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.

Answer 1

We first calculate the average of the ten data points. 9 + 7 + 6.5 + 7.5 + 7 + 8 + 5 + 6 + 7.5 + 8 = 71.5, and 71.5 / 10 = 7.15.
Then we find the sum of squares of deviations from the mean:
(9 - 7.15)^2 + (7 - 7.15)^2 + (6.5 - 7.15)^2 + (7.5 - 7.15)^2 + (7 - 7.15)^2 + (8 - 7.15)^2 + (5 - 7.15)^2 + (6 - 7.15)^2 + (7.5 - 7.15)^2 + (8 - 7.15)^2 = 11.525
Then we divide by the number of terms = 10, so the variance of the sample will be 11.525 / 10 = 1.15.

Answer 2

Answer:

The variance of the data set is 1.28, approximately.

Step-by-step explanation:

The variance is defined by

$\sigma^{2} =\frac{\sum (x-\mu)^{2} }{N-1}$

Where $\mu$ is the mean and $N$ is the total number of rings.

First, we find the mean

$\mu = \frac{9+7+6.5+7.5+7+8+5+6+7.5+8}{10}=\frac{71.5}{10}=7.15$

Then, we subtract the means with each element

$9-7.15=1.85\\7-7.15=-0.15\\6.5-7.15=-0.65\\7.5-7.15=0.35\\7-7.15=-0.15\\8-7.15=0.85\\5-7.15=-2.15\\6-7.15=-1.15\\7.5-7.15=0.35\\8-7.15=0.85$

Now, we elevate each difference to the square power

$(1.85)^{2} =3.4225\\(-0.15)^{2} =0.0225\\(-0.65)^{2} =0.4225\\(0.35)^{2} =0.1225\\(-0.15)^{2} =0.0225\\(0.85)^{2} =0.7225\\(-2.15)^{2} =4.6225\\(-1.15)^{2} =1.3225\\(0.35)^{2} =0.1225\\(0.85)^{2} =0.7225$

Then, we sum all these results

$\sum (x-\mu)^{2}= 11.525$

Now, we replace in the formula

$\sigma^{2} =\frac{\sum (x-\mu)^{2} }{N-1}=\frac{11.525}{10-1} \approx 1.28$

Therefore, the variance of the data set is 1.28, approximately.