## Short Note in English.

Now we understood that median of a distribution is that value which divides the distribution in to two equal parts. We can also find three values which divide the distribution in to four equal parts. These three values are called the quartiles. Likewise we can find nine values which divide the distribution in to ten equal parts. These nine values are called deciles. In a similar manner we can define percentiles. Percentiles are ninety nine values which divide the distribution in to hundred equal parts.

## QUARTILES

Quartiles are those values which divide the series into four equal parts. There are only three quartiles. We denote them by Q1, Q2, and Q3. Q1 is called the lower quartile, which is the value such that (1/4)th of the total observations fall below it and (3/4)th above it. Q2 is the same as median. Q3 is called the upper quartile, which is the value such that (3/4)th of the total observations fall below it and (1/4)th above it. The method of determining quartiles is same as that of median.

### INDIVIDUAL SERIES

STEPS

1. Arrange the data in ascending order
2. $${\mathbf Q_1 = \Biggl[\frac {{N} + {1}}{4}}\Biggl]^{th}$$ item
3. $${\mathbf Q_1 = \Biggl[\frac {3({N} + {1})}{4}}\Biggl]^{th}$$ item
DATA OF ODD NUMBERS

• Find the first quartile Q1, the median (Q2) and the third quartile Q3 of the following series.

• 12, 6, 21, 15, 9, 20, 16, 13, 8, 7, 9, 11, 5, 8, 10

Let us arrange the data in ascending order

5, 6, 7, 8, 8, 9, 9, 10, 11, 12, 13, 15, 16, 20, 21

Here N = 15

$${\mathbf Q_1 = \Biggl[\frac {{N} + {1}}{4}}\Biggl]^{th}$$ item

$${ = value\, of\, the\, \Biggl[\frac {{15} + {1}}{4}}\Biggl]^{th}$$ item

= value of the 4th item = 8.

$${\mathbf Q_2 = \Biggl[\frac {{N} + {1}}{2}}\Biggl]^{th}$$ item

$${ = value\, of\, the\, \Biggl[\frac {{15} + {1}}{2}}\Biggl]^{th}$$ item

= value of the 8th item = 10.

$${\mathbf Q_3 = \Biggl[\frac {3({N} + {1})}{4}}\Biggl]^{th}$$ item

$${ = value\, of\, the\, \Biggl[\frac {({15} + {1})}{4}}\Biggl]^{th}$$ item

= value of the 12th item = 15.

DATA OF EVEN NUMBERS

• Find the first quartile Q1, the median (Q2) and the third quartile Q3 of the following series.

• 166, 167, 171, 160, 133, 175, 156, 158, 175, 142, 120, 133

Let us arrange the data in ascending order

120, 133, 133, 142, 156, 158, 175, 166, 160, 167, 171, 175, 175

$${\mathbf Q_1 = \Biggl[\frac {{N} + {1}}{4}}\Biggl]^{th}$$ item

$${ = value\, of\, the\, \Biggl[\frac {{12} + {1}}{4}}\Biggl]^{th}$$ item

= value of the 3.25th item

= value of the 3rd item + 0.25 (4th item - 3rd item)

= 133 + 0.25 (142 - 133)

= 133 + 0.25(9)

= 135.25.

$${\mathbf Q_2 = \Biggl[\frac {{N} + {1}}{2}}\Biggl]^{th}$$ item

$${ = value\, of\, the\, \Biggl[\frac {{12} + {1}}{2}}\Biggl]^{th}$$ item

= value of the 6.5th item

= value of the 6th item + 0.5 (7th item - 6rd item)

= 158 + 0.5 (166 - 158)

= 158 + 0.5(8)

= 162.

$${\mathbf Q_3 = \Biggl[\frac {3({N} + {1})}{4}}\Biggl]^{th}$$ item

$${ = value\, of\, the\, \Biggl[\frac {({12} + {1})}{4}}\Biggl]^{th}$$ item

= value of the 9.75th item

= value of the 9th item + 0.75 (10th item - 9th item)

= 167 + 0.75 (171 - 167)

= 167 + 0.75(4)

= 170.

### DISCRETE SERIES

STEPS

1. Arrange the data in ascending order
2. Find the cumulative frequencies
3. $$\mathbf {Q_1 = \Biggl[\frac {{N} + {1}}{4}}\Biggl]^{th}$$ item
4. $$\mathbf {Q_3 = \Biggl[\frac {3({N} + {1})}{4}}\Biggl]^{th}$$ item
• Let us find the first quartile Q1, the median (Q2) and the third quartile Q3 of the following series.

