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. 0.72 1.15 0.40 2.14

Explanation:

Step1: Calculate the mean

First, we find the mean (\(\bar{x}\)) of the data set. The data set is \(9, 7, 6.5, 7.5, 7, 8, 5, 6, 7.5, 8\) and there are \(n = 10\) data points.

The sum of the data points:
\[

$$\begin{align*} 9 + 7 + 6.5 + 7.5 + 7 + 8 + 5 + 6 + 7.5 + 8&=(9 + 7)+(6.5 + 7.5)+(7 + 8)+(5 + 6)+(7.5 + 8)\\ &= 16+14 + 15+11+15.5\\ &=16 + 14=30; 30+15 = 45; 45+11 = 56; 56+15.5 = 71.5 \end{align*}$$

\]

The mean \(\bar{x}=\frac{\text{Sum}}{n}=\frac{71.5}{10} = 7.15\)

Step2: Calculate the squared differences from the mean

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

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

Step3: Calculate the sum of squared differences

Sum of \((x_i - \bar{x})^2\):

\[

$$\begin{align*} &3.4225+0.0225 + 0.4225+0.1225+0.0225+0.7225+4.6225+1.3225+0.1225+0.7225\\ &=(3.4225+0.0225)+(0.4225+0.1225)+(0.0225+0.7225)+(4.6225+1.3225)+(0.1225+0.7225)\\ &=3.445+0.545+0.745+5.945+0.845\\ &=3.445+0.545 = 3.99; 3.99+0.745 = 4.735; 4.735+5.945 = 10.68; 10.68+0.845 = 11.525 \end{align*}$$

\]

Step4: Calculate the variance

The variance \(s^2=\frac{\sum_{i = 1}^{n}(x_i-\bar{x})^2}{n}\) (for population variance, if we consider this as a population of 10 customers) or \(s^2=\frac{\sum_{i = 1}^{n}(x_i-\bar{x})^2}{n - 1}\) (for sample variance). But since the store manager is looking at all the last 10 customers (the entire population in this context), we use \(n = 10\).

\[
s^2=\frac{11.525}{10}=1.1525\approx1.15
\]

Answer:

1.15