Breath Math

 

1. Calculate the range and coefficient of range from the following data.

a) The heights of 10 children in cm: 122, 144, 154, 101, 168, 118, 155, 133, 160, 140

Solution:

Heights of 10 children in cm: 122, 144, 154, 101, 168, 118, 155,133,

160,140.

Range: H – L = 168 – 101 = 67

Coefficient of Range: ^{H – L}/_{H + L}

= ⁶⁷/_{168 + 101} = ⁶⁷/₂₆₉ = 0.249

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b) Marks scored by 12 students in a test: 31, 18, 27, 19, 25, 28, 49, 14, 41, 22, 33, 13.

Solution:

 Marks scored by 12 students

31, 18, 27, 19, 25, 28, 49, 14, 41, 22, 33, 13.

H = 49; L = 13

Range: H – L = 49 – 13 = 36

Coefficient of Range: ^{H – L}/_{H + L} = ³⁶/_{49 + 13} = ³⁶/₆₂ = 0.58.

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c) Number of trees planted in 6 months: 186, 234, 465, 361, 290, 142.

Solution:

No .of trees planted in 6 months:

186, 234, 465, 361, 290, 142.

H = 465; L = 142

Range: H – L = 465 – 142 = 323

Coefficient of Range: ^{H – L}/_{H + L} = ³²³/_465+142 = ³²³/₆₀₇ = 0.532

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2. State quartile deviation for the following data:

a) 30, 18, 23, 15, 11, 29, 37, 42, 10, 21.

Solution:

 Arrange the scores in ascending order:

n=10,

10, 11, 15, 18, 21, 23, 29, 30, 37, 42.

(a) First Q1 = ⁿ⁺¹/₄

= ¹⁰⁺¹/₄ = ¹¹/₄ = 2.75 = 3rd score = 15

(ii) Third Quartile

Q3 = ³⁽ⁿ⁺¹⁾/₄ = ^3×11/₄ = ³³/₄ = 8.25 = 9th = 37

(iii) Quartile Deviation:

= (Q3−Q1)/₂ = ³⁷⁻¹⁵/₂ = ²²/₂ = 11

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b) 3, 5, 8, 10, 12, 7, 5.

Solution:

3,5,5,7,8,10,12

n = 7

(i) Quartile Q1 = ⁿ⁺¹/₄ = ⁷⁺¹/₄ = ⁸/₄ = 2nd score

Q1 = 5

(ii) Quartile Q3 = 3^{(n + 1)}/₄ = 3[(8)/₄] = 6th score

Q3 = 10

(iii) Quartile Deviation:

= [Q3 − Q1]/₂ = ^{10 – 5}/₂ = ⁵/₂ = 2.5

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(c)

Age	3	6	9	12	15
No. of children	4	8	11	7	12

Solution:

x	f	commutative frequency
3	4	4
6	8	12
9	11	23
12	7	30
15	12	42

n = 42

Q1 = ⁿ/₄ ; score = ⁴²/₄ = 10.5;11th score

From fc  ∴Q1 = 6

Q3 = ³ⁿ/₄ = ^{3 ×42}/₄ = 31.5; 32nd score ∴Q3 = 15

Q.D = [Q3−Q1]/₂ = ¹⁵⁻⁶/₂ = ⁹/₂ = 4.5

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d)

Marks scored	10	20	30	40	50	60
No. of students	12	7	16	08	18	22

Solution:

x	f	fc
10	12	12
20	07	19
30	16	35
40	08	43
50	18	61
60	22	83

n = 83

Q1= ⁿ/₄ = ⁸³/₄ = 20.75; 21st score

∴Q1 = 30

Q3 = ³ⁿ/₄ = ^{3 ×83}/₄ = 3X20.75 = 62.25; 62nd score

∴Q3 = 60

Q.D = [Q3−Q1]/2 = ^{60 – 30}/₂ = ³⁰/₂ = 15

∴QD = 15

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3. Compute quartile deviation for each of the following tables.

a)

Class interval	Frequency	fc
5 – 15	11	11
15 – 25	5	16
25 – 35	15	31
35  – 45	9	40
45 – 55	22	62
55 – 65	8	70
65 – 75	17	87

Solution:

n = 87

Q1= ⁿ/₄ = ⁸⁷/₄ = 21.75

22nd score CI = 25 – 35

∴LRC = 25

fc = 31; i = 10

Q2 = LRL +( [^N/₄−fc]/_fm)i

= 25 + [(⁸⁷/₄−16)/₁₅] 10

= 25 + [^{21.75 −16}/₁₅] ×10

Q2 = 28.83

Q3 = LRL + [(^3N/₄ – fc)/_fm]*i

^3N/₄ = ^{3 × 87}/₄ = 65.25  class interval 55 – 65

L = 55, fc = 62, fm = 8, CI = 10

LRL = 55 +(^{65.25 – 62})/₈ ×10

= 55 + 4.06

LRL = 59.06

Quartile Deviation = [Q3 − Q1]/₂

= ^{59.06 – 28.83}/₂

= ^30.23/₂

Q.D = 15.11

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(b)

class interval	frequency	fc
1 – 9	4	4
10 – 19	3	7
20 – 29	20	27
30 – 39	12	39
40 – 49	5	44
50 – 59	8	52
60 – 69	14	66
70 – 79	27	93
80 – 89	2	95
90 – 99	5	100

H = 100

Solution:

n = 100

¹⁰⁰/₄ = 25th Score 20 – 29; LRL = 19.5

Fc = 7; fm = 20

Q1 = LRL+[ (^N/₄−fc)/_fm]*i

= 19.5 + [^{25 – 7}/₂₀]× 10

= 19.5 + ¹⁸/₂₀*10

Q1 = 28.5

^3N/₄ = ^{3 × 100}/₄ = 3 × 25 = 75th score cl 70 – 79

LRL = 69.5; fc = 66; fm = 14

Q3 = 69.5 + [^{(75 – 66)}/₁₄] × 10

= 69.5 + 3.33

Q3 = 72.83

Quartile Deviation = (Q3 − Q1)/₂

= ^{72.83 – 28.5} /₂

= ^44.33/₂

Q.D = 22.16

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STATISTICS – EXERCISE 1.5.2 – Class 9

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