1. Measures of Dispesion Dr. Amjad El-Shanti MD, PMH,Dr PH University of Palestine 2016

2. Measures of Dispersion After the central tendency of a frequency distribution is determined. The next step is to determine how spread out (Dispersed) the numbers are from each others or from the average. This can be done by calculating measures of dispersion. They are also called measures of spread or variation. Standard Measures of Dispersion Range Variance Standard Deviation Coefficient of variation

3. 1. Range It is the simplest measure of dispersion Definition: It is the difference between the highest and lowest values. Advantages: It is quick and easy to calculate. Disadvantages: - It does not use directly the majority of the observations. - It is very sensitive to extreme values. Calculation: A)Ungrouped Data: Largest observation- Smallest observation B)Grouped Data: Upper limit of last interval – Lowest limit of first interval Example (1): The following data represent the height of 10 persons: 20 – 60 – 53 – 80 – 89 – 56 – 42 – 46 – 88 – 95 cm -Find the range Answer: Largest observation = 95, Smallest observation = 20 The range= 95- 20= 75 cm

4. 1. Range Example (2): The following data represent the age of 50 persons: -Compute the range Answer: Upper limit of last interval = 65 Lower limit of first interval= 15 The range= 65- 15= 50 years Age (Years)No. of persons (Frequency) 15-4 25-8 35-26 45-8 55-644 Total50

5. 2. Variance (S 2 ) It is the average of the squared deviation of observations- from the mean. Formula based on definition: ∑(x i -x) 2 S 2 = n-1 Computation Formula: ∑x 2 - (∑x) 2 S 2 = n n-1 Where: ∑x 2 = Sum of squared observations (∑x) 2 = Square of sum observations N = No. of observations It is the average of the squared deviation of observations- from the mean. Formula based on definition: ∑(x i -x) 2 S 2 = n-1 Computation Formula: ∑x 2 - (∑x) 2 S 2 = n n-1 Where: ∑x 2 = Sum of squared observations (∑x) 2 = Square of sum observations N = No. of observations

6. 3. Standard Deviation (S) Definition: It is a measure of the spread of data around their mean It is the positive square root of the variance  The value of standard deviation and variance are always positive. Advantages: -It is the preferred measure of dispersion. - It uses all of the measurements in the set. Disadvantages: - It is influenced by a few (or even only one) extreme values. - It is not calculated in open table. Calculation: A)Ungrouped Data Steps for calculation: 1- Determine the sum of observations (∑x i ) 2- Find (∑x i ) 2 3- Find the sure of each observation x i 2 4- Find the sum of the squared observation (∑x i 2 ) ∑x i 2 - (∑x i ) 2 S = n √ n-1

7. 3. Standard Deviation (S) Example: The following data represent the weight of 5 persons. 14 – 15 – 16 – 17 – 18 kg - Find the standard deviation Answer: (∑x i )= 14+15+16+17+18= 80 (∑x i ) 2 = (80) 2 = 6400 (x i 2 ) = 196+225+256+289+324 = 1290 1290- 6400 S = 5 = 1.6 kg √ 5-1

8. 3. Standard Deviation (S) Calculation: B) Grouped Data Steps for calculation: 1-Determine the mid-point of each interval x j 2- Find the product f j x j for each interval, then find the sum of these product∑ f j x j 3- Find the product f j x j 2 by multiplying the mid point x j by the corresponding product f j x j 4- Find the sum of the squared observation ∑f j x j 2 ∑f j x j 2 - (∑f j x j ) 2 S = ∑ f j √ ∑ f j -1

9. 3. Standard Deviation (S) Example(1): The following table represents the weight of 20 persons. - Compute the standard deviation Weight (kg)Frequency 15-3 25-6 35-8 45-2 55-641 Total20

10. 3. Standard Deviation (S) Answer(1): 28000 - (720) 2 S = 20 = 10.5 kg √ 20-1 Weight (kg)fjfj XjXj F j X j F j X j 2 15-320601200 25-6301805400 35-84032012800 45-2501005000 55-64160 3600 Total ∑ f j= 20 720= ∑f j x j 28000=∑f j x j 2

11. 3. Standard Deviation (S) Example(2): The following table represents the number of attacks of diarrhea among 30 children. - Compute the standard deviation Attacks of diarrheaFrequency 1-8 4-7 7-10 9-125 Total30

12. 3. Standard Deviation (S) Answer(2): 1320.75 - (178.5) 2 S = 30 = 2.98 = 3 attacks √ 30-1 Attacks of diarrheafjfj XjXj F j X j F j X j 2 1-821632 4-7535175 7-107.575562.5 9-12510.552.5551.25 Total ∑ f j= 30 178.5=∑ f j x j 1320.75=∑f j x j 2

13. 4. Semi-inter-quartile-range(S.I.Q.R) Definition: It equal to half distance between third and first quartiles.. Advantages: -It can be calculated for closed or open –ended tables. - It is the preferable measure of variation when we want to avoid outlying observations. Disadvantages: - It does not use directly the majority of the observations. Calculation: A)Ungrouped Data Steps for calculation: 1- Arrange Data 2- Find Q1 (first quartile) and Q3 (third quartile) 3- Rank Q1= n+1/4, Rank Q3= 3( n+1)/4 4- Compute S.I.Q.R given by: Q3 - Q1 S.I.Q.R= 2

