IBA, JU WMBA Program 2 (continued)

Course Instructor: Dr Swapan Kumar Dhar

Quartiles, Deciles and PercentilesMedian divides the set of observations into two equal parts so that the number of observations less than median is equal to the number of observations greater than median. Similarly we can divide the set of observations into a fixed number of equal parts. These are called partitions. The values that divide the set of observations into different partitions are called partition values or Quantiles. Some of the important partition values are Quartiles, Deciles and Percentiles. Quartiles: There are three Quartiles, which divide the set of observations into four equal

parts. They are named as first quartile , second quartile and third quartile Of them is

the median and and are also called lower and upper quartiles respectively.

(a) From Ungrouped Data:

Value of the term if N is odd, i = 1, 2, 3.

= Value of observation, if N is even, i= 1, 2, 3.

Example 1: Find the quartiles of the following numbers.

10 72 18 45 32 56 64 27 60

Solution: Arranging the numbers in ascending order of magnitude, we get

10 18 27 32 45 56 60 64 72

Here,

1st quartile = value of 2.5th term .

Value of 4.75 th term = 45. Value of 7th term = 60.

Example 2: Consider the temperature (in Celsius) of several days during a summer season.

33 33 32 32.5 32 32 31.5 31.5 30.8 31 30.8 30.5 30.8 29.5 29 30 Find (i) maximum temperature of the first 25% lowest temperature days (ii) minimum temperature of the last 25% high temperature days (iii) median temperature.Solution: Arranging the numbers in ascending order of magnitude, we get

29 29.5 30 30.5 30.8 30.8 30.8 31 31.5 31.5 32 32 32 32.5 33 33

(i) Maximum temperature of the first 25% lowest temperature days is given by first quartile , where

= Value of observation as n = 16 and it is even.

= Value of observation = = .

(ii) Minimum temperature of the last 25% highest temperature days is given by third quartile , where

= Value of observation as n = 16 and it is even.

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= Value of observation = = .

(iii) Median temperature is

Median = = Value of observation as n = 16 and it is even.

= Value of observation = = .

(b) From Grouped Data or from Frequency Table:

Example 3: Find the quartiles of the following distribution.Height (in inches) 58 59 60 61 62 63 64 65 66Number of Students 15 20 32 35 33 22 20 10 08

Also find the medianSolution:

Height (in inches) Frequency(f) Cumulative frequency58 15 1559 20 3560 32 6761 35 10262 33 13563 22 15764 20 17765 10 18766 8 195

= Size of th item = Size of th item = 49 th item = 60.

= Size of 2 th item = Size of 2 th item = 98 th item = 61.

= Size of 3 th item = Size of 3 th item = 147 th item = 63.

Since = Median, we can write median = 61.

From Frequency distribution with class interval:The formula to calculate quartiles is

.Where

= Lower limit of the quartile class= Total frequency

= Cumulative frequency for the class just above the quartile class

Frequency of the quartile class.

= Class interval of the quartile class. = 1, 2, 3.

The quartile class is that class for which cumulative frequency .

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Example 4: The following distribution represents the monthly salary of a group of industrial workers:

Salary (in Taka) Number of Workers (f) Cumulative Frequency< 1500 18 18

1500 – 1700 42 601700 – 1900 65 1251900 – 2100 150 2752100 – 2300 70 3452300 – 2500 45 390

2500 > 20 410

(i) Find the maximum salary of the first 25% low paid workers.(ii) Find maximum salary of the last 25% high paid workers.(iii) Represent the above two values by an appropriate diagram.(iv) How many workers have income less than 2100 Taka?Solution: (i) The maximum salary of the first 25% low paid workers is calculated by . For calculating 1st

quartile it is necessary to obtain the quartile class. Since =102.5 lies in the class 1700 – 1900

(Cumulative frequency 102.5), the quartile class is 1700 – 1900. So, it is given by

Taka.

(ii) The minimum salary of the last 25% high paid workers is calculated by and it is given by

Taka. Here lies in the class 2100 –

2300, because cumulative frequency of the class is greater than or equal to = 307.5.

(iii) The two values of and can be well represented by Box – and – Whisker plot as given below:

The value of is Taka.

<1500 1700 1900 2100 2300 2500>

Figure: Box – and – Whisker plot to represent the values of quartiles.

