Question

grace’s 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:
σ = \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.
options: $4.08, $50.00, $10.00, $5.00

Explanation:

Step1: Find the mean of the data set

The data points are: \( \$50, \$55, \$60, \$50, \$55, \$60, \$50, \$60, \$55 \)
First, count the number of data points, \( n = 9 \)
Sum of the data points: \( 50 + 55 + 60 + 50 + 55 + 60 + 50 + 60 + 55 \)
\( = (50\times3)+(55\times3)+(60\times3) \)
\( = 150 + 165 + 180 \)
\( = 495 \)
Mean \( (\mu) = \frac{495}{9} = 55 \)

Step2: Calculate the squared differences from the mean

For each data point \( x_i \), calculate \( (x_i - \mu)^2 \)

• For \( x_i = 50 \): \( (50 - 55)^2 = (-5)^2 = 25 \) (there are 3 such points, so total contribution: \( 3\times25 = 75 \))
• For \( x_i = 55 \): \( (55 - 55)^2 = 0^2 = 0 \) (there are 3 such points, so total contribution: \( 3\times0 = 0 \))
• For \( x_i = 60 \): \( (60 - 55)^2 = 5^2 = 25 \) (there are 3 such points, so total contribution: \( 3\times25 = 75 \))

Sum of squared differences: \( \sum_{i = 1}^{n}(x_i - \mu)^2 = 75 + 0 + 75 = 150 \)

Step3: Calculate the variance

Variance \( (\sigma^2) = \frac{\sum_{i = 1}^{n}(x_i - \mu)^2}{n} = \frac{150}{9} = \frac{50}{3} \approx 16.67 \)

Step4: Calculate the standard deviation

Standard deviation \( (\sigma) = \sqrt{\frac{50}{3}} \approx \sqrt{16.67} \approx 4.08 \)

Answer:

\( \$4.08 \)