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# Measures of Dispersion

Different Measures of Dispersion are 1. Range 2. Mean deviation 3. Quartile deviation 4. Standard deviation 5. Variance 6. Coefficient of Variation

Measures of Dispersion

The following data provide the runs scored by two batsmen in the last 10 matches.

Batsman A: 25, 20, 45, 93, 8, 14, 32, 87, 72, 4

Batsman B: 33, 50, 47, 38, 45, 40, 36, 48, 37, 26

The mean of both datas are same (40), but they differ significantly.

From the above diagrams, we see that runs of batsman B are grouped around the mean. But the runs of batsman A are scattered from 0 to 100, though they both have same mean.

Thus, some additional statistical information may be required to determine how the values are spread in data. For this, we shall discuss Measures of Dispersion.

Dispersion is a measure which gives an idea about the scatteredness of the values.

Measures of Variation (or) Dispersion of a data provide an idea of how observations spread out (or) scattered throughout the data.

Different Measures of Dispersion are

1. Range

2. Mean deviation

3. Quartile deviation

4. Standard deviation

5. Variance

6. Coefficient of Variation

## 1. Range

The difference between the largest value and the smallest value is called Range.

Range R = L – S

Coefficient of range = (LS) / (L + S)

where L - Largest value; S - Smallest value

Example 8.1 Find the range and coefficient of range of the following data: 25, 67, 48, 53, 18, 39, 44.

Solution Largest value L = 67; Smallest value S =18

Range R = LS = 67 −18 = 49

Coefficient of range = (LS) / (L + S)

Coefficient of range = (67 – 18 ) / (67 +18) = 49/85

= 0.576

Example 8.2 Find the range of the following distribution.

Solution  Here Largest value L = 28

Smallest value S = 18

Range R = LS

R = 28 −18 = 10 Years

Example 8.3 The range of a set of data is 13.67 and the largest value is 70.08. Find the smallest value.

Solution

Range R = 13.67

Largest value L = 70.08

Range R = LS

13.67 = 70.08 – S

S = 70.08 −13.67 = 56.41

Therefore, the smallest value is 56.41.

## 2. Deviations from the mean

For a given data with n observations x 1 , x2xn , the deviations from the mean  are

## 3. Squares of deviations from the mean

The squares of deviations from the mean  of the observations x1, x2, . . . . , xn are

Note

We note that (x i    )2  0 for all observations xi , i = 1,2,3,…,n. If the deviations from the mean (x i  −  ) are small, then the squares of the deviations will be very small.

## 4. Variance

The mean of the squares of the deviations from the mean is called Variance.

It is denoted by σ2 (read as sigma square).

Variance = Mean of squares of deviations

## 5. Standard Deviation

The positive square root of Variance is called Standard deviation. That is, standard deviation is the positive square root of the mean of the squares of deviations of the given values from their mean. It is denoted by σ.

Standard deviation gives a clear idea about how far the values are spreading or deviating from the mean.

## 1. Calculation of Standard Deviation for ungrouped data

### (i) Direct Method

Note

·        While computing standard deviation, arranging data in ascending order is not mandatory.

·        If the data values are given directly then to find standard deviation we can use the formula σ =

·        If the data values are not given directly but the squares of the deviations from the mean of each observation is given then to find standard deviation we can use the formula σ =

Example 8.4 The number of televisions sold in each day of a week are 13, 8, 4, 9, 7, 12, 10.

Find its standard deviation.

Solution

### (ii) Mean method

Another convenient way of finding standard deviation is to use the following formula.

Standard deviation (by mean method) σ =

If di = xi –  are the deviations, then

Example 8.5 The amount of rainfall in a particular season for 6 days are given as 17.8 cm, 19.2 cm, 16.3 cm, 12.5 cm, 12.8 cm and 11.4 cm. Find its standard deviation.

Solution Arranging the numbers in ascending order we get, 11.4, 12.5, 12.8, 16.3, 17.8, 19.2.

Number of observations n = 6

### (iii) Assumed Mean method

When the mean value is not an integer (since calculations are very tedious in decimal form) then it is better to use the assumed mean method to find the standard deviation.

Let x 1 , x2, x 3 , ..., xn  be the given data values and let  be their mean.

Let di be the deviation of xi from the assumed mean A, which is usually the middle value or near the middle value of the given data.

di= xi − A gives, xi  = di + A ...(1)

Σdi= Σ(xi −A)

= Σxi −(A + A + A + . . . to n times)

Σdi = ΣxiA × n

Example 8.6 The marks scored by 10 students in a class test are 25, 29, 30, 33, 35, 37, 38, 40, 44, 48. Find the standard deviation.

Solution The mean of marks is 35.9 which is not an integer. Hence we take assumed mean, A = 35, n = 10 .

### (iv) Step deviation method

Let x 1 , x2, x 3 ,...xn  be the given data. Let A be the assumed mean.

Let c be the common divisor of x i - A .

Note

We can use any of the above methods for finding the standard deviation

Example 8.7 The amount that the children have spent for purchasing some eatables in one day trip of a school are 5, 10, 15, 20, 25, 30, 35, 40. Using step deviation method, find the standard deviation of the amount they have spent.

Solution We note that all the observations are divisible by 5. Hence we can use the step deviation method. Let the Assumed mean A = 20, n = 8.

Example 8.8 Find the standard deviation of the following data 7, 4, 8, 10, 11. Add 3 to all the values then find the standard deviation for the new values.

Solution Arranging the values in ascending order we get, 4, 7, 8, 10, 11 and n = 5

When we add 3 to all the values, we get the new values as 7,10,11,13,14.

From the above, we see that the standard deviation will not change when we add some fixed constant to all the values.

Example 8.9 Find the standard deviation of the data 2, 3, 5, 7, 8. Multiply each data by 4. Find the standard deviation of the new values.

Solution  Given, n = 5

When we multiply each data by 4, we get the new values as 8, 12, 20, 28, 32.

From the above, we see that when we multiply each data by 4 the standard deviation also get multiplied by 4.

Example 8.10 Find the mean and variance of the first n natural numbers.

### (i) Mean method

Example 8.11

48 students were asked to write the total number of hours per week they spent on watching television. With this information find the standard deviation of hours spent for watching television.

### (ii) Assumed Mean Medthod

Let x 1 , x2, x 3 , ...xn be the given data with frequencies f1 , f2, f3 , ... fn  respectively. Let x be their mean and A be the assumed mean..

Example 8.12

The marks scored by the students in a slip test are given below.

Find the standard deviation of their marks.

Solution

Let the assumed mean, A = 8

## 2. Calculation of Standard deviation for continuous frequency distribution

### (i) Mean method

Standard deviation   where   xi = Middle value of the i th class.

fi = Frequency of the i th class.

### (ii) Shortcut method (or) Step deviation method

To make the calculation simple, we provide the following formula. Let A be the assumed mean, xi be the middle value of the ith class and c is the width of the class interval.

### Example 8.13

Marks of the students in a particular subject of a class are given below.

Find its standard deviation.

### Solution

Let the assumed mean, A = 35, c = 10

Example 8.14

The mean and standard deviation of 15 observations are found to be 10 and 5 respectively. On rechecking it was found that one of the observation with value 8 was incorrect. Calculate the correct mean and standard deviation if the correct observation value was 23?

Solution

Wrong observation value = 8, Correct observation value = 23.

Correct total = 150 − 8 + 23 = 165

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