Long Questions

Here,
Tabulating given data for calculation of Quartile deviation (QD),

Position of first quartile = $(\frac N4 )^{th}$ item = $(\frac{30}{4})^{th}$ item = 7.5^th 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 = 22.5^th 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

Here,
Tabulating given data for calculation of Quartile deviation (QD),

Position of first quartile = $(\frac N4 )^{th}$ item = $(\frac{46}{4})^{th}$ item = 11.5^th 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 = 26.25^th 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

Here,
Tabulating given data for calculation of Quartile deviation (QD),

Position of first quartile = $(\frac N4 )^{th}$ item = $(\frac{35}{4})^{th}$ item = 8.75^th 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 = 26.25^th 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

Here,
Tabulating given data for calculation of Quartile deviation (QD),

Position of first quartile = $(\frac N4 )^{th}$ item = $(\frac{112}{4})^{th}$ item = 28^th 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 = 84^th 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

Here,
Tabulating given data for calculation of Quartile deviation (QD),

Position of first quartile = $(\frac N4 )^{th}$ item = $(\frac{39}{4})^{th}$ item = 9.75^th 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 = 29.25^th 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

Here,
Tabulating given data for calculation of Quartile deviation (QD),

Position of first quartile = $(\frac N4 )^{th}$ item = $(\frac{44}{4})^{th}$ item = 11^th 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 = 33^th 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

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),

Position of first quartile = $(\frac N4 )^{th}$ item = $(\frac{29}{4})^{th}$ item = 7.25^th 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 = 21.75^th 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

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),

Position of first quartile = $(\frac N4 )^{th}$ item = $(\frac{28}{4})^{th}$ item = 7^th 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 = 21^th 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