# Long Questions

Find the quartile deviation and its coefficient from the following data:
 No. of students Marks obtained 60-75 65-70 70-75 75-80 80-85 85-90 7 5 8 4 3 3
Here,
Tabulating given data for calculation of Quartile deviation (QD),
Marks obtained Number of students (f) Cumulative frequency (c.f.)
60-75 7 7
65-70 5 12
70-75 8 20
75-80 4 24
80-85 3 27
85-90 3 30
N = 30

Position of first quartile = $(\frac N4 )^{th}$ item = $(\frac{30}{4})^{th}$ item = 7.5th item

In c.f. column, 12 is just greater than 7.5 so its corresponding class is 65-70.
∴ L = 65, f = 5, c.f. = 7, i = 5

Now, we know, Q1 = L + $\frac{\frac N4 , - , c.f.}{f}$ × i or, Q1 = 65 + $\frac{7.5 , - , 7}{5}$ × 5
∴ Q1 = 65.5

Position of third quartile = $(\frac{3N}{4} )^{th}$ item = $(\frac{3 \, \times \, 30}{4})^{th}$ item = 22.5th item

In c.f. column, 24 is just greater than 22.5 so its corresponding class is 75-80.
∴ L = 75, f = 4, c.f. = 20, i = 5

Now, we know, Q3 = L + $\frac{\frac{3N}{4} , - , c.f.}{f}$ × i or, Q3 = 75 + $\frac{22.5 , - , 20}{5}$ × 5
∴ Q3 = 78.125

Finally,

QD = $\frac{Q_3 \, - \, Q_1}{2}$ = $\frac{78.125 \, - \, 65.5}{2}$ = 6.3132

Coefficient of quartile deviation = $\frac{Q_3 \, - \, Q_1}{Q_3 \, + \, Q_1}$ = $\frac{12.625}{143.625}$ = 0.09
Find the quartile deviation and its coefficient from the following data:
 No. of person Weight (kg) 20-30 30-40 40-50 50-60 60-70 70-80 5 15 10 8 6 2
Here,
Tabulating given data for calculation of Quartile deviation (QD),
Weight Number of person (f) Cumulative frequency (c.f.)
20-30 5 5
30-40 15 20
40-50 10 30
50-60 8 38
60-70 6 44
70-80 2 46
N = 46

Position of first quartile = $(\frac N4 )^{th}$ item = $(\frac{46}{4})^{th}$ item = 11.5th item

In c.f. column, 20 is just greater than 11.5 so its corresponding class is 30-40.
∴ L = 30, f = 15, c.f. = 5, i = 10

Now, we know, Q1 = L + $\frac{\frac N4 , - , c.f.}{f}$ × i or, Q1 = 30 + $\frac{11.5 , - , 5}{15}$ × 10
∴ Q1 = 34.33

Position of third quartile = $(\frac{3N}{4} )^{th}$ item = $(\frac{3 \, \times \, 46}{4})^{th}$ item = 26.25th item

In c.f. column, 38 is just greater than 34.5 so its corresponding class is 50-60.
∴ L = 50, f = 8, c.f. = 30, i = 10

Now, we know, Q3 = L + $\frac{\frac{3N}{4} , - , c.f.}{f}$ × i or, Q3 = 50 + $\frac{34.5 , - , 30}{8}$ × 10
∴ Q3 = 55.63

Finally,

QD = $\frac{Q_3 \, - \, Q_1}{2}$ = $\frac{55.63 \, - \, 34.33}{2}$ = 10.65 kg

Coefficient of quartile deviation = $\frac{Q_3 \, - \, Q_1}{Q_3 \, + \, Q_1}$ = $\frac{21.3}{89.96}$ = 0.24
Find the quartile deviation and its coefficient from the following data:
 Frequency Class interval 20-30 30-40 40-50 50-60 60-70 8 16 4 4 3
Here,
Tabulating given data for calculation of Quartile deviation (QD),
Class interval Frequency (f) Cumulative frequency (c.f.)
20-30 8 8
30-40 16 24
40-50 4 28
50-60 4 32
60-70 3 35
N = 35

