If the total number of observations is 20, ∑ x i= 1000 and \(\sum {\rm{x}}_{\rm{i}}^2 = 84000\) , then what is the variance of the distribution?
1700
The question asks us to calculate the variance of a distribution given the total number of observations, the sum of the observations, and the sum of the squared observations. Variance is a key measure of dispersion or spread in a dataset. It tells us how much the individual data points deviate from the mean of the dataset.
There are different formulas for variance depending on whether you are calculating the population variance (\(\sigma^2\)) or the sample variance (\(s^2\)). When the question refers to "the distribution" without specifying a sample, the population variance formula is often used. The formula that is most convenient to use with the given data (\(\sum x_i\) and \(\sum x_i^2\)) is:
Population Variance (\(\sigma^2\)) = \(\frac{\sum x_i^2}{n} - \left(\frac{\sum x_i}{n}\right)^2\)
This formula requires us to first calculate the mean (\(\bar{x}\)), which is \(\frac{\sum x_i}{n}\).
Let's calculate the variance using the given values.
The mean is the sum of observations divided by the number of observations.
\(\bar{x} = \frac{\sum x_i}{n} = \frac{1000}{20}\)
\(\bar{x} = 50\)
Now, plug the values of n, \(\sum x_i\), and \(\sum x_i^2\) into the population variance formula:
\(\sigma^2 = \frac{\sum x_i^2}{n} - \left(\frac{\sum x_i}{n}\right)^2\)
\(\sigma^2 = \frac{84000}{20} - \left(\frac{1000}{20}\right)^2\)
\(\sigma^2 = 4200 - (50)^2\)
\(\sigma^2 = 4200 - 2500\)
\(\sigma^2 = 1700\)
The calculated variance of the distribution is 1700.
| Parameter | Value |
|---|---|
| Number of observations (n) | 20 |
| Sum of x (\(\sum x_i\)) | 1000 |
| Sum of x squared (\(\sum x_i^2\)) | 84000 |
| Mean (\(\bar{x}\)) | 50 |
| Variance (\(\sigma^2\)) | 1700 |
Based on the calculations, the variance of the distribution is 1700.
| Concept | Definition | Formula (Population) |
|---|---|---|
| Mean (\(\mu\) or \(\bar{x}\)) | The average of all data points. | \(\frac{\sum x_i}{n}\) |
| Variance (\(\sigma^2\)) | Average of the squared differences from the Mean. Measures data spread. | \(\frac{\sum (x_i - \mu)^2}{n}\) or \(\frac{\sum x_i^2}{n} - \mu^2\) |
| Standard Deviation (\(\sigma\)) | The square root of the variance. Measures typical distance from the Mean. | \(\sqrt{\sigma^2}\) |
| Sum of Squares (\(\sum x_i^2\)) | Sum of the squares of each observation. | Given in this problem |
Variance is a fundamental concept in statistics. A higher variance indicates that the data points are more spread out from the mean, while a lower variance indicates that they are clustered closer to the mean.
When dealing with a sample of data (not the entire population), the sample variance (\(s^2\)) is calculated using a slightly different formula, dividing by \(n-1\) instead of \(n\):
\(s^2 = \frac{\sum (x_i - \bar{x})^2}{n-1}\)
or
\(s^2 = \frac{\sum x_i^2 - (\sum x_i)^2/n}{n-1}\)
The division by \(n-1\) is known as Bessel's correction and is used to provide an unbiased estimate of the population variance when only a sample is available. However, in this specific problem, the phrasing "variance of the distribution" given the complete sum and sum of squares for 20 observations typically implies using the population variance formula.
Variance is used in many statistical methods, such as ANOVA (Analysis of Variance), regression analysis, and hypothesis testing. Understanding how to calculate and interpret variance is crucial for analyzing data effectively.
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