1) Variance measures how spread out a particular data set is. It does this by measuring the average distance each data point is from the true mean. Variance has a formula. It is the following:
\(\sigma^2=\frac{\sum_{i=1}^{N}(X_{i}-\mu)^2}{N}\)
Yes, this formula looks very complicated, but it is first important to define a few variables.
\(N = \text{sample size}\\ \mu = \text{mean of population}\\ X_i=\text{term of data}\\ \sigma^2=\text{variance}\)
What is this formula really trying to tell you to do? Well, here is an elementary breakdown:
1. Find the mean of the data.
2. Subtract each individual element from the mean
3. Square all results from the previous step and find the sum of the squared difference.
4. Divide by the sample size.
Let's get to it!
1) Find the mean of the data.
In order to find the mean, we must first find the sum of all the data points.
\(x_i\) | |
\(x_1\) | 182 |
\(x_2\) | 274 |
\(x_3\) | 207 |
\(x_4\) | 216 |
\(x_5\) | 285 |
\(x_6\) | 190 |
\(x_7\) | 138 |
\(x_8\) | 240 |
\(x_9\) | 291 |
\(x_{10}\) | 288 |
\(x_{11}\) | 176 |
\(x_{12}\) | 186 |
\(x_{13}\) | 190 |
\(x_{14}\) | 213 |
\(x_{15}\) | 236 |
\(x_{16}\) | 242 |
\(x_{17}\) | 186 |
\(x_{18}\) | 259 |
\(x_{19}\) | 164 |
\(x_{20}\) | 172 |
\(x_{21}\) | 233 |
\(x_{22}\) | 268 |
\(x_{23}\) | 152 |
\(x_{24}\) | 274 |
\(\Sigma\) | 5262 |
If the sum of all the data points is 5262, then the mean is this number divided by the number of data points, 24.
\(\mu=\frac{5262}{24}=219.25\)
2) Subtract each individual element from the mean and square that result. I will replicate the table above.
\(x_i\) | \(\mu\) | \(x_i-\mu\) | |
\(x_1\) | 182 | -37.25 | |
\(x_2\) | 274 | 54.75 | |
\(x_3\) | 207 | -12.25 | |
\(x_4\) | 216 | -3.25 | |
\(x_5\) | 285 | 65.75 | |
\(x_6\) | 190 | -29.25 | |
\(x_7\) | 138 | -81.25 | |
\(x_8\) | 240 | 20.75 | |
\(x_9\) | 291 | 71.75 | |
\(x_{10}\) | 288 | 68.75 | |
\(x_{11}\) | 176 | -43.25 | |
\(x_{12}\) | 186 | -33.25 | |
\(x_{13}\) | 190 | -29.25 | |
\(x_{14}\) | 213 | -6.25 | |
\(x_{15}\) | 236 | 16.75 | |
\(x_{16}\) | 242 | 22.75 | |
\(x_{17}\) | 186 | -33.25 | |
\(x_{18}\) | 259 | 39.75 | |
\(x_{19}\) | 164 | -55.25 | |
\(x_{20}\) | 172 | -47.25 | |
\(x_{21}\) | 233 | 13.75 | |
\(x_{22}\) | 268 | 48.75 | |
\(x_{23}\) | 152 | -67.25 | |
\(x_{24}\) | 274 | 54.75 | |
\(\Sigma\) | 5262 | 219.25 |
3) Square all results from the previous step and find the sum of the squared difference.
