NCERT Solutions for class 10 Maths Chapter 14- Statistics Exercise 14.1 have been provided here for students to prepare well for the Board exam. It has been prepared by our subject experts and is in accordance with the NCERT syllabus and guidelines prescribed by CBSE.
These solutions of 10th class Maths, are helpful for students, to score well in the term 1 and term 2 examination, as it works for them as a reference tool to do the revision. The materials are in the context of the upcoming session of 2021-2022.
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1. A survey was conducted by a group of students as a part of their environment awareness program, 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.
Number of Plants | 0-2 | 2-4 | 4-6 | 6-8 | 8-10 | 10-12 | 12-14 |
Number of Houses | 1 | 2 | 1 | 5 | 6 | 2 | 3 |
Which method did you use for finding the mean, and why?
Solution:
In order to find the mean value, we will use direct method because the numerical value of f_{i}Â and x_{i}Â are small.
Find the midpoint of the given interval using the formula.
Midpoint (x_{i}) = (upper limit + lower limit)/2
No. of plants
(Class interval) |
No. of houses
Frequency (f_{i}) |
Mid-point (x_{i}) | f_{i}x_{i} |
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 |
Sum f_{iÂ }= 20 | Sum f_{i}x_{i}Â = 162 |
The formula to find the mean is:
Mean = xÌ„ = âˆ‘f_{i }x_{i}Â /âˆ‘f_{iÂ }
= 162/20
= 8.1_{}
Therefore, the mean number of plants per house is 8.1
2. Consider the following distribution of daily wages of 50 workers of a factory.
Daily wages (in Rs.) | 100-120 | 120-140 | 140-160 | 160-180 | 180-200 |
Number of workers | 12 | 14 | 8 | 6 | 10 |
Find the mean daily wages of the workers of the factory by using an appropriate method.
Solution:
Find the midpoint of the given interval using the formula.
Midpoint (x_{i}) = (upper limit + lower limit)/2
In this case, the value of mid-point (x_{i}) is very large, so let us assume the mean value, A = 150 and class interval is h = 20.
So, u_{iÂ }= (x_{i}Â â€“ A)/h = u_{i Â }= (x_{i}Â â€“ 150)/20
Substitute and find the values as follows:
Daily wages
(Class interval) |
Number of workers
frequency (f_{i}) |
Mid-point (x_{i}) | u_{iÂ }= (x_{i}Â â€“ 150)/20 | f_{i}u_{i} |
100-120 | 12 | 110 | -2 | -24 |
120-140 | 14 | 130 | -1 | -14 |
140-160 | 8 | 150 | 0 | 0 |
160-180 | 6 | 170 | 1 | 6 |
180-200 | 10 | 190 | 2 | 20 |
Total | Sum f_{iÂ }= 50 | Sum f_{i}u_{i}Â = -12 |
So, the formula to find out the mean is:
Mean = xÌ„ = A + hâˆ‘f_{i}u_{i}Â /âˆ‘f_{iÂ }=150 + (20 Ã— -12/50) = 150 â€“ 4.8 = 145.20
Thus, mean daily wage of the workers = Rs. 145.20
3. 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.
Daily Pocket Allowance(in c) | 11-13 | 13-15 | 15-17 | 17-19 | 19-21 | 21-23 | 23-35 |
Number of children | 7 | 6 | 9 | 13 | f | 5 | 4 |
Solution:
To find out the missing frequency, use the mean formula.
Here, the value of mid-point (x_{i})Â meanÂ xÌ„Â = 18
Class interval | Number of children (f_{i}) | Mid-point (x_{i}) | Â Â Â f_{i}x_{iÂ Â Â Â } |
11-13 | 7 | 12 | 84 |
13-15 | 6 | 14 | 84 |
15-17 | 9 | 16 | 144 |
17-19 | 13 | 18 = A | 234 |
19-21 | f | 20 | 20f |
21-23 | 5 | 22 | 110 |
23-25 | 4 | 24 | 96 |
Total | f_{i}Â = 44+f | Sum f_{i}x_{i}Â = 752+20f |
The mean formula is
Mean = xÌ„ = âˆ‘f_{i}x_{i}Â /âˆ‘f_{iÂ }= (752+20f)/(44+f)
Now substitute the values and equate to find the missing frequency (f)
â‡’ 18 = (752+20f)/(44+f)
â‡’ 18(44+f) = (752+20f)
â‡’ 792+18f = 752+20f
â‡’ 792+18f = 752+20f
â‡’ 792 â€“ 752 = 20f â€“ 18f
â‡’ 40 = 2f
â‡’ f = 20
So, the missing frequency, f = 20.
