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

First, sum all the values and divide by the number of data points ($n=10$):
$\text{Mean} = \frac{9 + 7 + 6.5 + 7.5 + 7 + 8 + 5 + 6 + 7.5 + 8}{10} = \frac{71.5}{10} = 7.15$

Step2: Find squared deviations

Subtract the mean from each value, then square the result:
$(9-7.15)^2=3.4225$, $(7-7.15)^2=0.0225$, $(6.5-7.15)^2=0.4225$, $(7.5-7.15)^2=0.1225$, $(7-7.15)^2=0.0225$,
$(8-7.15)^2=0.7225$, $(5-7.15)^2=4.6225$, $(6-7.15)^2=1.3225$, $(7.5-7.15)^2=0.1225$, $(8-7.15)^2=0.7225$

Step3: Calculate sample variance

Sum the squared deviations and divide by $n-1$ (since this is sample data for past customer purchases):
$\text{Variance} = \frac{3.4225+0.0225+0.4225+0.1225+0.0225+0.7225+4.6225+1.3225+0.1225+0.7225}{9}$
$\text{Variance} = \frac{11.505}{9} \approx 1.278$
Note: If using population variance (divide by $n$), $\frac{11.505}{10}=1.1505$, which matches the option. Since the question does not specify sample/population, and 1.15 is an option, we use population variance here.

Answer:

1.15