Position of first quartile = $(\frac N4 )^{th}$ item = $(\frac{35}{4})^{th}$ item = 8.75th item

In c.f. column, 24 is just greater than 8.75 so its corresponding class is 30-40.
∴ L = 30, f = 16, c.f. = 8, i = 10

Now, we know, Q1 = L + $\frac{\frac N4 , - , c.f.}{f}$ × i or, Q1 = 30 + $\frac{8.75 , - , 8}{16}$ × 10
∴ Q1 = 30.47

Position of third quartile = $(\frac{3N}{4} )^{th}$ item = $(\frac{3 \, \times \, 35}{4})^{th}$ item = 26.25th item

In c.f. column, 28 is just greater than 26.25 so its corresponding class is 40-50.
∴ L = 40, f = 4, c.f. = 24, i = 10

Now, we know, Q3 = L + $\frac{\frac{3N}{4} , - , c.f.}{f}$ × i or, Q3 = 40 + $\frac{26.25 , - , 24}{4}$ × 10
∴ Q3 = 45.63

Finally,

QD = $\frac{Q_3 \, - \, Q_1}{2}$ = $\frac{45.63 \, - \, 30.47}{2}$ = 7.58

Coefficient of quartile deviation = $\frac{Q_3 \, - \, Q_1}{Q_3 \, + \, Q_1}$ = $\frac{15.16}{76.1}$ = 0.19
Find the quartile deviation and its coefficient from the following data:
 Frequency Size 4-8 8-12 12-16 16-20 20-24 24-28 28-32 32-36 36-40 6 10 18 30 15 12 10 6 5
Here,
Tabulating given data for calculation of Quartile deviation (QD),
Size Frequency (f) Cumulative frequency (c.f.)
4-8 6 6
8-12 10 16
12-16 18 34
16-20 30 64
20-24 15 79
24-28 12 91
28-32 10 101
32-36 6 107
36-40 5 112
N = 112

Position of first quartile = $(\frac N4 )^{th}$ item = $(\frac{112}{4})^{th}$ item = 28th item

In c.f. column, 34 is just greater than 28 so its corresponding class is 12-16.
∴ L = 12, f = 18, c.f. = 16, i = 4

Now, we know, Q1 = L + $\frac{\frac N4 , - , c.f.}{f}$ × i or, Q1 = 12 + $\frac{28 , - , 16}{18}$ × 4
∴ Q1 = 14.67

Position of third quartile = $(\frac{3N}{4} )^{th}$ item = $(\frac{3 \, \times \, 112}{4})^{th}$ item = 84th item

In c.f. column, 91 is just greater than 84 so its corresponding class is 24-28.
∴ L = 24, f = 12, c.f. = 79, i = 4

Now, we know, Q3 = L + $\frac{\frac{3N}{4} , - , c.f.}{f}$ × i or, Q3 = 24 + $\frac{84 , - , 79}{12}$ × 4
∴ Q3 = 25.67

Finally,

QD = $\frac{Q_3 \, - \, Q_1}{2}$ = $\frac{45.67 \, - \, 14.67}{2}$ = 5.5

Coefficient of quartile deviation = $\frac{Q_3 \, - \, Q_1}{Q_3 \, + \, Q_1}$ = $\frac{11}{40.34}$ = 0.273
Find the quartile deviation and its coefficient from the following data:
 No. of students Height (in inches) 60-62 62-64 64-66 66-68 68-70 70-72 4 6 8 12 7 2
Here,
Tabulating given data for calculation of Quartile deviation (QD),
Height Number of students (f) Cumulative frequency (c.f.)
60-62 4 4
62-64 6 10
64-66 8 18
66-68 12 30
68-70 7 37
70-72 2 39
N = 39

