If the sum of 10 observations is 12 and the sum of their squares is 18, then the standard deviation is:
3/5
Calculating Standard Deviation from Sum and Sum of Squares
The question asks us to find the standard deviation of 10 observations given the sum of the observations and the sum of their squares.
We are given:
- Number of observations, \(n = 10\)
- Sum of observations, \(\Sigma x = 12\)
- Sum of squares of observations, \(\Sigma x^2 = 18\)
The formula for the standard deviation (\(\sigma\)) is:
\(\sigma = \sqrt{\frac{\Sigma x^2}{n} - \left(\frac{\Sigma x}{n}\right)^2}\)
Alternatively, using the mean (\(\bar{x} = \frac{\Sigma x}{n}\)), the formula is:
\(\sigma = \sqrt{\frac{\Sigma x^2}{n} - (\bar{x})^2}\)
Step-by-Step Calculation of Standard Deviation
Let's calculate the required terms:
1. Calculate the Mean (\(\bar{x}\)):
The mean is the sum of observations divided by the number of observations.
\(\bar{x} = \frac{\Sigma x}{n} = \frac{12}{10} = 1.2\)
2. Calculate the Mean of Squares (\(\frac{\Sigma x^2}{n}\)):
This is the sum of squares divided by the number of observations.
\(\frac{\Sigma x^2}{n} = \frac{18}{10} = 1.8\)
3. Calculate the Square of the Mean \((\bar{x})^2\):
Square the value of the mean calculated in step 1.
\((\bar{x})^2 = (1.2)^2 = 1.44\)
4. Substitute Values into the Standard Deviation Formula:
Now, plug the values from steps 2 and 3 into the standard deviation formula:
\(\sigma = \sqrt{\frac{\Sigma x^2}{n} - (\bar{x})^2}\)
\(\sigma = \sqrt{1.8 - 1.44}\)
\(\sigma = \sqrt{0.36}\)
5. Calculate the Square Root:
Find the square root of 0.36.
\(\sigma = 0.6\)
6. Express the Standard Deviation as a Fraction:
The decimal 0.6 can be written as a fraction:
\(0.6 = \frac{6}{10} = \frac{3}{5}\)
So, the standard deviation is \(\frac{3}{5}\).
Let's summarize the values:
| Parameter | Value |
|---|---|
| Number of observations (n) | 10 |
| Sum of observations (\(\Sigma x\)) | 12 |
| Sum of squares (\(\Sigma x^2\)) | 18 |
| Mean (\(\bar{x}\)) | 1.2 |
| Mean of squares (\(\frac{\Sigma x^2}{n}\)) | 1.8 |
| Square of Mean \((\bar{x})^2\) | 1.44 |
| Variance (\(\sigma^2\)) | 0.36 |
| Standard Deviation (\(\sigma\)) | 0.6 or \(\frac{3}{5}\) |
The calculated standard deviation is \(\frac{3}{5}\), which corresponds to option 3.
Revision Table: Standard Deviation and Variance
| Measure | Definition | Formula (Population) | Formula (Sample) |
|---|---|---|---|
| Variance (\(\sigma^2\) or \(s^2\)) | Average of the squared differences from the Mean. | \(\sigma^2 = \frac{\Sigma (x_i - \mu)^2}{N}\) or \(\frac{\Sigma x^2}{N} - (\mu)^2\) | \(s^2 = \frac{\Sigma (x_i - \bar{x})^2}{n-1}\) or \(\frac{\Sigma x^2 - n(\bar{x})^2}{n-1}\) |
| Standard Deviation (\(\sigma\) or \(s\)) | Square root of the Variance. Indicates the spread of data points around the Mean. | \(\sigma = \sqrt{\frac{\Sigma x^2}{N} - (\mu)^2}\) | \(s = \sqrt{\frac{\Sigma x^2 - n(\bar{x})^2}{n-1}}\) |
In this problem, we used the population standard deviation formula as we were given the sum of all observations and their squares directly, implying this data represents the entire set being considered, not a sample from a larger population, or the question doesn't specify it's a sample, so we use the simpler population formula.
Additional Information on Measures of Dispersion
Standard deviation is a key measure of dispersion, indicating how spread out the numbers in a data set are from their average value (the mean). A low standard deviation means most of the numbers are close to the average, while a high standard deviation means the numbers are more spread out.
- Dispersion: This refers to the extent to which a distribution is stretched or squeezed. Measures of dispersion include range, interquartile range, variance, and standard deviation.
- Variance: It is the average of the squared differences from the mean. Variance gives a measure of the spread but is in squared units, making it less intuitive than standard deviation.
- Standard Deviation: By taking the square root of the variance, the standard deviation is brought back to the original units of the data, making it easier to interpret the spread relative to the mean.
Understanding standard deviation is crucial in statistics for analysing data variability, comparing data sets, and in various statistical tests and models.
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