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 formula for the mean \(\bar{x}=\frac{\sum_{i = 1}^{n}x_{i}}{n}\).
Here \(n = 10\), and \(\sum_{i=1}^{10}x_{i}=9 + 7+6.5 + 7.5+7 + 8+5 + 6+7.5 + 8=71.5\).
So \(\bar{x}=\frac{71.5}{10}=7.15\).

Step2: Calculate the squared - differences from the mean

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

Step3: Calculate the variance

The formula for the variance \(s^{2}=\frac{\sum_{i = 1}^{n}(x_{i}-\bar{x})^{2}}{n}\).
\(\sum_{i = 1}^{10}(x_{i}-\bar{x})^{2}=3.4225+0.0225 + 0.4225+0.1225+0.0225+0.7225+4.6225+1.3225+0.1225+0.7225 = 11.5\).
\(s^{2}=\frac{11.5}{10}=1.15\)

Answer:

1.15