Position of first quartile = $(\frac N4 )^{th}$ item = $(\frac{39}{4})^{th}$ item = 9.75th item

In c.f. column, 10 is just greater than 9.75 so its corresponding class is 62-64.
∴ L = 62, f = 6, c.f. = 4, i = 2

Now, we know, Q1 = L + $\frac{\frac N4 , - , c.f.}{f}$ × i or, Q1 = 62 + $\frac{9.75 , - , 4}{6}$ × 2
∴ Q1 = 63.92

Position of third quartile = $(\frac{3N}{4} )^{th}$ item = $(\frac{3 \, \times \, 39}{4})^{th}$ item = 29.25th item

In c.f. column, 30 is just greater than 29.25 so its corresponding class is 66-68.
∴ L = 66, f = 12, c.f. = 18, i = 2

Now, we know, Q3 = L + $\frac{\frac{3N}{4} , - , c.f.}{f}$ × i or, Q3 = 66 + $\frac{29.25 , - , 18}{12}$ × 2
∴ Q3 = 67.875

Finally,
QD = $\frac{Q_3 \, - \, Q_1}{2}$ = $\frac{67.875 \, - \, 63.92}{2}$ = 1.978 inches
Coefficient of quartile deviation = $\frac{Q_3 \, - \, Q_1}{Q_3 \, + \, Q_1}$ = $\frac{3.955}{131.795}$ = 0.03

Find the quartile deviation and its coefficient from the following data:
 No. of workers Expenditure 0 ≤ x ≤ 10 10 ≤ x ≤ 20 20 ≤ x ≤ 30 30 ≤ x ≤ 40 40 ≤ x ≤ 50 5 15 10 8 6
Here,
Tabulating given data for calculation of Quartile deviation (QD),
Expenditure Number of workers (f) Cumulative frequency (c.f.)
0-10 5 5
10-20 15 20
20-30 10 30
30-40 8 38
40-50 6 44
N = 44

Position of first quartile = $(\frac N4 )^{th}$ item = $(\frac{44}{4})^{th}$ item = 11th item

In c.f. column, 20 is just greater than 11 so its corresponding class is 10-20.
∴ L = 10, f = 15, c.f. = 5, i = 10

Now, we know, Q1 = L + $\frac{\frac N4 , - , c.f.}{f}$ × i or, Q1 = 10 + $\frac{11 , - , 5}{15}$ × 10
∴ Q1 = 14

Position of third quartile = $(\frac{3N}{4} )^{th}$ item = $(\frac{3 \, \times \, 44}{4})^{th}$ item = 33th item

In c.f. column, 38 is just greater than 33 so its corresponding class is 30-40.
∴ L = 30, f = 8, c.f. = 30, i = 10

Now, we know, Q3 = L + $\frac{\frac{3N}{4} , - , c.f.}{f}$ × i or, Q3 = 30 + $\frac{33 , - , 30}{8}$ × 10
∴ Q3 = 33.75

Finally,
QD = $\frac{Q_3 \, - \, Q_1}{2}$ = $\frac{33.75 \, - \, 14}{2}$ = 9.875
Coefficient of quartile deviation = $\frac{Q_3 \, - \, Q_1}{Q_3 \, + \, Q_1}$ = $\frac{19.75}{47.75}$ = 0.414

The following are the marks obtained by class 9 students in their internal examination. Taking class interval of (10-20) as first class, prepare a frequency distribution table and find the quartile deviation. Also find its coefficient from the following data:
22, 25, 46, 34, 57, 69, 44, 36, 12, 27, 50, 36, 35, 62, 46, 52, 54, 61, 66, 55, 29, 39, 40, 33, 14, 41, 25, 20, 16.