4. Thirty women were examined in a hospital by a doctor and the number of heart beats per minute were recorded and summarized as follows. Find the mean heart beats per minute for these women, choosing a suitable method.
Number of heart beats per minute | 65-68 | 68-71 | 71-74 | 74-77 | 77-80 | 80-83 | 83-86 |
Number of women | 2 | 4 | 3 | 8 | 7 | 4 | 2 |
Solution:
From the given data, let us assume the mean as A = 75.5
x_{iÂ }= (Upper limit + Lower limit)/2
Class size (h) = 3
Now, find the u_{i }and f_{i}u_{i} as follows:
Class Interval | Number of women (f_{i}) | Mid-point (x_{i}) | u_{i}Â = (x_{i}Â â€“ 75.5)/h | f_{i}u_{i} |
65-68 | 2 | 66.5 | -3 | -6 |
68-71 | 4 | 69.5 | -2 | -8 |
71-74 | 3 | 72.5 | -1 | -3 |
74-77 | 8 | 75.5 | 0 | 0 |
77-80 | 7 | 78.5 | 1 | 7 |
80-83 | 4 | 81.5 | 3 | 8 |
83-86 | 2 | 84.5 | 3 | 6 |
Sum f_{i}= 30 | Sum f_{i}u_{iÂ }= 4 |
Mean = xÌ„ = A + hâˆ‘f_{i}u_{i}Â /âˆ‘f_{iÂ }
= 75.5 + 3Ã—(4/30)
= 75.5 + 4/10
= 75.5 + 0.4
= 75.9
Therefore, the mean heart beats per minute for these women is 75.9
5. 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.
Number of mangoes | 50-52 | 53-55 | 56-58 | 59-61 | 62-64 |
Number of boxes | 15 | 110 | 135 | 115 | 25 |
Find the mean number of mangoes kept in a packing box. Which method of finding the mean did you choose?
Solution:
Since, the given data is not continuous so we add 0.5 to the upper limit and subtract 0.45 from the lower limit as the gap between two intervals are 1
Here, assumed mean (A) = 57
Class size (h) = 3
Here, the step deviation is used because the frequency values are big.
Class Interval | Number of boxes (f_{i}) | Mid-point (x_{i}) | d_{i}Â = x_{i}Â â€“ A | f_{i}d_{i} |
49.5-52.5 | 15 | 51 | -6 | 90 |
52.5-55.5 | 110 | 54 | -3 | -330 |
55.5-58.5 | 135 | 57 = A | 0 | 0 |
58.5-61.5 | 115 | 60 | 3 | 345 |
61.5-64.5 | 25 | 63 | 6 | 150 |
Sum f_{i}Â = 400 | Sum f_{i}d_{i}Â = 75 |
The formula to find out the Mean is:
Mean = xÌ„ = A +h âˆ‘f_{i}d_{i}Â /âˆ‘f_{iÂ }
= 57 + 3(75/400)
= 57 + 0.1875
= 57.19
Therefore, the mean number of mangoes kept in a packing box is 57.19
6. The table below shows the daily expenditure on food of 25 households in a locality. Find the mean daily expenditure on food by a suitable method.
Daily expenditure(in c) | 100-150 | 150-200 | 200-250 | 250-300 | 300-350 |
Number of households | 4 | 5 | 12 | 2 | 2 |
Solution:
Find the midpoint of the given interval using the formula.
Midpoint (x_{i}) = (upper limit + lower limit)/2
Let is assume the mean (A) = 225
Class size (h) = 50
Class Interval | Number of households (f_{i}) | Mid-point (x_{i}) | d_{i}Â = x_{i}Â â€“ A | u_{i }= d_{i}/50 | f_{i}u_{i} |
100-150 | 4 | 125 | -100 | -2 | -8 |
150-200 | 5 | 175 | -50 | -1 | -5 |
200-250 | 12 | 225 | 0 | 0 | 0 |
250-300 | 2 | 275 | 50 | 1 | 2 |
300-350 | 2 | 325 | 100 | 2 | 4 |
Sum f_{i}Â = 25 | Sum f_{i}u_{i}Â = -7 |
Mean = xÌ„ = A +hâˆ‘f_{i}u_{i}Â /âˆ‘f_{i}
_{Â }= 225+50(-7/25)
= 225-14
= 211
Therefore, the mean daily expenditure on food is 211
7. To find out the concentration of SO_{2}Â 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:
Concentration of SO_{2} ( in ppm) | Frequency |
0.00 â€“ 0.04 | 4 |
0.04 â€“ 0.08 | 9 |
0.08 â€“ 0.12 | 9 |
0.12 â€“ 0.16 | 2 |
0.16 â€“ 0.20 | 4 |
0.20 â€“ 0.24 | 2 |
Find the mean concentration of SO_{2}Â in the air.
