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MORE ON CALCULATING THE VARIANCE


In previous posts, you were introduced to the concept of variance. Now, we are learning more about it.

Population variance
Formula 1: $\sigma^2 = \frac{\sum_{i=1}^{n}  (X_{i} - \bar{X})^2}{n}$
Formula 2: $\sigma^2 = \frac{\sum_{i=1}^{n} {X_{i}}^2}{n} - (\bar{X})^2$
The two formulae give the same result.

Example 1
Mr. Ahmad had 6 cell phone counters. In July 2016, the net profits obtained from each counter were: 4, 7, 5, 3, 5, 6 (in millions rupiahs). What was the variance of the net profit of the six counters in July 2016?

Answer
In this example, we are asked to calculate the variance of the net profits of the six counters owned by Mr. Ahmad in July 2016. The given data are the net profit data of all the counters. So, they are the data of all population members. To solve this problem, firstly calculate the mean.
$\bar{X} = Rp \frac{4+7+5+3+5+6}{6} million = Rp \: 5 \:  million$
Using Formula 1, calculate the variance.
$\sigma^2 = \frac{(4-5)^2+(7-5)^2+(5-5)^2+(3-5)^2+(5-5)^2+(6-5)^2}{6} (million \: Rupiahs)^2$
$\sigma^2 \approx 1.67 (million \: Rupiahs)^2$
Alternatively, we may use Formula 2 as follows.
$\sigma^2 = (\frac{4^2+7^2+5^2+3^2+5^2+6^2}{6} - 5^2) (million \: Rupiahs)^2$
$\sigma^2 \approx (26.67 - 25) (million \: Rupiahs)^2 = 1.67 (million \:  Rupiahs)^2$

Sample variance
Formula 3: $s^2 = \frac{\sum_{i=1}^{n} (X_{i} - \bar{X})^2}{n-1}$
Formula 4: $s^2 = \frac{\sum_{i=1}^{n} {X_{i}}^2}{n-1} - \frac{n}{n-1} (\bar{X})^2$

Example 2
Mr. Ahmad had hundreds of cell phone counters. He wanted to know the magnitude of variability of the net profits from all the counters he had in July 2016. To get the answer fast, he got the net profit data from 6 counters as samples, and obtained the following results: 4, 7, 5, 3, 5, 6 (in millions Rupiahs). What was variance of these data?

Answer
In this example, the six data are not data of all counters, but only some parts of them. Therefore, the variance that will be obtained is the sample variance.
As shown in Example 1, $\bar{X} = Rp \frac{4+7+5+3+5+6}{6} million = Rp \: 5 \:  million$. The use of Formula 3 will result in:
$s^2 = \frac{(4-5)^2+(7-5)^2+(5-5)^2+(3-5)^2+(5-5)^2+(6-5)^2}{6-1} (million \: Rupiahs)^2$
$s^2 = 2 (million \: Rupiahs)^2$
Alternatively, we may use Formula 4 as follows.
$s^2 = (\frac{4^2+7^2+5^2+3^2+5^2+6^2}{6-1} - \frac{6}{6-1} \cdot 5^2) (million \: Rupiahs)^2$
$s^2 = (32 - 30) (million \: Rupiahs)^2 = 2 (million \: Rupiahs)^2$


EXERCISE PART I
For the following data sets, calculate the population variance using Formula 1 and Formula 2. Compare the results.
Data set A: 15, 11, 20, 27, 17
Data set B: 142, 150, 146, 145, 153, 127, 145
Data set C: 0.53, 0.92, 0.47, 0.65, 0.71, 0.44, 0.54, 0.70

EXERCISE PART II
Use the data sets in Exercise Part I, but assume that they are sample data. Use Formula 3 and 4. Compare the results.

EXERCISE PART III
  1. The Quality Control Division of a company consists of a Division Head and four staffs. The following are data on the number of total working hours of the Division members in August 2016: 205, 195, 217, 243, 190. Determine the variance of the Division members’ total working hours in August 2016.
  2. The daily net profits for the first 7 days in August 2018, in thousands Rupiahs, were as follows: 214, 230, 195, 204, 184, 210, 156. Calculate the variance of daily net profits in the first week of August 2018.
  3. Mr. Andi had a small coffee shop. He would like to estimate the variance of daily net profits achieved in August 2018 by calculating the sample variance of daily net profits. He selected at random 10 days from which the net profits data were obtained. He got the following data,  in thousands Rupiahs: 230, 195, 204, 184, 210, 156, 214, 177, 160, 180. Which variance was appropriate in this case? Calculate it.




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