STATISTICS – EXERCISE 1.5.2 – Class 9

STATISTICS – EXERCISE 1.5.2 – Class 9


  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


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.


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


  1. 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


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


(c)

Age3691215
No. of children4811712

Solution:

xfcommutative frequency
344
6812
91123
12730
151242

\ 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


d)

Marks scored102030405060
No. of students12716081822

Solution:

xffc
101212
200719
301635
400843
501861
602283

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


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

a)

Class intervalFrequencyfc
5 – 151111
15 – 25516
25 – 351531
35  – 45940
45 – 552262
55 – 65870
65 – 751787

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


(b)

class intervalfrequencyfc
1 – 944
10 – 1937
20 – 292027
30 – 391239
40 – 49544
50 – 59852
60 – 691466
70 – 792793
80 – 89295
90 – 995100

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


STATISTICS – EXERCISE 1.5.2 – Class 9


 

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