Solution:
To find out the mean, first find the midpoint of the given frequencies as follows:
Concentration of SO_{2Â }(in ppm) | Frequency (f_{i}) | Mid-point (x_{i}) | f_{i}x_{i} |
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 | 0.20 | 0.40 |
Total | Sum f_{i}Â = 30 | Sum (f_{i}x_{i}) = 2.96 |
The formula to find out the mean is
Mean = xÌ„ = âˆ‘f_{i}x_{i}Â /âˆ‘f_{i}
= 2.96/30
= 0.099 ppm
Therefore, the mean concentration of SO_{2} in air is 0.099 ppm.
8. 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.
Number of days | 0-6 | 6-10 | 10-14 | 14-20 | 20-28 | 28-38 | 38-40 |
Number of students | 11 | 10 | 7 | 4 | 4 | 3 | 1 |
Solution:
Find the midpoint of the given interval using the formula.
Midpoint (x_{i}) = (upper limit + lower limit)/2
Class interval | Frequency (f_{i}) | Mid-point (x_{i}) | f_{i}x_{i} |
0-6 | 11 | 3 | 33 |
6-10 | 10 | 8 | 80 |
10-14 | 7 | 12 | 84 |
14-20 | 4 | 17 | 68 |
20-28 | 4 | 24 | 96 |
28-38 | 3 | 33 | 99 |
38-40 | 1 | 39 | 39 |
SumÂ f_{i}Â = 40 | SumÂ f_{i}x_{i}Â = 499 |
The mean formula is,
Mean = xÌ„ = âˆ‘f_{i}x_{i}Â /âˆ‘f_{i}
= 499/40
= 12.48 days
Therefore, the mean number of days a student was absent = 12.48.
9. The following table gives the literacy rate (in percentage) of 35 cities. Find the mean
literacy rate.
Literacy rate (in %) | 45-55 | 55-65 | 65-75 | 75-85 | 85-98 |
Number of cities | 3 | 10 | 11 | 8 | 3 |
Solution:
Find the midpoint of the given interval using the formula.
Midpoint (x_{i}) = (upper limit + lower limit)/2
In this case, the value of mid-point (x_{i}) is very large, so let us assume the mean value, A = 70 and class interval is h = 10.
So, u_{iÂ }= (x_{i}-A)/h = u_{iÂ }= (x_{i}-70)/10
Substitute and find the values as follows:
Class Interval | Frequency (f_{i}) | (x_{i}) | d_{i}Â =Â x_{i}Â â€“ a | u_{i}Â = d_{i}/h | f_{i}u_{i} |
45-55 | 3 | 50 | -20 | -2 | -6 |
55-65 | 10 | 60 | -10 | -1 | -10 |
65-75 | 11 | 70 | 0 | 0 | 0 |
75-85 | 8 | 80 | 10 | 1 | 8 |
85-95 | 3 | 90 | 20 | 2 | 6 |
SumÂ f_{i}Â Â = 35 | SumÂ f_{i}u_{i}Â = -2 |
So, Mean = xÌ„ = A+(âˆ‘f_{i}u_{i}Â /âˆ‘f_{i})Ã—h
= 70+(-2/35)Ã—10
= 69.42
Therefore, the mean literacy part = 69.42
In this exercise, students will solve questions based on finding the mean number based upon the statistics given, with the help of :
- Direct Method
- Assumed Mean Method
- Step-deviation Method
A group of collection of data will be provided to the students to find the mean of them and also finding the missing frequency. A more advanced version, you will be able to study in upcoming exercises. Reach us to practise the exercise problems and solutions for class 10 Maths, chapter 14. Also, we are collectively providing here some more materials such as notes, books, questions papers, etc. for students to help them practise well.
The questions in Ex.14.1 are solved as per the pattern explained in the example questions before the exercise. NCERT solutions are one of the best learning material, where problems are solved in a detailed way, following each and every step and method.
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