NCERT Solutions For Class 10 Maths Exercise 14.1 Statistics
NCERT Solutions For Class 10 Maths Exercise 14.1 Statistics has been brought to you by Future Study Point to help you in your homework, assignments, and forthcoming examination of class 10 CBSE board exam. NCERT Solutions For Class 10 Maths Exercise 14.1 Statistics is one of the greatest input study materials through which you can clear your concepts on Statistics of 10 class CBSE. All questions are solved by an expert in maths by a stepbystep method.
NCERT Solutions For Class 10 Maths Exercise 14.1 Statistics
Q1.A survey was conducted by a group of students as a part of their environmental awareness programm in which they collected the following data regarding the number of plants in 20 houses in a locality . Find the mean number of plants per house.
Which method did you use for finding the mean , and why ?
Ans.
No. of Plants  Class Mark (x_{i})  No. of Houses (f_{i })  f_{i}x_{i} 
0 – 2 2 – 4 4 – 6 6 – 8 8 – 10 10 – 12 12 – 14
 1 3 5 7 9 11 13
 1 2 1 5 6 2 3 ∑f_{i }_{ }= 20  1 6 5 35 54 22 39 ∑f_{i}x_{i}= 162 
We have
Hence mean number of the plants is 8.1
The direct method is proper to use here because the values of observations and their frequencies are small.
NCERT Solutions For Class 10 Maths Exercise 14.1 Statistics
Q2.Consider the following distribution of daily wages of 50 workers of a factory. Find the mean daily wages of the workers of the factory by using an appropriate method.
Ans.
Class Interval  Class Mark (x_{i})  Frequency(f_{i})  u_{i}=(x_{i}A)/h  f_{i}x_{i} 
100 – 120  110  12  2  24 
120 – 140  130  14  1  14 
140 – 160  150 = A  8  0  0 
160 – 180  170  6  1  6 
180 – 200  190  10  2  20 
∑f_{i }_{ }= 50  ∑f_{i}u_{i}= 12 
Since the class interval is of the size,h= 20 and sum of frequencies is 50, both of them are large therefore applying the step deviation method
Hence the mean daily wages of the workers of the factory is 145.20
NCERT Solutions For Class 10 Maths Exercise 14.1 Statistics
Q3. The following distribution shows the daily pocket allowance of children of a locality. The mean pocket allowance is Rs 18. Find the missing frequency f.
Ans.
Daily pocket allowance (in Rs)  Class mark(x_{i})  Frequency(f_{i})  d_{i}= x_{i}A  f_{i}d_{i} 
11 – 13 13 – 15 15 – 17 17 – 19 19 – 21 21 – 23 23 – 25  12 14 16 18 = A 20 22 24  7 6 9 13 f 5 4  6 4 2 0 2 4 6  42 24 18 0 2f 20 24 
∑f_{i }_{ }= 44 + f  ∑f_{i}d_{i}= 2f – 40 
Applying the shortcut method of mean
Where A = 18, ∑f_{i}d_{i}= 2f – 40 , ∑f_{i }_{ }= 44 + f and
2f – 40 = 0⇒ f = 20
Hence missing frequency is 20
NCERT Solutions For Class 10 Maths Exercise 14.1 Statistics
Q4.Thirty women were examined in a hospital by a doctor and the number of heartbeats per minute were recorded and summarized as follows. Find the mean heartbeats per minute for these women, choosing a suitable method.
Ans.
No of heart beats per minute  Class mark(x_{i})  Frequency(f_{i})  f_{i}x_{i}

65 – 68  66.5  2  133 
68 – 71  69.5  4  278 
71 – 74  72.5  3  217.5 
74 – 77  75.5  8  604 
77 – 80  78.5  7  549.5 
80 – 83  81.5  4  326 
83 – 86  84.5  2  169 
∑f_{i}= 30  ∑f_{i}x_{i}= 2277 
Applying the direct method for getting the mean of the data
Hence mean of the heartbeats per minute of the women is 75.9
NCERT Solutions For Class 10 Maths Exercise 14.1 Statistics
Q5. In a retail market ,fruit vendors were selling mangoes kept in packing boxes.These boxes contained varying number of mangoes .The following was the distribution of mangoes according to the number of boxes.