• Values Frequencies
5 8
8 13
12 20
15 22
17 11
20 6

Here N = 80,

$$\mathbf {Q_1 = \Biggl[\frac {{N} + {1}}{4}}\Biggl]^{th}$$ = 20.25

$$\mathbf {Q_2 = \Biggl[\frac {{N} + {1}}{2}}\Biggl]^{th}$$ = 40.5

$$\mathbf {Q_3 = \Biggl[\frac {3({N} + {1})}{4}}\Biggl]^{th}$$ = 60.75

Now we can create a table showing cumulative frequencies

Values Frequencies cf
5 8 8
8 13 21 (Q1 lies here)
12 20 41 (Q2 lies here)
15 22 63 (Q3 lies here)
17 11 74
20 6 80
N = 80

$$\mathbf {Q_1 = \Biggl[\frac {{N} + {1}}{4}}\Biggl]^{th}$$ item

= value of the 21st item = 8

$$\mathbf {Q_2 = \Biggl[\frac {{N} + {1}}{2}}\Biggl]^{th}$$ item

= value of the 41st item = 12

$$\mathbf {Q_3 = \Biggl[\frac {3({N} + {1})}{4}}\Biggl]^{th}$$ item

= value of the 61st item = 15

### CONTINUOUS SERIES

STEPS

1. Find the cumulative frequency
2. Find Q1 and Q3 classes
3. $${ \mathbf Q_1 =\, size\, of\,\frac {{N}}{4}}^{th}$$ item
4. $${ \mathbf Q_3 =\, size\, of\,\frac {3{N}}{4}}^{th}$$ item

After locating these classes, the values of Q1 and Q3 can be interpolated.

$$\mathbf Q_1 \,= \,{ L + \frac{\frac{N}{4} - {cf}}{f} × h}$$

Where,

L = the lower limit of the lower quartile class

f = freqwuency of the lower quartile class

h = magnitude of the lower quartile class

cf = the cumulative frequency of the class preceding the lower quartile class $$\mathbf Q_3 \,= \,{ L + \frac{\frac{3N}{4} - {cf}}{f} × h}$$

Where,

L = the lower limit of the upper quartile class

f = freqwuency of the upper quartile class

h = magnitude of the upper quartile class

cf = the cumulative frequency of the class preceding the upper quartile class

• Let us find the first quartile Q1, the median (Q2) and the third quartile Q3 of the following series.

• Age Frequencies
0-5 2
5-10 5
10-15 9
15-20 13
20-25 18
25-30 24
30-35 19
35-40 17
40-45 10
45-50 3

We can create a table with cumulative frequency to find Q1, Q2, and Q3.

Age Frequencies cf
0-5 2 2
5-10 5 7
10-15 9 16
15-20 13 29
20-25 18 47
25-30 24 71
30-35 19 90
35-40 17 107
40-45 10 117
45-50 3 120
N = 120

N = 120, $${ \frac {{N}}{4}}$$ = 30

Lower Quartile Class = 20 - 25

L = 20, $${\frac {{N}}{4}}$$ = 30, cf = 29, f = 18, h = 5

$$\mathbf Q_1 \,= \,{ L + \frac{\frac{N}{4} - {cf}}{f} × h}$$

= $${ 20 + \frac{{30} - {29}}{18} × 5}$$

= 20.28

N = 120, $${ \frac {{N}}{2}}$$ = 60

Median Class = 25 - 30

L = 25, $${\frac {{N}}{2}}$$ = 60, cf = 47, f = 24, h = 5

$$\mathbf Q_2 \,= \,{ L + \frac{\frac{N}{2} - {cf}}{f} × h}$$

= $${ 25 + \frac{{60} - {47}}{24} × 5}$$

= 27.7

N = 120, $${ \frac {3{N}}{4}}$$ = 90

Upper Quartile Class = 30 - 35

L = 30, $${\frac {3{N}}{4}}$$ = 90, cf = 71, f = 19, h = 5

$$\mathbf Q_3 \,= \,{ L + \frac{\frac{3{N}}{4} - {cf}}{f} × h}$$

= $${ 30 + \frac{{90} - {71}}{19} × 5}$$

= 35