14. 4. Semi-inter-quartile-range(S.I.Q.R) Example : From the following set of observations compute semi-inter-quartile range 15 – 34 – 7 – 12 – 18 -9 – 1 – 42 – 56 – 28 – 13 – 24 – 35 Answer : Arrange Data 1 -7 – 9 – 12 – 13 – 15 – 18 – 24 – 28 – 34 – 35 – 42 – 56 Q1 rank = (13+1)/ 4 = 14/ 4 = 3.5 Q1= 9+12= 10.5 Q3 rank= 3(13+1)/4= 42/4= 10.5 Q3= 34+35= 34.5 34.5 - 10.5 S.I.Q.R= = 12 2

15. 4. Semi-inter-quartile-range(S.I.Q.R) Calculation: B) Grouped Data Steps for calculation: 1- compute the general rank of Q1= n/4 and general rank of Q3 = 3n/4 2- Determine the Q1 by comparing the general rank of Q1 with CF values from above downwards and the first interval met with whose CF is equal to or greater than the general rank of the Q1 is the Q1 interval. 3- Determine the Q3 by same way of Q1 but for Q3. 4- Determine the special rank of Q1 and Q3 given by: special rank = general rank – CF of pre Q interval 5- Determine the value of Q1 and Q3 given by: Q i = LL of Q i interval + special rank X width of Q i interval frequency of Q i interval 6- S.I.Q.R = Q3 - Q1 2

16. 4. Semi-inter-quartile-range(S.I.Q.R) B) Grouped Data Example: The following table represents data about age of 40 persons: - Compute the semi-inter-quartile range? Age (Years)Frequency 15-3 25-5 35-8 45-12 55-10 65-742 Total40

17. 4. Semi-inter-quartile-range(S.I.Q.R) B) Grouped Data Answer: Find the cumulative frequency Age (Years)FrequencyCF 15-33 25-58 35-816 45-1228 55-1038 65-74240 Total40

18. 4. Semi-inter-quartile-range(S.I.Q.R) B) Grouped Data Answer: Q1: General Rank= n/4= 40/4= 10 Q1 Interval= 35-45 Special rank= 10- 8= 2 Q1= 35+ 2X 10 = 37.5 years 8 Q3: General Rank= 3n/4= 120/4= 30 Q1 Interval= 55-65 Special rank= 30-2 8= 2 Q1= 55+ 2X 10 = 57 years 10 S.I.Q.R= 57 - 37.5 = 9.75 years 2

19. 5. Coefficient of Variation(CV) Characteristics: It is a measure of relative variation. It is independent of the units used. Therefore can be used for comparing sets of measurements of different units.  Suppose we want to compare dispersion of two sets of data, use the standard deviation for comparison, it may lead to fallacious results. It may be that two variables involved are measured in different units for example, we may wish to know. For example, we may wish to know, for a certain population, whether serum cholesterol levels (measured in milligram per 100 ml) are more variable than body weight (measured in pounds). The coefficient variation is a measure that is independent of the units of measurement and it is the solution for such comparison. It is expressed as percentage.

20. 5. Coefficient of Variation(CV) Calculation Ungrouped Data or Closed- ended tablesOpen or Closed-ended tables CV= S X 100 X C V = SIQR X 100 Median The standard deviation is expressed as percentage of mean. The semi-inter-quartile range is expressed as a percentage of the median.

21. 5. Coefficient of Variation(CV) Example(1): The following are the age of 5 children: 5 – 3 – 4 – 7 – 6 - Find the coefficient of variation Answer (1): X = 5+3+4+7+6 =25 5 (∑ X i ) 2 = (25) 2 = 625 (∑ X i 2 )= 25+9+16+49+36= 135 135 - 625 S= 5 = 1.58 years √ 5 - 1 CV= (1.58/5) X100 = 31.6%

22. 5. Coefficient of Variation(CV) Example(2): The following table represents the data about age of 80 person: - Find the coefficient of variation. Age (Years)Frequency 5-5 15-10 25-20 35-25 45-13 55+7 Total80

23. 5. Coefficient of Variation(CV) Answer(2): Age (Years)FrequencyCF 5-55 15-1015 25-2035 35-2560 45-1373 55+780 Total80 Q1Q2 = MedianQ3 General rank= n/4= 80/4= 20 n/2= 80/2=40=3n/4= 3(80)/4= 60 Interval25-3535-45 Special rank= 20-15=5=40-35=5=60-35=25 Value25+ 5X10 20 =27.5 years 35+ 5X10 25 =37 years 35+ 25X10 25 = 45 years

24. 5. Coefficient of Variation(CV) Answer(2): S. I. Q. R= 45 – 27.5 = 8.75 2 CV= 8.75 X 100 = 23.6% 37