Telangana SCERT Solution Class X (10) Maths Chapter 14 Statistics Exercise 14.1
Statistics
Exercise: 14.1
1> Solution
No of plants | No of Houses (fi) | Class marks xi | Fixi |
0-2 | 1 | 1 | 1 |
2-4 | 2 | 3 | 6 |
4-6 | 1 | 5 | 5 |
6-8 | 5 | 7 | 35 |
8-10 | 6 | 9 | 54 |
10-12 | 2 | 11 | 22 |
12-14 | 3 | 13 | 39 |
∑Fi – 20 | ∑Fixi = 162 |
Mean x̄ = ∑fixi/∑fi
= 162/20
= 81/10
Therefore, plants per house = 8.1
(2) Solution
Daily wages in rupees (C.I) | No of workers fi | Class marks xi | Di = (xi – a)
Xi – 275 |
fidi |
200-250 | 12 | 275 | -50 | -600 |
250-300 | 14 –> | 275a | 0 | 0 |
300-350 | 8 | 325 | 50 | 400 |
350-400 | 6 | 375 | 100 | 600 |
400-450 | 10 | 425 | 150 | 1500 |
∑fi = 50 | ∑fidi = 1900 |
Mean is assumed mean method
x̄ = a + ∑fidi
∑ fi
= 275 + 1900/50
= 275+38
x̄ = 313
Therefore, The mean of daily wages of workers is 313.
(3) Solution
Daily pocket allowance (in Rs) | No. of children fi | Class marks xi | Fixi |
11-13 | 7 | 12 | 84 |
13-15 | 6 | 14 | 84 |
15-17 | 9 | 16 | 144 |
17-19 | 13 | 18 | 234 |
19-21 | F | 20 | 20f |
21-23 | 5 | 22 | 110 |
23-25 | 4 | 24 | 96 |
∑fi = 44+f | ∑fixi = 752+20f |
= Mean = 18 (given)
Mean x̄ = ∑fixi/∑fi
18 = 752+20f/44+f
= 18 (44+f) = 752+20f
= 792 + 18f = 752 + 20f
= 792 – 752 = 20f – 18f
= 40 = 20f
=> f = 40/2 = 20
Therefore, f = 20
(4) Solution
No of heart beats/min (C.I) | No of women fi | Class marks xi | Di = xi – a
= xi– 75.5 |
Fidi |
65-68 | 2 | 66.5 | -9 | -18 |
68-71 | 4 | 69.5 | -6 | -24 |
71-74 | 3 | 72.5 | -3 | -9 |
74-77 | 8 à | 75.5 | 0 | 0 |
77-80 | 7 | 78.5a | 3 | 21 |
80-83 | 4 | 78.5 | 6 | 24 |
83-86 | 2 | 81.5 | 9 | 18 |
∑fi = 30 | 84.5 | ∑fidi = 12 |
A = 75.5
∑fi = 30
∑fidi = 12
Mean = x̄ = a + ∑fidi
∑fi
= 75.5 + 12/30
= 75.5 + 0.4
x̄ = 75.9
Therefore, Average heart beat of the women per minute = 75.9
(5) Solution
No of oranges (C.I) | No of basket fi | Class marks xi | di = xi – a | Ui = xi – a/h | Fiui |
10-14 | 15 | 12 | -10 | -2 | -30 |
15-19 | 110 | 17 | -5 | -1 | -110 |
20-24 | 135 à | 22a | 0 | 0 | 0 |
25-29 | 115 | 27 | 5 | 1 | 115 |
30-34 | 25 | 32 | 10 | 2 | 50 |
∑fi = 400 | ∑fiui = 25 |
A = 22, h = 5
∑fi = 400, ∑fiui = 25
Mean x̄ = a + [∑fiui/∑fi] x h
= 22 + 25/400 x 5
= 22+5/16
x̄ = 22.31
Therefore, The mean number of oranges kept in each basket = 22.31
(6) Solution
Daily expenditure (in Rs) (C.I) | No of house hold (fi) | Class marks xi | di = xi –a = xi – 225 | U = xi-a/h | Fiui |
100-150 | 4 | 125 | -100 | -2 | -8 |
150-200 | 5 | 175 | -50 | -1 | -5 |
200-250 | 12 | 225 a | 0 | 0 | 0 |
250-300 | 2 | 275 | 50 | 1 | 2 |
300-350 | 2 | 325 | 100 | 2 | 4 |
∑fi = 25 | ∑fiui = 7 |
A = 225, h = 50
∑fi = 25 ∑fiui = -7
Mean x̄ = a + (∑fiui/∑fi) x h
= 225 + (-7/25) x 502
= 225-14
x̄ = 211
Therefore, The mean daily expenditure of the each family = Rs 211
(7) Solution
Concentration of so2 in p p m (C.I) | Frequency fi | Class marks xi | fixi |
0.00-0.04 | 4 | 0.02 | 0.08 |
0.04-0.08 | 9 | 0.06 | 0.54 |
0.08-0.12 | 9 | 0.10 | 0.90 |
0.12-0.16 | 2 | 0.14 | 0.28 |
0.16-0.20 | 4 | 0.18 | 0.72 |
0.20-0.24 | 2 | 0.22 | 0.44 |
∑fi = 30 | ∑fixi = 2.96 |
∑fi = 30
∑fixi = 2.96
Mean x̄ = ∑fixi/∑fi
= 2.96/30
= 0.296/3
x̄ = 0.09866
Therefore, The mean concentration of so2 in the air = 0.099 p p m
(8) Solution
No of days (C.I) | No of students fi | Class marks xi | di = xi – a | fidi |
35-38 | 1 | 36.5 | -18 | -18 |
38-41 | 3 | 39.5 | -15 | -45 |
41-44 | 4 | 42.5 | -12 | -48 |
44-47 | 4 | 45.5 | -9 | -36 |
47-50 | 7 | 48.5 | -6 | -42 |
50-53 | 10 | 51.5 | -3 | -30 |
53-56 | 11 | 54.5a | 0 | 0 |
∑fi = 40 | ∑fidi = -219 |
Therefore, a = 54.5
∑fi = 40, ∑fidi = 219
Mean x̄ = a + ∑fidi/∑fi
= 54.5 + (-219/40)
= 54 – 5.47
x̄ = 49.03
x̄ = 49 days
Therefore, The mean number of days the student was present = 49 days
(9) Solution
Literacy rate in y | No of cities (fi) | Class marks | Fi xi |
45-55 | 3 | 50 | 150 |
55-65 | 10 | 60 | 600 |
65-75 | 11 | 70 | 770 |
75-85 | 8 | 80 | 640 |
85-95 | 3 | 90 | 270 |
∑fi = 35, ∑fixi = 2430
Mean x̄ = ∑fixi/∑fi
= 2430/35
= 69.42
Therefore, Mean literacy rate of the cities = 69.42 %
Here is your solution of Telangana SCERT Class 10 Math Chapter 14 Statistics Exercise 14.1
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