- 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
- 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)
| 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
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
- 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
(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
STATISTICS – EXERCISE 1.5.2 – Class 9
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