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 sum of the data values is \(9 + 7+6.5 + 7.5+7 + 8+5 + 6+7.5+8=71.5\). There are \(n = 10\) data - points. The mean \(\bar{x}=\frac{71.5}{10}=7.15\).

Step2: Calculate the squared differences

\((9 - 7.15)^2=(1.85)^2 = 3.4225\), \((7 - 7.15)^2=(-0.15)^2 = 0.0225\), \((6.5 - 7.15)^2=(-0.65)^2 = 0.4225\), \((7.5 - 7.15)^2=(0.35)^2 = 0.1225\), \((7 - 7.15)^2=(-0.15)^2 = 0.0225\), \((8 - 7.15)^2=(0.85)^2 = 0.7225\), \((5 - 7.15)^2=(-2.15)^2 = 4.6225\), \((6 - 7.15)^2=(-1.15)^2 = 1.3225\), \((7.5 - 7.15)^2=(0.35)^2 = 0.1225\), \((8 - 7.15)^2=(0.85)^2 = 0.7225\).

Step3: Calculate the sum of squared differences

The sum of the squared differences is \(3.4225+0.0225 + 0.4225+0.1225+0.0225+0.7225+4.6225+1.3225+0.1225+0.7225 = 11.5\).

Step4: Calculate the variance

The variance \(s^{2}=\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n}\). Substituting the sum of squared differences and \(n = 10\), we get \(s^{2}=\frac{11.5}{10}=1.15\).

Answer:

1.15