Chapter 14 – Statistics
Exercise 14.1 Page: 270
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 fi and xi are small.
Find the midpoint of the given interval using the formula.
Midpoint (xi) = (upper limit + lower limit)/2
| No. of plants (Class interval) | No. of houses Frequency (fi) | Mid-point (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 |
| Sum fi = 20 | Sum fixi = 162 |
The formula to find the mean is:
Mean = x̄ = ∑fi xi /∑fi
= 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 (xi) = (upper limit + lower limit)/2
In this case, the value of mid-point (xi) is very large, so let us assume the mean value, A = 150 and class interval is h = 20.
So, ui = (xi – A)/h = ui = (xi – 150)/20
Substitute and find the values as follows:
| Daily wages (Class interval) | Number of workers frequency (fi) | Mid-point (xi) | ui = (xi – 150)/20 | fiui |
| 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 fi = 50 | Sum fiui = -12 |
So, the formula to find out the mean is:
Mean = x̄ = A + h∑fiui /∑fi =150 + (20 × -12/50) = 150 – 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 (xi) mean x̄ = 18
| Class interval | Number of children (fi) | Mid-point (xi) | fixi |
| 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 | fi = 44+f | Sum fixi = 752+20f |
The mean formula is
Mean = x̄ = ∑fixi /∑fi = (752+20f)/(44+f)
Now substitute the values and equate to find the missing frequency (f)
⇒ 18 = (752+20f)/(44+f)
⇒ 18(44+f) = (752+20f)
⇒ 792+18f = 752+20f
⇒ 792+18f = 752+20f
⇒ 792 – 752 = 20f – 18f
⇒ 40 = 2f
⇒ 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
xi = (Upper limit + Lower limit)/2
Class size (h) = 3
Now, find the ui and fiui as follows:
| Class Interval | Number of women (fi) | Mid-point (xi) | ui = (xi – 75.5)/h | fiui |
| 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 fi= 30 | Sum fiui = 4 |
Mean = x̄ = A + h∑fiui /∑fi
= 75.5 + 3×(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 (fi) | Mid-point (xi) | di = xi – A | fidi |
| 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 fi = 400 | Sum fidi = 75 |
The formula to find out the Mean is:
Mean = x̄ = A +h ∑fidi /∑fi
= 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 (xi) = (upper limit + lower limit)/2
Let is assume the mean (A) = 225
Class size (h) = 50
| Class Interval | Number of households (fi) | Mid-point (xi) | di = xi – A | ui = di/50 | fiui |
| 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 fi = 25 | Sum fiui = -7 |
Mean = x̄ = A +h∑fiui /∑fi
= 225+50(-7/25)
= 225-14
= 211
Therefore, the mean daily expenditure on food is 211
7. 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:
| Concentration of SO2 ( in ppm) | Frequency |
| 0.00 – 0.04 | 4 |
| 0.04 – 0.08 | 9 |
| 0.08 – 0.12 | 9 |
| 0.12 – 0.16 | 2 |
| 0.16 – 0.20 | 4 |
| 0.20 – 0.24 | 2 |
Find the mean concentration of SO2 in the air.
Solution:
To find out the mean, first find the midpoint of the given frequencies as follows:
| Concentration of SO2 (in ppm) | Frequency (fi) | Mid-point (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 | 0.20 | 0.40 |
| Total | Sum fi = 30 | Sum (fixi) = 2.96 |
The formula to find out the mean is
Mean = x̄ = ∑fixi /∑fi
= 2.96/30
= 0.099 ppm
Therefore, the mean concentration of SO2 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 (xi) = (upper limit + lower limit)/2
| Class interval | Frequency (fi) | Mid-point (xi) | fixi |
| 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 |
| Sum fi = 40 | Sum fixi = 499 |
The mean formula is,
Mean = x̄ = ∑fixi /∑fi
= 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 (xi) = (upper limit + lower limit)/2
In this case, the value of mid-point (xi) is very large, so let us assume the mean value, A = 70 and class interval is h = 10.
So, ui = (xi-A)/h = ui = (xi-70)/10
Substitute and find the values as follows:
| Class Interval | Frequency (fi) | (xi) | di = xi – a | ui = di/h | fiui |
| 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 |
| Sum fi = 35 | Sum fiui = -2 |
So, Mean = x̄ = A+(∑fiui /∑fi)×h
= 70+(-2/35)×10
= 69.42
Therefore, the mean literacy part = 69.42
NCERT Solutions for Class 10 Maths Chapter 14 Statistics
Class 10 Maths Chapter 14, Statistics, is one of the most important of all the chapter present in the textbook. The weightage of this chapter in the CBSE Term II exam is around 11 to 12 marks. On average, there will be 3 questions which could be asked from this chapter and marks will be distributed in a manner of 3+4+4( it could vary as per question).
Topics covered in Chapter 14, Statistics are;
- Mean of Grouped Data
- Mode of Grouped Data
- Median of Grouped Data
- Graphical Representation of Cumulative Frequency Distribution
List of Exercises in class 10 Maths Chapter 14 :
Exercise 14.1 Solutions 9 Question ( 9 long)
Exercise 14.2 Solutions 6 Question ( 6 long)
Exercise 14.3 Solutions 7 Question ( 7 long)
Exercise 14.4 Solutions 3 Question ( 3 long)
NCERT solutions for Class 10 Maths Chapter 14- Statistics are made available for students who want to obtain good marks in this chapter. The methods and procedure to solve the questions have been explained clearly in these NCERT Solutions, such that, students find it easy to understand the fundamentals quickly.
The world is highly data-oriented, in fact, each and every field has a group of data, which represents the relevant information. Statistics is the branch of mathematics which deals with the representation of data in a meaningful way.
You will face many real-life scenarios where the fundamentals of statistics are used to represent a set of data in tabular form or in graphs or in pie charts. There are a number of methods you will learn from this chapter such as, step deviation methods, finding mode and median of grouped data, converting frequency distribution and the relation between mode, mean and median methods, etc. 10th Class NCERT solutions are the best study materials to prepare for the CBSE Term II exam.
Key Features of NCERT Solutions for Class 10 Maths Chapter 14- Statistics
- The solutions for the statistics chapter works as a reference for the students.
- It will help students to score marks against the questions asked from the statistics chapter.
- Students can prepare and do the revision for chapter 14 with this source.
- The questions of statistics are solved by subject experts.
- The content of the material is as per the term-wise CBSE Syllabus (2021-22) and guidelines.
Statistics can also be understood in a much better way by using the other solutions which are provided at RAJ EDUCATION HUB. The solutions are prepared to help students perform well in the CBSE exams.