*According to the CBSE Syllabus 2023-24, this chapter has been renumbered as Chapter 13.
NCERT Solutions for Class 10 Maths Chapter 14 – Statistics Exercise 14.2 have been provided here to help students prepare for their CBSE Exam. The problems in NCERT Solutions have been answered by our subject experts, covering the NCERT syllabus and guidelines prescribed by the Central Board of Secondary Education.
The Class 10 Maths Solutions work as a reference tool for students, with the help of which they can do a quick revision of all the topics. BYJU’S has provided updated study materials, i.e. NCERT Class 10 Maths Solutions, for all the students in the context of the upcoming session of 2023-24.
NCERT Solutions Class 10 Maths Chapter 14 Statistics Exercise 14.2
Access Other Exercise Solutions of Class 10 Maths Chapter 14 – Statistics
Exercise 14.1 Solutions 9 Question (9 long)
Exercise 14.3 Solutions 7 Question (7 long)
Exercise 14.4 Solutions 3 Question (3 long)
Access Answers to NCERT Class 10 Maths Chapter 14 – Statistics Exercise 14.2
1. The following table shows the ages of the patients admitted to a hospital during a year:
| Age (in years) | 5-15 | 15-25 | 25-35 | 35-45 | 45-55 | 55-65 |
| Number of patients | 6 | 11 | 21 | 23 | 14 | 5 |
Find the mode and the mean of the data given above. Compare and interpret the two
measures of central tendency.
Solution:
To find out the modal class, let us the consider the class interval with high frequency.
Here, the greatest frequency = 23, so the modal class = 35 – 45,
Lower limit of modal class = l = 35,
class width (h) = 10,
fm = 23,
f1 = 21 and f2 = 14
The formula to find the mode is
Mode = l + [(fm – f1)/ (2fm – f1 – f2)] × h
Substitute the values in the formula, we get
Mode = 35+[(23-21)/(46-21-14)]×10
= 35 + (20/11)
= 35 + 1.8
= 36.8 years
So the mode of the given data = 36.8 years
Calculation of Mean:
First find the midpoint using the formula, xi = (upper limit +lower limit)/2
| Class Interval | Frequency (fi) | Mid-point (xi) | fixi |
| 5-15 | 6 | 10 | 60 |
| 15-25 | 11 | 20 | 220 |
| 25-35 | 21 | 30 | 630 |
| 35-45 | 23 | 40 | 920 |
| 45-55 | 14 | 50 | 700 |
| 55-65 | 5 | 60 | 300 |
| Sum fi = 80 | Sum fixi = 2830 |
The mean formula is
Mean = x̄ = ∑fixi /∑fi
= 2830/80
= 35.375 years
Therefore, the mean of the given data = 35.375 years
2. The following data gives the information on the observed lifetimes (in hours) of 225
electrical components:
| Lifetime (in hours) | 0-20 | 20-40 | 40-60 | 60-80 | 80-100 | 100-120 |
| Frequency | 10 | 35 | 52 | 61 | 38 | 29 |
Determine the modal lifetimes of the components.
Solution:
From the given data the modal class is 60–80.
Lower limit of modal class = l = 60,
The frequencies are:
fm = 61, f1 = 52, f2 = 38 and h = 20
The formula to find the mode is
Mode = l+ [(fm – f1)/(2fm – f1 – f2)] × h
Substitute the values in the formula, we get
Mode = 60 + [(61 – 52)/ (122 – 52 – 38)] × 20
Mode = 60 + [(9 × 20)/32]
Mode = 60 + (45/8) = 60 + 5.625
Therefore, modal lifetime of the components = 65.625 hours.
3. The following data gives the distribution of total monthly household expenditure of 200
families of a village. Find the modal monthly expenditure of the families. Also, find the
mean monthly expenditure:
| Expenditure (in Rs.) | Number of families |
| 1000-1500 | 24 |
| 1500-2000 | 40 |
| 2000-2500 | 33 |
| 2500-3000 | 28 |
| 3000-3500 | 30 |
| 3500-4000 | 22 |
| 4000-4500 | 16 |
| 4500-5000 | 7 |
Solution:
Given data:
Modal class = 1500-2000,
l = 1500,
Frequencies:
fm = 40 f1 = 24, f2 = 33 and
h = 500
Mode formula:
Mode = l + [(fm – f1)/ (2fm – f1 – f2)] × h
Substitute the values in the formula, we get
Mode = 1500 + [(40 – 24)/ (80 – 24 – 33)] × 500
Mode = 1500 + [(16 × 500)/23]
Mode = 1500 + (8000/23) = 1500 + 347.83
Therefore, modal monthly expenditure of the families = Rupees 1847.83
Calculation for mean:
First find the midpoint using the formula, xi =(upper limit +lower limit)/2
Let us assume a mean, (a) be 2750.
