Problems Based on Measures of Dispersion

What does Dispersion mean?

Dispersion refers to the distribution of a particular data set. The point to which a given set of data is likely to differ around a mean value is referred to as statistical dispersion. In other words, dispersion aids in the comprehension of data distribution.

Measures of Dispersion

Measures of dispersion are used in statistics to interpret data variability. It indicates whether the variable is undistributed or distributed.

In statistics, there are two primary dispersion methods:

• Absolute Measure of  Dispersion
• Relative Measure of Dispersion

Absolute Measure of Dispersion 

The original unit as the data set present is used in an absolute measure of dispersion. The absolute dispersion approach expresses changes as the average of observed deviations, such as standard or means deviations. It includes terms such as range, standard deviation, and quartile deviation.

The following are examples of absolute metrics of dispersion:

Range: A data set’s range is just the difference between the two extreme values in the set. Range = 7 -1= 6 in the following example: 1, 3,5, 6, 7.

Variance: The variance is calculated by calculating the mean and later subtracting it from each value given in the set, then squaring the values, adding each square, and finally dividing by the total number of values in the data set. variance =𝝨 ( x – µ )²/N

Standard Deviation: The standard deviation is defined as the square root of the variance, i.e., S.D. = √(𝝨 ( x – µ )²/N)

Quartile Deviation: The quartiles are values that divide a list of integers into quarters. Half of the distance between the third and first quartiles is the quartile deviation.

Mean and Mean Deviation: The mean is simply the average of the data set provided. The mean deviation is the arithmetic mean of the absolute difference of the values from a measure of central tendency.

Problems based on Various Measures of Dispersion

• Question: Find the Variance and Standard Deviation of the Following Numbers:  3, 5, 6, 10, 9, 10, 1, 3

Solution: 

The mean = (3 + 5 + 6 + 10 + 9 + 10 + 1 + 3)/8 = 46/ 8 = 5.75

Step 1: Subtract the average from each individual data point

 (3 – 5.75), (5 – 5.75), (6 – 5.75), (9 – 5.75), (9 – 5.75), (10 – 5.75)(1 – 5.75), (3 – 5.75),

= -2.75, -0.75, 0.25, 3.25, 3.25, 4.25, -4.75, -2.75

Step 2: Squaring the above values we get, 22.563, 7.563, 7.563, 0.563, 0.063, 10.563, 10.563, 18.063

Step 3: 7.563 + 0.563 + 0.063 + 10.563 + 10.563 + 18.063 + 22.563 + 7.563 

= 77.504

Step 4: n = 8, therefore variance (σ2) = 77.504/ 8 = 9.688

Thus, Standard deviation (σ) = 3.112

• Find the quartiles and quartile deviation of the following data:

17, 2, 7, 27, 15, 15, 14, 8, 10, 24, 48, 10, 18, 7, 18, 28

Solution:

Given data:

17, 2, 7, 27, 15, 15, 14, 10, 24, 48, 10, 18, 7, 18, 28, 8

The given data in ascending order is:

2, 7, 7, 8, 10, 10, 14, 15, 15, 17, 18, 18, 24, 27, 28, 48

Number of observations(n) = 16

 the upper half of the observations is:

  15, 17, 18, 18, 24, 27, 28, 48 (number of observations=8(even))

Q3 =Median of the upper half of theobservations

= (1/2)[4th observation + 5th observation]

= (18 + 24)/2

= 42/2

= 21

Now, the lower half of the observations is:

2, 7, 7, 8, 10, 10, 14,15 (number of observations=8(even))

Q1 = Median of the lower half of the observations

= (1/2)[4th observation + 5th observation]

= (10 + 8)/2

= 18/2

= 9

Q2 = Median of the given data set

As n is even, median = (1/2) [(n/2)th observation+ (n/2 + 1)th observation]

= (1/2)[8th observation + 9th observation]

= (15 + 15)/2

= 30/2

= 15

Q3 = 15

Quartile deviation = (Q1 – Q2)/2

= (21 – 9)/2

= 12/2

= 6

Hence, the quartile deviation for the given data set is 6.

Coefficient of Dispersion

When two series with significantly differing averages are compared, the coefficients of dispersion (along with the measure of dispersion) are determined. When two series with different measurement units are compared, the dispersion coefficient is also employed. C.D. is the abbreviation for it.

The following are some of the most prevalent dispersion coefficients:

Solved examples on Coefficient of Dispersion

• The variance of the data 2, 4, 5, 6, 8, 17 is 23.33. 

Then the variance of 6, 12, 15, 18, 24, 51 will be 

(A)23.23  (B)25.33  (C)46.66  (D)69.99

Solution:  (D) is the correct answer. When each observation is multiplied by 3, then the variance is also multiplied by 3.

•  If the arithmetic means of 14 observations 26, 12, 14, 15, x, 17, 9, 11, 18, 16, 28, 20, 22, 8 is 17. Find the missing observation.

Solution:

Given 14 observations are: 26, 12, 14, 15, x, 17, 9, 15, 18, 16, 20, 20, 22, 8

Arithmetic mean = 17

We know that,

Arithmetic mean = Sum of observations/Total number of observations

18 = (212 + x)/14

18 x 14 = 212 + x

212 + x = 252

x = 252 – 212

x = 40

Hence, 40 is the missing observation.

• A set of n values x1, x2, …, xn has a standard deviation of 6.

 The standard deviation of n values x1  + k, x2  + k, …, xn  + k will be 

(A)σ  (B)σ + k  (C)σ – k  (D)kσ 

Solution  

(A) is the correct answer. Irrespective of the increase in the value of the constant, the standard deviation will not be affected.

Conclusion

The Measures of Dispersion is an important topic in Statistics. Several problems based on these measures may be asked in exams. It is used in statistics to interpret data variability, i.e. to determine how homogeneous or heterogeneous the data is. In simple words, it indicates whether the variable is undistributed or distributed. Examples of measures of dispersion include range, variance, standard deviation, mean, mean deviation, quartile, and quartile deviation. Regularly practicing the problems based on measures of dispersion may help you score optimum marks in your entrance exams.