(iv) From the cumulative frequency it is observed that 275 workers’ salary is less than 2100 Taka.

Example 5: Marks obtained by 25 students are given below:

Marks obtained 0-10 10-20 20-30 30-40 40-50Number of students 3 4 8 6 4

Find the quartiles of the above distribution.

Solution: Calculations of Quartiles

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Marks Frequency (f) Cumulative frequency0-10 3 3

10-20 4 7

20-30 8 15

30-40 6 21

40-50 4 25

Total N = 25 -

Calculation of

lies between C.F. 3 and C.F. 7 and the corresponding class is 10 - 20.

Calculation of

lies between C.F. 7 and C.F. 15 and their corresponding upper class is 20 - 30.

Calculation of

lies between C.F. 15 and C.F. 21 and the corresponding upper classes are 30 - 40.

Deciles: Deciles divide the set of observations into ten equal parts and there are nine deciles, denoted by

(a) For Ungrouped Data: First arrange the data in ascending or descending order of magnitude.

Value of the term if N is odd, i = 1, 2,…,9.

= Value of observation, if N is even, i= 1, 2, 3,…,9.

Example 6: Repeat the Example 2. (i) Find maximum temperature of the first 10 per cent lowest temperature days (ii) Find the minimum temperature of the last 20% high temperature days.

Solution: (i) Maximum temperature of the first 10% lowest temperature days is given by first decile ,

where

= Value of observation as N = 16 and it is even.

= Value of observation = = .

(ii) Minimum temperature of the last 20% high temperature days is given by third decile , where

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= Value of observation as N = 16 and it is even.

= Value of observation = = .

(b) Grouped frequency distribution:Example 7: The following data represent the distribution of Jack fruit trees by number of Jack fruits in the tree:

No. of Jack fruit 10 15 18 19 20 25 30 32 40 45 50No of trees 5 18 22 36 15 42 12 18 27 08 10Cumulative Frequency 5 23 45 81 96 138 150 168 195 203 213

(i) Find the maximum number of jack fruits in first 30% lower producing trees.(ii) Find the minimum number of jack fruits in last 30% high producing trees.

Solution: (i) Maximum number of jack fruits in first 30% lower producing trees is obtained from , where

= Value of th observation = Value of 64.2 th observation = 19 [C.F. > 64.2 or 65 is 81]

(ii) Minimum number of jack fruits in last 30% high producing trees is calculated by , where

= Value of th observation = Value of 149.8 th observation = 30 [C.F. > 149.8 is 150].

From Frequency distribution with class interval:The formula to calculate deciles is

.Where

= Lower limit of the decile class= Total frequency

= Cumulative frequency for the class just above the decile class

Frequency of the decile class.

= Class interval of the decile class. = 1, 2, …,9.

The decile class is that class for which cumulative frequency .

Percentiles: Percentiles divide the set of observations into 100 equal parts and there are 99 percentiles,

denoted by

(a) Ungrouped Data:

Value of the term if N is odd, i = 1, 2, …,99.

= Value of observation, if N is even, i= 1, 2, …,99.

(b) Grouped Data without Class Intervals:

value of the term

value of the term

. . . . . .. . .

value of the term

(b) Grouped frequency distribution with class interval:

The formula to calculate percentiles is

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.Where

= Lower limit of the percentile class= Total frequency

= Cumulative frequency for the class just above the percentile class

Frequency of the percentile class.

= Class interval of the percentile class. = 1, 2, …,99.

The decile class is that class for which cumulative frequency .

Example 8: The following are the marks obtained by 50 students in Statistics:

Marks Number of students10 marks and less 220 marks and less 730 marks and less 1740 marks and less 2950 marks and less 3860 marks and less 4570 marks and less 50

If 54% of the students pass the test, find the minimum pass marks.

Solution: 54% students pass the test, i.e. 46% students fail in the test. In other words 46% of the student

get less than pass marks, which is the 46th percentile. So, we have to calculate 46th percentile

=

Calculation of Percentile

Marks Frequency Cumulative frequency00 - 10 2 210 - 20 5 720 - 30 10 1730 - 40 12 2940 - 50 09 3850 - 60 07 4560 - 70 05 50

To find the percentile class, we have to calculate , which is 23. So blocked row is the percentile

class, because 23 lies in this class. According to the formula given,

= = = 35.

Hence, 35 is the pass marks.

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