Question 1165659
The sample size needed to estimate the mean monthly earnings is **77**.

To find the required sample size ($n$), we use the formula for estimating a population mean:

$$n = \left(\frac{Z \cdot \sigma}{ME}\right)^2$$

### 1. Identify the Variables

* **Standard Deviation ($\sigma$):** $\$589$
* **Margin of Error ($ME$):** $\$132$
* **Confidence Level:** $95\%$
* **Z-score ($Z$):** For a $95\%$ confidence level, the critical $Z$-score is $\mathbf{1.96}$. 

### 2. Calculate the Sample Size

Substitute the values into the formula:

$$n = \left(\frac{1.96 \times 589}{132}\right)^2$$

First, calculate the term inside the parenthesis:

$$n = \left(\frac{1154.44}{132}\right)^2$$
$$n \approx (8.745757...)^2$$
$$n \approx 76.487$$

### 3. Round Up

Since the sample size must be a whole number and you must err on the side of caution (to ensure the margin of error is *no more* than \$132), you must **always round up** to the next whole number.

$$n = \mathbf{77}$$

The researcher needs a sample size of **77** graduates.