| Class Interval | fi | xi | di = xi – a | ui = di/h | fiui |
| 1000-1500 | 24 | 1250 | -1500 | -3 | -72 |
| 1500-2000 | 40 | 1750 | -1000 | -2 | -80 |
| 2000-2500 | 33 | 2250 | -500 | -1 | -33 |
| 2500-3000 | 28 | 2750 = a | 0 | 0 | 0 |
| 3000-3500 | 30 | 3250 | 500 | 1 | 30 |
| 3500-4000 | 22 | 3750 | 1000 | 2 | 44 |
| 4000-4500 | 16 | 4250 | 1500 | 3 | 48 |
| 4500-5000 | 7 | 4750 | 2000 | 4 | 28 |
| fi = 200 | fiui = -35 |
The formula to calculate the mean,
Mean = x̄ = a +(∑fiui /∑fi) × h
Substitute the values in the given formula
= 2750 + (-35/200) × 500
= 2750 – 87.50
= 2662.50
So, the mean monthly expenditure of the families = Rs. 2662.50
4. The following distribution gives the state-wise teacher-student ratio in higher secondary schools of India. Find the mode and mean of this data. Interpret the two measures
| No of students per teacher | Number of states / U.T |
| 15-20 | 3 |
| 20-25 | 8 |
| 25-30 | 9 |
| 30-35 | 10 |
| 35-40 | 3 |
| 40-45 | 0 |
| 45-50 | 0 |
| 50-55 | 2 |
Solution:
Given data:
Modal class = 30 – 35,
l = 30,
Class width (h) = 5,
fm = 10, f1 = 9 and f2 = 3
Mode Formula:
Mode = l + [(fm – f1)/ (2fm – f1 – f2)] × h
Substitute the values in the given formula
Mode = 30 + [(10 – 9)/ (20 – 9 – 3)] × 5
= 30 + (5/8)
= 30 + 0.625
= 30.625
Therefore, the mode of the given data = 30.625
Calculation of mean:
Find the midpoint using the formula, xi =(upper limit +lower limit)/2
| Class Interval | Frequency (fi) | Mid-point (xi) | fixi |
| 15-20 | 3 | 17.5 | 52.5 |
| 20-25 | 8 | 22.5 | 180.0 |
| 25-30 | 9 | 27.5 | 247.5 |
| 30-35 | 10 | 32.5 | 325.0 |
| 35-40 | 3 | 37.5 | 112.5 |
| 40-45 | 0 | 42.5 | 0 |
| 45-50 | 0 | 47.5 | 0 |
| 50-55 | 2 | 52.5 | 105.0 |
| Sum fi = 35 | Sum fixi = 1022.5 |
Mean = x̄ = ∑fixi /∑fi
= 1022.5/35
= 29.2 (approx)
Therefore, mean = 29.2
5. The given distribution shows the number of runs scored by some top batsmen of the world in one- day international cricket matches.
| Run Scored | Number of Batsman |
| 3000-4000 | 4 |
| 4000-5000 | 18 |
| 5000-6000 | 9 |
| 6000-7000 | 7 |
| 7000-8000 | 6 |
| 8000-9000 | 3 |
| 9000-10000 | 1 |
| 10000-11000 | 1 |
Find the mode of the data.
Solution:
Given data:
Modal class = 4000 – 5000,
l = 4000,
class width (h) = 1000,
fm = 18, f1 = 4 and f2 = 9
Mode Formula:
Mode = l + [(fm – f1)/ (2fm – f1 – f2)] × h
Substitute the values
Mode = 4000 + [(18 – 4)/ (36 – 4 – 9)] × 1000
= 4000 + (14000/23)
= 4000 + 608.695
= 4608.695
= 4608.7 (approximately)
Thus, the mode of the given data is 4608.7 runs.
6. A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarized it in the table given below. Find the mode of the data:
| Number of cars | Frequency |
| 0-10 | 7 |
| 10-20 | 14 |
| 20-30 | 13 |
| 30-40 | 12 |
| 40-50 | 20 |
| 50-60 | 11 |
| 60-70 | 15 |
| 70-80 | 8 |
Solution:
Given Data:
Modal class = 40 – 50, l = 40,
Class width (h) = 10, fm = 20, f1 = 12 and f2 = 11
Mode = l + [(fm – f1)/(2fm – f1 – f2)] × h
Substitute the values
Mode = 40 + [(20 – 12)/ (40 – 12 – 11)] × 10
= 40 + (80/17)
= 40 + 4.7
= 44.7
Thus, the mode of the given data is 44.7 cars.
In this exercise of NCERT Class 10 Solutions, students will solve questions based on finding the mode of grouped data and the mean value. In a grouped frequency distribution, we cannot find the mode with the help of given frequencies, but we can locate a class with the maximum frequency called the modal class. The mode is a value inside the modal class, and its formula is,
where l = lower limit of the modal class,
h = size of the class interval (assuming all class sizes to be equal),
f1 = frequency of the modal class,
f0= frequency of the class preceding the modal class,
f2 = frequency of the class succeeding the modal class.
Approach us to practise the exercise problems with their solutions for class 10 Maths, Chapter 14. Also, get some more materials, such as notes, books, questions papers, etc., to make practice more effective.
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