For the following sample:
2 2 4 1 3 2 1 2
Calculate the range and the standard deviation.
Given is a sample.
No. | Sample |
1 | 2 |
2 | 2 |
3 | 4 |
4 | 1 |
5 | 3 |
6 | 2 |
7 | 1 |
8 | 2 |
Now here we need to determine the range and standard deviation of the given sample.
Before we go on to solve the problem let us know a bit about range and standard deviation.
Range
Range is the simplest measure of dispersion. Range gives us a first hand idea about the spread of the data. It is given by,
Sample Standard Deviation
Sample Standard deviation is a commonly used measure of dispersion. It is calculated using,
where n is the number of observations.
Coming back to our problem,
First we calculate the range of the data,
Now we calculate the sample standard deviation.
The table of calculations is provided below.
No. | Sample(xi) | (xi-x̄)^2 |
1 | 2 | 0.0156 |
2 | 2 | 0.0156 |
3 | 4 | 3.5156 |
4 | 1 | 1.2656 |
5 | 3 | 0.7656 |
6 | 2 | 0.0156 |
7 | 1 | 1.2656 |
8 | 2 | 0.0156 |
Total | 17 | 6.8748 |
Hence,
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