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Mean Median Mode Formula:
Nowadays, you must be aware of the use of data through statistics almost everywhere. Take an example of cricket match. While watching the match you see several graphical representations of different types of data such as sixes per over or wickets in an inning of different matches or many more. There is use of mean median mode formula. It has a wide range of applications.
Measures of dispersion:
Based on the observations and the types of the measure of central tendency the dispersion or scatter in a data is measured. Thus we calculate mean median mode. There are following measures of dispersion:
- Range
- Quartile deviation (Excluded)
- Mean deviation
- Standard deviation.
S. No. | Data form | Formula |
1 | Range | Maximum Value – Minimum Value |
2 | Mean Deviation for Ungrouped Data | M.D.(\(\bar{x}\) = \(\frac{\sum{f_{i}(x_{i} – \bar{x})}}{N}\) |
3 | Mean deviation for Grouped Data | M.D.(\(\bar{x}\) = \(\frac{\sum{(x_{i} – \bar{x})}}{n}\), where N = \(\sum{f_{i}}\) |
4 | Median | If n is odd, then M = \((\frac{n+1}{2})^{th}\), term
If n is even, then M = \(\frac{(\frac{n}{2})^{th}term+(\frac{n}{2}+1)^{th}term}{2}\) |
5 | Mode | The value which occurs most frequently. |
6 | Variance | \(\sigma ^{2}\) = \(\frac{\sum (x- \bar{x})^{2}}{n}\) |
7 | Standard Deviation | \(S = \sigma = \sqrt{\frac{\sum (x- \bar{x})^{2}}{n}}\) |
8 | Coefficient of variation (C.V.) | \(\frac{σ}{x}\) x 100, \(\bar{x}\) ≠ 0 |
Where,
x= number of items present
n = Total number of items
\(\bar{x}\) = Mean
Short trick to find Variance and standard deviation:
σ ^{2} = \(\frac{h^{2}}{N^{2}}\) [N∑f_{i}y_{i}^{2} – (∑f_{i}y_{i})^{2} ],
σ = \(\frac{h}{N}\) \(\sqrt{ N∑f_{i}{y_{i}}^{2} – (∑f_{i}y_{i})^{2}}\)
Mean Median Mode Examples:
- Find the mean deviation about the median for the following data: 3, 9, 5, 3, 12, 10, 18, 4, 7, 19, 21.
Solution:
as we see there is 11 obsevations and it is absolutely odd. First let’s arrange them in ascending order, we get 3, 3, 4, 5, 7, 9, 10, 12, 18, 19, 21.
Median = \((\frac{11 + 1}{2})^{th}\)
or 6th observation = 9
The absolute values of the deviations from the median respectively, i.e., |x_{i} – M| are 6, 6, 5, 4, 2, 0, 1, 3, 9, 10, 12
\(\sum_{i=1}^{11}\) |x_{i} – M| = 58
Thus,
D. (M) = \(\frac{1}{11}\) = \(\sum_{i=1}^{11}\) |x_{i} – M| = \(\frac{1}{11}\) x 58 = 5.27 - Find the variance and standard deviation for the following data
x_{i} | 4 | 8 | 11 | 17 | 20 | 24 | 32 |
f_{i} | 3 | 5 | 9 | 5 | 4 | 3 | 1 |
Solution:
In tabular form it is:
x_{i} | f_{i} | f_{i}x_{i} | x_{i} – \(\bar{x}\) | (x_{i} – \(\bar{x}\) )^{2} | F_{i}(x_{i} – \(\bar{x}\) )^{2} |
4 | 3 | 12 | -10 | 100 | 300 |
8 | 5 | 40 | -6 | 36 | 180 |
11 | 9 | 99 | -3 | 9 | 81 |
17 | 5 | 85 | 3 | 9 | 45 |
20 | 4 | 80 | 6 | 36 | 144 |
24 | 3 | 72 | 10 | 100 | 300 |
32 | 1 | 32 | 18 | 324 | 324 |
30 | 420 | 1374 |
N = 30,
\(\sum_{i=1}^{7}\) f_{i}x_{i} = 420,
\(\sum_{i=1}^{7}\)f_{i}(x_{i} – \(\bar{x}\) )^{2} = 1374
so, \(\bar{x} = \frac{\sum_{i=1}^{7} f_{i}x_{i}}{N}\)
= \(\frac{1}{30}\) x 1374 = 45.8
Standard deviation (σ ) = \(\sqrt{45.8}\) = 6.77
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