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.

Explanation:

Step1: Calculate the mean

The data set is \(50, 55, 60, 50, 55, 60, 50, 60, 55\). There are \(n = 9\) data - points.
The mean \(\bar{x}=\frac{50 + 55+60 + 50+55+60+50+60+55}{9}=\frac{495}{9}=55\).

Step2: Calculate the squared differences

For \(x_1 = 50\): \((x_1-\bar{x})^2=(50 - 55)^2=(-5)^2 = 25\)
For \(x_2 = 55\): \((x_2-\bar{x})^2=(55 - 55)^2=0^2 = 0\)
For \(x_3 = 60\): \((x_3-\bar{x})^2=(60 - 55)^2=5^2 = 25\)
We have 3 values of \(x = 50\), 3 values of \(x = 55\) and 3 values of \(x = 60\).
The sum \(\sum_{i = 1}^{n}(x_i-\bar{x})^2=3\times25+3\times0+3\times25=75 + 0+75 = 150\).

Step3: Calculate the standard deviation

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

Answer:

\(4.08\)