Find the mean number of the mangoes kept in a packing box. Which method of finding the mean did you choose.
Ans.
No of mangoes  Class mark(x_{i})  Frequency (f_{i})  u_{i=}(x_{i}A)/h  f_{i}u_{i} 
50 – 52 53 – 55 56 – 58 59 – 61 62 – 64  51 54 57= A 60 63  15 110 135 115 25  2 1 0 1 2  30 110 0 115 50 
∑f_{i}= 400  ∑f_{i}u_{i }= 25 
Here we have to apply step deviation because class intervals are not continuous
Hence the mean number of the mangoes kept in a packing box is 57.19
NCERT Solutions For Class 10 Maths Exercise 14.1 Statistics
Q6.The table below shows the daily expenditure on food of 25 households in a locality.
Find the mean daily expenditure on food by suitable method.
Ans..
Daily expenditure(in Rs)  Class marks(x_{i})  Frequency (f_{i})  u_{i}= (x_{i }A)/h  f_{i}u_{i} 
100 – 150  125  4  2  8 
150 – 200  175  5  1  5 
200 – 250  225  12  0  0 
250 – 300  275  2  1  2 
300 – 350  325  2  2  4 
∑f_{i}=25  ∑f_{i}u_{i}= 7 
Hence the mean daily expenditure on food is 211
NCERT Solutions For Class 10 Maths Exercise 14.1 Statistics
Q7. To find out the concentration of SO2 in the air (in parts per million,i.e PPM) ,the data was collected for 30 localities in a certain city and is presented below.
Ans.
Concentration of SO2(PPM)  Class Mark(x_{i})  Frequency(f_{i})  u_{i}=(x_{i} A)/h  f_{i}x_{i} 
o.oo – 0.04  0.02  4  2  8 
0.04 – 0.08  0.06  9  1  9 
0.08 – 0.12  0.10  9  0  0 
0.12 – 0.16  0.14  2  1  2 
0.16 – 0.20  0.18  4  2  8 
0.20 – 0.24
 0.22
 2 ∑f_{i}=30  3
 6 ∑f_{i}u_{i}=1 
Applying the step deviation method for getting the mean
Hence the average concentration of SO2 is 0.0987
NCERT Solutions For Class 10 Maths Exercise 14.1 Statistics
Q8. A class teacher has the following absentee record of 40 students of a class for the whole term. Find the mean number of days a student was absent.
Ans.
No. of days  Class Mark(x_{i})  Frequency (f_{i})  f_{i}x_{i} 
0 – 6  3  11  33 
6 – 10  8  10  80 
10 – 14  12  7  84 
14 20  17  4  68 
20 – 28  24  4  96 
28 – 38  33  3  99 
38 – 40  39  1  39 
∑f_{i}= 40  ∑f_{i}x_{i}= 499 
Applying direct method of getting the mean
=12 .48
The mean number of days a student was absent is 12.48
NCERT Solutions For Class 10 Maths Exercise 14.1 Statistics
Q9.The following table gives the literacy rate (in percentage) of 35 cities. Find the mean literacy rate.
Ans.
Litracy rate (%)  Class mark(x_{i})  Frequency(f_{i})  u_{i}= (x_{i}– A)/h  f_{i}x_{i} 
45 – 55  50  3  2  6 
55 – 65  60  10  1  10 
65 – 75  70 = A  11  0  0 
75 – 85  80  8  1  8 
85 – 95
 90
 3 ∑f_{i}= 35  2
 6 ∑f_{i}u_{i}=2 
Applying the step deviation method
Where A = 70, h = 10
= 70 – 0.57 = 69.43 %
Hence the mean litracy rate is 69.43 %
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