU

Solution:

The probability that a household has an annual income of less than $\$50,000$ is $0.45$.

Explanation of Solution

Given information:

The following table contains the data of a survey of annual incomes in $100$ households.

$\begin{array}{cccccc}\text{Income}& \$0-24,999& \$25,000-49,999& \$50,000-74,999& \$75,000-99,999& \$100,000\text{or}\text{more}\\ \begin{array}{l}\text{Number }\text{of }\\ \text{households}\end{array}& 22& 23& 17& 12& 26\end{array}$

Formula used:

If $E$ is any event and $S$ is the sample space then, the probability of event $E$ is, $P\left(E\right)=\frac{n\left(E\right)}{n\left(S\right)}$.

Calculation:

Total number of households are $100$, so $n\left(S\right)=100$

Let event $E$ denotes households with an annual income of lessthan $\$50,000$.

According to given table, an income of less than $\$50,000$ represents two income groups. $\$0-\$24,999$ and $\$25,000-\$49,999$ or more.

There are $22$ households with an income group of $\$0-\$24,999$.

There are $23$ households with an income group of $\$25,000-\$49,999$.

Therefore the total number of households with an income of less than $\$50,000$ is $n\left(E\right)=22+23=45$.

Using the formula for probability of an event, $P\left(E\right)=\frac{n\left(E\right)}{n\left(S\right)}$.

$\Rightarrow P\left(E\right)=\frac{45}{100}=0.45$

Therefore, the probability that a household has an annual income of less than $\$50,000$ is $0.45$.