Question

a store manager is looking at past jewelry sales to determine what sizes of rings to keep in stock. the list shows the ring sizes purchased by the last ten jewelry customers.
9, 7, 6.5, 7.5, 7, 8, 5, 6, 7.5, 8
what is the variance of the data set? round to the nearest hundredths.
○ 0.40
○ 0.72
○ 1.15
○ 2.14

Explanation:

Step1: Calculate the mean

The data set is \( 9, 7, 6.5, 7.5, 7, 8, 5, 6, 7.5, 8 \). The number of data points \( n = 10 \).
Mean \( \bar{x}=\frac{9 + 7 + 6.5 + 7.5 + 7 + 8 + 5 + 6 + 7.5 + 8}{10} \)
\( \bar{x}=\frac{72}{10}=7.2 \)

Step2: Calculate squared deviations

For each data point \( x_i \), compute \( (x_i - \bar{x})^2 \):

• \( (9 - 7.2)^2 = 3.24 \)
• \( (7 - 7.2)^2 = 0.04 \)
• \( (6.5 - 7.2)^2 = 0.49 \)
• \( (7.5 - 7.2)^2 = 0.09 \)
• \( (7 - 7.2)^2 = 0.04 \)
• \( (8 - 7.2)^2 = 0.64 \)
• \( (5 - 7.2)^2 = 4.84 \)
• \( (6 - 7.2)^2 = 1.44 \)
• \( (7.5 - 7.2)^2 = 0.09 \)
• \( (8 - 7.2)^2 = 0.64 \)

Step3: Sum of squared deviations

Sum these values:
\( 3.24 + 0.04 + 0.49 + 0.09 + 0.04 + 0.64 + 4.84 + 1.44 + 0.09 + 0.64 = 11.5 \)

Step4: Calculate variance

Variance \( s^2=\frac{\sum (x_i - \bar{x})^2}{n - 1} \) (sample variance, as it's from a sample of customers)
\( s^2=\frac{11.5}{9}\approx1.28 \)? Wait, no—wait, maybe it's population variance? Wait, the problem says "data set"—if it's population (all ten customers), then \( \sigma^2=\frac{11.5}{10}=1.15 \). Ah, that matches the option. So we use population variance (since it's all ten customers' data).

Answer:

1.15