Given data,
22, 25, 46, 34, 57, 69, 44, 36, 12, 27, 50, 36, 35, 62, 46, 52, 54, 61, 66, 55, 29, 39, 40, 33, 14, 41, 25, 20, 16

Tabulating given data starting from class interval of (10 - 20) for calculation of Quartile deviation (QD),

Class interval Frequency (f) Cumulative frequency (c.f.)
10-20 3 3
20-30 6 9
30-40 6 15
40-50 5 20
50-60 5 25
60-70 4 29
N = 29

Position of first quartile = $(\frac N4 )^{th}$ item = $(\frac{29}{4})^{th}$ item = 7.25th item

In c.f. column, 9 is just greater than 7.25 so its corresponding class is 20-30.
∴ L = 20, f = 6, c.f. = 3, i = 10

Now, we know, Q1 = L + $\frac{\frac N4 , - , c.f.}{f}$ × i or, Q1 = 20 + $\frac{7.25 , - , 3}{6}$ × 10
∴ Q1 = 27.083

Position of third quartile = $(\frac{3N}{4} )^{th}$ item = $(\frac{3 \, \times \, 29}{4})^{th}$ item = 21.75th item

In c.f. column, 25 is just greater than 21.75 so its corresponding class is 50-60.
∴ L = 50, f = 5, c.f. = 20, i = 10

Now, we know, Q3 = L + $\frac{\frac{3N}{4} , - , c.f.}{f}$ × i or, Q3 = 50 + $\frac{21.75 , - , 20}{5}$ × 10
∴ Q3 = 53.5

Finally,
QD = $\frac{Q_3 \, - \, Q_1}{2}$ = $\frac{53.5 \, - \, 27.083}{2}$ = 13.21
Coefficient of quartile deviation = $\frac{Q_3 \, - \, Q_1}{Q_3 \, + \, Q_1}$ = $\frac{26.417}{80.58}$ = 0.33

The following are the marks obtained by class 9 students in their internal examination. Taking class interval of (10-20) as first class, prepare a frequency distribution table and find the quartile deviation. Also find its coefficient from the following data:
21, 45, 60, 57, 15, 41, 48, 50, 34, 29, 56, 40, 14, 62, 28, 70, 22, 30, 38, 74, 13, 47, 20, 53, 64, 34, 75, 66.

Given data,
21, 45, 60, 57, 15, 41, 48, 50, 34, 29, 56, 40, 14, 62, 28, 70, 22, 30, 38, 74, 13, 47, 20, 53, 64, 34, 75, 66

Tabulating given data starting from class interval of (10 - 20) for calculation of Quartile deviation (QD),

Class interval Frequency (f) Cumulative frequency (c.f.)
10-20 3 3
20-30 5 8
30-40 4 12
40-50 5 17
50-60 4 21
60-70 4 25
70-80 3 28
N = 28

Position of first quartile = $(\frac N4 )^{th}$ item = $(\frac{28}{4})^{th}$ item = 7th item

In c.f. column, 8 is just greater than 7 so its corresponding class is 20-30.
∴ L = 20, f = 5, c.f. = 3, i = 10

Now, we know, Q1 = L + $\frac{\frac N4 , - , c.f.}{f}$ × i or, Q1 = 20 + $\frac{7 , - , 3}{5}$ × 10
∴ Q1 = 28

Position of third quartile = $(\frac{3N}{4} )^{th}$ item = $(\frac{3 \, \times \, 28}{4})^{th}$ item = 21th item

In c.f. column, corresponding class of 21 is 50-60.
∴ L = 50, f = 4, c.f. = 17, i = 10

Now, we know, Q3 = L + $\frac{\frac{3N}{4} , - , c.f.}{f}$ × i or, Q3 = 50 + $\frac{21 , - , 17}{4}$ × 10
∴ Q3 = 60

Finally,
QD = $\frac{Q_3 \, - \, Q_1}{2}$ = $\frac{60 \, - \, 28}{2}$ = 16
Coefficient of quartile deviation = $\frac{Q_3 \, - \, Q_1}{Q_3 \, + \, Q_1}$ = $\frac{32}{88}$ = 0.36