Question

graces entertainment - expenses for the year were: $50, $55, $60, $50, $55, $60, $50, $60, $55. what is the standard deviation of her expenses? use the following formula to calculate standard deviation: $sigma=sqrt{\frac{sum_{i = 1}^{n}(x_{i}-\text{mean})^{2}}{n}}$ where $x_{i}$ is each data point, and $n$ is the number of data points. $10.00 $ $50.00 $ $4.08 $ $5.00 $

Explanation:

Step1: Calculate the mean

The data set is $50, 55, 60, 50, 55, 60, 50, 60, 55$.
The sum of the data points is $50\times3 + 55\times3+60\times3=150 + 165+180 = 495$.
The number of data points $n = 9$.
The mean $\bar{x}=\frac{495}{9}=55$.

Step2: Calculate the squared - differences

For $x_1 = 50$: $(50 - 55)^2=(-5)^2 = 25$.
For $x_2 = 55$: $(55 - 55)^2=0^2 = 0$.
For $x_3 = 60$: $(60 - 55)^2=5^2 = 25$.
Since there are 3 of each value, the sum of squared - differences $\sum_{i = 1}^{n}(x_i-\bar{x})^2=3\times25+3\times0 + 3\times25=150$.

Step3: Calculate the standard deviation

Using the formula $\sigma=\sqrt{\frac{\sum_{i = 1}^{n}(x_i - \bar{x})^2}{n}}$, we substitute $\sum_{i = 1}^{n}(x_i - \bar{x})^2 = 150$ and $n = 9$.
$\sigma=\sqrt{\frac{150}{9}}=\sqrt{\frac{50}{3}}\approx4.08$.

Answer:

$4.08$