In applied mathematics, statistics is a branch that deals with collection, organization and interpretation of data. It is similar to the study of probability of events occurring based on collection of data or the known quantities of data. In statistics class 11, gives the importance of studying the measures of dispersion and the methods of calculating the grouped and ungrouped data. Some of the concepts and examples involved in statistics class 11 are given for reference.
Measures of Dispersion
The dispersion or scatter in the data is measured on the basis of the observations and the
types of the measure of central tendency. There are following measures of dispersion:
- Range
- Quartile deviation
- Mean deviation
- Standard deviation.
In Statistics class 11, you get a clear knowledge on all the measures of dispersion except quartile deviation which will be studied in your higher classes.
Range
Range is the difference between the maximum value and the minimum value
Range = Maximum Value – Minimum Value
Mean Deviation
Mean deviation is the basic measure of deviations from a value and the value is generally a mean value or a median value. In order to find out the mean deviation, first take the mean of deviation for the observations from a value is d = x – a Here x is the observation and a is the fixed value.
The basic formula to find out the mean deviation is
Mean Deviation = Sum of absolute values of deviations from ‘a’ / Number of observations
Mean Deviation for Ungrouped Data
Calculation of mean deviation for ungrouped data involves the following steps :
Let us assume the observations x_{1}, x_{2}, x_{3}, …..x_{n}
Step 1 : Calculate the central tendency measures about to find the mean deviation and let it be ‘a’.
Step 2 : Find the deviation of x_{i }from a, i.e., x_{1} – a, x_{2}– a, x_{3} – a,. . . , x_{n}– a
Step 3 : Find the absolute values of the deviations, i.e., | x_{1} – a |, | x_{2}– a |, |x_{3} – a|,. . . ,|x_{n}– a| and the drop the minus sign (–), if it is there,
Step 4 : calculate the mean of the absolute values of the deviations . This mean obtained is the mean deviation about a, i.e.,
\(M.D (a)= \frac{\sum_{i=1}^{n}\left |x_{i}-a \right |}{n}\) \(M.D (\bar{x})= \frac{\sum_{i=1}^{n}\left |x_{i}-\bar{x} \right |}{n}\), where \(\bar{x}\) = Mean \(M.D (M)= \frac{\sum_{i=1}^{n}\left |x_{i}-M \right |}{n}\), Where M = MedianMean Deviation for Grouped Data
The data can be grouped into two ways namely,
- Discrete frequency distribution
- Continuous frequency distribution
The methods of finding the mean deviation for both the types are given below
Discrete Frequency Distribution
Here, the given data consist of n distinct values x_{1}, x_{2}, x_{3},….x_{n }has frequencies f_{1}, f_{2}, f_{3},….f_{n }respectively. This data is represented in the tabular form as and is called discrete frequency distribution and the data are given below
x |
x_{1} |
x_{2} |
x_{3} |
…… |
x_{n} |
f |
f_{1} |
f_{2} |
f_{3} |
…… |
f_{n} |
Mean Deviation About Mean
First find the mean \(\bar{x}\), using the given formula :
\(\bar{x}=\frac{\sum_{i=1}^{n}x_{i}f_{i}}{\sum_{i=1}^{n}f_{i}}=\frac{1}{N}\sum_{i=1}^{n}x_{i}f_{i}\)Where the numerator denotes the sum of products of observations x_{i} with the respective frequencies f_{i} and the denominator denotes the sum of frequencies.
Now take the absolute values, \(\left | x_{i}-\bar{x} \right |\) , for all i = 1, 2, 3, ..n
Therefore, the required mean deviation about the mean is given by
\(M.D(\bar{x})=\frac{\sum_{i=1}^{n}f_{i}\left | x_{i}-\bar{x} \right |}{\sum_{i=1}^{n}f_{i}}=\frac{1}{N}\sum_{i=1}^{n}f_{i}\left | x_{i}-\bar{x} \right |\)Mean Deviation About Median
To find the mean deviation about median for the given discrete frequency distribution. First arrange the observation in ascending order to get cumulative frequency which is equal or greater than N/2, where N is the sum of the frequencies.
Therefore, the mean deviation for median is given by,
\(M.D(M)= \frac{1}{N}\sum_{i=1}^{n}f_{i}\left | x_{i}-M \right |\)Continuous Frequency Distribution
To find the mean deviation for the continuous frequency distribution, assume that the frequency in each class is centered at its midpoint. After finding the midpoint, proceed further to find the mean deviation similar to the discrete frequency distribution.
Standard Deviation
The positive square root of variance is called standard deviation (S.D) and it is denoted by the symbol, 𝝈 and the formula to find S.D is given by
\(\sigma =\sqrt{\frac{1}{n}\sum_{i=1}^{n}(x_{i}-\bar{x})^{2}}\)Where the variance is denoted as 𝝈^{2} and it is given by
\(\sigma^{2} =\frac{1}{n}\sum_{i=1}^{n}(x_{i}-\bar{x})^{2}\)Sample Example
Here, statistics class 11 solved problem is given below. Go through once to get a clear idea to solve the problem.
Question :
Obtain 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 :
Data in tabular form :
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 8 11 17 20 24 32 |
3 5 9 5 4 3 1 |
12 40 99 85 80 72 32 |
-10 -6 -3 3 6 10 18 |
100 38 9 9 36 100 324 |
300 180 81 45 144 300 324 |
30 |
420 |
1374 |
Here, N = 30 = f_{i}
\(\sum_{i=1}^{7}f_{i}x_{i}=420\) \(\sum_{i=1}^{7}f_{i}(x_{i}-\bar{x})^{2}=1374\)Therefore, \(\bar{x}=\frac{\sum_{i=1}^{7}f_{i}x_{i}}{N}\)
= 420/30
= 14
\(\bar{x}\)= 14Variance = \(\sigma^{2} =\frac{1}{n}\sum_{i=1}^{n}(x_{i}-\bar{x})^{2}\) \(\sigma^{2} =\frac{1}{n}\sum_{i=1}^{7}f_{i}(x_{i}-\bar{x})^{2}\)
= 1374/30 = 45.8
Variance = 45.8
Standard Deviation (𝝈) = \(\sqrt{45.8}\) = 6.77
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