What is the standard deviation of the observations \(-\sqrt{6}, -\sqrt{5},- \sqrt{4}, -1, 1, \sqrt{4}, \sqrt{5}, \sqrt{6} \ ?\)
2
Calculating Standard Deviation for Observations
The question asks us to find the standard deviation of a given set of observations. The observations are: \(-\sqrt{6}, -\sqrt{5},- \sqrt{4}, -1, 1, \sqrt{4}, \sqrt{5}, \sqrt{6}\).
First, let's simplify the observations:
\(-\sqrt{6}, -\sqrt{5},-2, -1, 1, 2, \sqrt{5}, \sqrt{6}\)
The number of observations, \(N\), is 8.
Steps to Calculate Standard Deviation
To find the standard deviation, we follow these steps:
- Calculate the mean (\(\mu\)) of the observations.
- Calculate the variance (\(\sigma^2\)), which is the average of the squared differences from the mean.
- Calculate the standard deviation (\(\sigma\)), which is the square root of the variance.
Step 1: Calculate the Mean
The mean (\(\mu\)) is the sum of all observations divided by the number of observations.
Sum of observations = \((-\sqrt{6}) + (-\sqrt{5}) + (-2) + (-1) + 1 + 2 + \sqrt{5} + \sqrt{6}\)
Sum of observations = \((-\sqrt{6} + \sqrt{6}) + (-\sqrt{5} + \sqrt{5}) + (-2 + 2) + (-1 + 1)\)
Sum of observations = \(0 + 0 + 0 + 0 = 0\)
Mean (\(\mu\)) = \(\frac{\text{Sum of observations}}{N} = \frac{0}{8} = 0\)
The mean of the given observations is 0.
Step 2: Calculate the Variance
The variance (\(\sigma^2\)) is the average of the squared differences from the mean. Since the mean is 0, the variance is the average of the squared observations.
Let's find the square of each observation:
- \((-\sqrt{6})^2 = 6\)
- \((-\sqrt{5})^2 = 5\)
- \((-2)^2 = 4\)
- \((-1)^2 = 1\)
- \(1^2 = 1\)
- \(2^2 = 4\)
- \((\sqrt{5})^2 = 5\)
- \((\sqrt{6})^2 = 6\)
Sum of squared observations = \(6 + 5 + 4 + 1 + 1 + 4 + 5 + 6 = 32\)
Variance (\(\sigma^2\)) = \(\frac{\text{Sum of squared observations}}{N} = \frac{32}{8} = 4\)
The variance of the observations is 4.
Step 3: Calculate the Standard Deviation
The standard deviation (\(\sigma\)) is the square root of the variance.
Standard Deviation (\(\sigma\)) = \(\sqrt{\text{Variance}} = \sqrt{4} = 2\)
The standard deviation of the given observations is 2.
Let's present the observations and their squared values in a table:
| Observation (\(x\)) | \(x^2\) |
|---|---|
| \(-\sqrt{6}\) | 6 |
| \(-\sqrt{5}\) | 5 |
| -2 | 4 |
| -1 | 1 |
| 1 | 1 |
| 2 | 4 |
| \(\sqrt{5}\) | 5 |
| \(\sqrt{6}\) | 6 |
| Sum: 0 | Sum: 32 |
Mean = \(0/8 = 0\)
Variance = \(32/8 = 4\)
Standard Deviation = \(\sqrt{4} = 2\)
The standard deviation of the observations is 2.
Revision Table: Standard Deviation Calculation
| Concept | Formula | Calculation (for this problem) |
|---|---|---|
| Mean (\(\mu\)) | \(\frac{\Sigma x}{N}\) | \(\frac{0}{8} = 0\) |
| Variance (\(\sigma^2\)) | \(\frac{\Sigma (x - \mu)^2}{N}\) or \(\frac{\Sigma x^2}{N} - \mu^2\) | \(\frac{32}{8} - 0^2 = 4\) |
| Standard Deviation (\(\sigma\)) | \(\sqrt{\sigma^2}\) | \(\sqrt{4} = 2\) |
Additional Information: Understanding Standard Deviation
Standard deviation is a measure of the amount of variation or dispersion of a set of values. A low standard deviation indicates that the values tend to be close to the mean (also called the expected value) of the set, while a high standard deviation indicates that the values are spread out over a wider range.
- It is the square root of the variance.
- It is widely used in statistics and probability theory.
- It is denoted by the Greek letter sigma (\(\sigma\)) for the population standard deviation or by \(s\) for the sample standard deviation. In this case, we treated the given values as the entire population, so we used \(\sigma\).
- Standard deviation is in the same units as the data, making it easier to interpret than variance.
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