Question

the monthly incomes for 12 randomly selected people, each with a bachelors degree in economics, are shown on the right. complete parts (a) through (c) below. assume the population is normally distributed. 4450.97 4596.67 4366.39 4455.64 4151.92 3727.05 4283.53 4527.91 4407.37 3946.65 4023.24 4221.57 (a) find the sample mean. \\( \overline { x } = \square \\) (round to one decimal place as needed.)

Explanation:

Step1: Sum up all the data values

$$\begin{align*} &4450.97+4596.67+4366.39+4455.64+4151.92+3727.05+4283.53+4527.91+4407.37+3946.65+4023.24+4221.57\\ =&(4450.97 + 4596.67)+(4366.39+4455.64)+(4151.92 + 3727.05)+(4283.53+4527.91)+(4407.37+3946.65)+(4023.24+4221.57)\\ =&9047.64+8822.03+7878.97+8811.44+8354.02+8244.81\\ =&(9047.64+8822.03)+(7878.97+8811.44)+(8354.02+8244.81)\\ =&17869.67+16690.41+16598.83\\ =&(17869.67+16690.41)+16598.83\\ =&34560.08+16598.83\\ =&51158.91 \end{align*}$$

Step2: Calculate the sample mean

The formula for the sample mean $\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}$, where $n = 12$ (the number of data points) and $\sum_{i=1}^{n}x_{i}=51158.91$.
So, $\bar{x}=\frac{51158.91}{12}=4263.2425$

Step3: Round the result

Rounding $4263.2425$ to one decimal place gives $4263.2$

Answer:

$4263.2$