\(x_i\) | \(\mu\) | \(x_i-\mu\) | \((x_i-\mu)^2\) | |
\(x_1\) | 182 | -37.25 | 1387.5625 | |
\(x_2\) | 274 | 54.75 | 2997.5625 | |
\(x_3\) | 207 | -12.25 | 150.0625 | |
\(x_4\) | 216 | -3.25 | 10.5625 | |
\(x_5\) | 285 | 65.75 | 4323.0625 | |
\(x_6\) | 190 | -29.25 | 855.5625 | |
\(x_7\) | 138 | -81.25 | 6601.5625 | |
\(x_8\) | 240 | 20.75 | 430.5625 | |
\(x_9\) | 291 | 71.75 | 5148.0625 | |
\(x_{10}\) | 288 | 68.75 | 4726.5625 | |
\(x_{11}\) | 176 | -43.25 | 1870.5625 | |
\(x_{12}\) | 186 | -33.25 | 1105.5625 | |
\(x_{13}\) | 190 | -29.25 | 855.5625 | |
\(x_{14}\) | 213 | -6.25 | 39.0625 | |
\(x_{15}\) | 236 | 16.75 | 280.5625 | |
\(x_{16}\) | 242 | 22.75 | 517.5625 | |
\(x_{17}\) | 186 | -33.25 | 1105.5625 | |
\(x_{18}\) | 259 | 39.75 | 1580.0625 | |
\(x_{19}\) | 164 | -55.25 | 3052.5625 | |
\(x_{20}\) | 172 | -47.25 | 2232.5625 | |
\(x_{21}\) | 233 | 13.75 | 189.0625 | |
\(x_{22}\) | 268 | 48.75 | 2376.5625 | |
\(x_{23}\) | 152 | -67.25 | 4522.5625 | |
\(x_{24}\) | 274 | 54.75 | 2997.5625 | |
\(\Sigma\) | 5262 | 219.25 | 49356.5 |
4) Divide by the sample size
\(x_i\) | \(\mu\) | \(x_i-\mu\) | \((x_i-\mu)^2\) | \(\frac{(x_i-\mu)^2}{N}\) | |
\(x_1\) | 182 | -37.25 | 1387.5625 | ||
\(x_2\) | 274 | 54.75 | 2997.5625 | ||
\(x_3\) | 207 | -12.25 | 150.0625 | ||
\(x_4\) | 216 | -3.25 | 10.5625 | ||
\(x_5\) | 285 | 65.75 | 4323.0625 | ||
\(x_6\) | 190 | -29.25 | 855.5625 | ||
\(x_7\) | 138 | -81.25 | 6601.5625 | ||
\(x_8\) | 240 | 20.75 | 430.5625 | ||
\(x_9\) | 291 | 71.75 | 5148.0625 | ||
\(x_{10}\) | 288 | 68.75 | 4726.5625 | ||
\(x_{11}\) | 176 | -43.25 | 1870.5625 | ||
\(x_{12}\) | 186 | -33.25 | 1105.5625 | ||
\(x_{13}\) | 190 | -29.25 | 855.5625 | ||
\(x_{14}\) | 213 | -6.25 | 39.0625 | ||
\(x_{15}\) | 236 | 16.75 | 280.5625 | ||
\(x_{16}\) | 242 | 22.75 | 517.5625 | ||
\(x_{17}\) | 186 | -33.25 | 1105.5625 | ||
\(x_{18}\) | 259 | 39.75 | 1580.0625 | ||
\(x_{19}\) | 164 | -55.25 | 3052.5625 | ||
\(x_{20}\) | 172 | -47.25 | 2232.5625 | ||
\(x_{21}\) | 233 | 13.75 | 189.0625 | ||
\(x_{22}\) | 268 | 48.75 | 2376.5625 | ||
\(x_{23}\) | 152 | -67.25 | 4522.5625 | ||
\(x_{24}\) | 274 | 54.75 | 2997.5625 | ||
\(\Sigma\) | 5262 | 219.25 | 0 | 49356.5 | \(2056.5208\overline{3}\) |
I realize that this answer differs from Cphill's. I have reviewed my work many times, and I do not believe that I have done anything incorrect.
2) The standard deviation, thankfully, does not require this much computation. Take the square root of the previous result, \(2056.5208\overline{3}\).
\(\sigma=\sqrt{2056.5208\overline{3}}\approx45.3489\)
I realize that this is a lot of information to take in at once. If you have any questions, then do not hesitate to ask.