Statistics Class 11

In applied mathematics, statistics is a branch that deals with collection, organization and interpretation of data. It is similar to the study of the probability of events occurring based on the collection of data or the known quantities of data. In statistics class 11, gives the importance of statistics in the study of the measures of dispersion and the methods of calculating the grouped and ungrouped data. Some of the concepts and examples involved and scope of statistics class 11 are explained in detail

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 the following measures of dispersion:

  • Range
  • Quartile deviation
  • Mean deviation
  • Standard deviation.

In statistics class 11, you get a clear knowledge of all the measures of dispersion except quartile deviation which will be studied in your higher classes.

Range

The 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 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 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 x1, x2, x3, …..xn

Step 1: Calculate the central tendency measures about to find the mean deviation and let it be ‘a’.

Step 2: Find the deviation of xi from a, i.e., x1 – a, x2– a, x3 – a,. . . , xn– a

Step 3: Find the absolute values of the deviations, i.e., | x1 – a |, | x2– a |, |x3 – a|,. . . ,|xn– 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 = Median

Mean 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 x1, x2, x3,….xn has frequencies f1, f2, f3,….fn respectively. This data is represented in the tabular form as and is called discrete frequency distribution and the data are given below

x x1 x2 x3 …… xn
f f1 f2 f3 …… fn

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 xi with the respective frequencies fi 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 centred 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 𝝈2 and it is given by

\(\sigma^{2} =\frac{1}{n}\sum_{i=1}^{n}(x_{i}-\bar{x})^{2}\)

Statistics Class 11 Example Problems

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 :

xi 4 8 11 17 20 24 32
fi 3 5 9 5 4 3 1

Solution :

Data in tabular form :

xi fi fi xi \(x_{i}-\bar{x}\) \((x_{i}-\bar{x})^{2}\) \(f_{i}(x_{i}-\bar{x})^{2}\)
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 = fi

\(\sum_{i=1}^{7}f_{i}x_{i}=420\) \(\sum_{i=1}^{7}f_{i}(x_{i}-\bar{x})^{2}=1374\)

Therefore, \(\bar{x}=\frac{\sum_{i=1}^{7}f_{i}x_{i}}{N}\)

= 420/30

= 14

\(\bar{x}\)= 14

Variance = \(\sigma^{2} =\frac{1}{n}\sum_{i=1}^{n}(x_{i}-\bar{x})^{2}\) \(\sigma^{2} =\frac{1}{n}\sum_{i=1}^{7}f_{i}(x_{i}-\bar{x})^{2}\)

= 1374/30 = 45.8

Variance = 45.8

Standard Deviation (𝝈) = \(\sqrt{45.8}\) = 6.77

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