NCERT Solutions For Statistics Class 10 Chapter 14
NCERT Solutions for Class 10 Maths Chapter 14 Statistics are provided here. These NCERT Solutions are created by expert team at Careers360 keeping in mind latest syllabus and pattern of CBSE 202324. Students preparing for the Class 10 board exams must go through the Statistics Class 10 Maths chapter 14 NCERT solutions. This is one of the most important chapters in NCERT books for Class 10 Maths. The NCERT solutions for Class 10 Maths Chapter 14 Statistics covers each and every question of the exercise comprehensively and provide step by step solutions.
NCERT solutions for class 10 maths Chapter 14 Statistics Also See,
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NCERT Solutions For Statistics Class 10 Chapter 14  Important Formulae
Measures of Central Tendency  Mean, Median, and Mode:
>> Mean:
 Direct Method: The mean, represented as X, can be calculated directly using the formula:
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X = ∑(f_{i}x_{i}) / ∑f_{i}
Where 'f_{i}' denotes the frequency of the value 'x_{i}'.
X = a + ∑(f_{i}d_{i}) / ∑f_{i}
Where 'a' is an assumed mean and 'd_{i}' is the deviation of each value 'x_{i}' from the assumed mean.
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As per latest 2024 syllabus. Maths formulas, equations, & theorems of class 11 & 12th chapters
X = a + [∑(f_{i}u_{i}) / ∑f_{i}]h
Where 'u_{i}' represents the step deviations and 'h' is the class interval.
>> Median:
The median is the central value in a set of observations, and its calculation depends on the number of observations.
For an odd number of observations,
Median = [value of the (n+1)/2]th term in the ordered set
Median = average of the middle two values
>> Mode:
Mode = l + [(f_{1}  f_{0})/(2f_{1}  f_{0}  f_{2})] h
Where,
l = lower limit of the modal class
f_{1} = frequency of the modal class
f_{0} = frequency of the class before the modal class
f_{2} = frequency of the class after the modal class
H = Class interval
Free download NCERT Solutions for Class 10 Maths Chapter 14 Statistics PDF for CBSE Exam.
NCERT Solutions for Class 10 Maths Chapter 14 Statistics (Intext Questions and Exercise)
Answer:
Let the assumed mean be a = 75.5
No. of heartbeats per minute  Number of women  Classmark   
6568  2  66.5  9  18 
6871  4  69.5  6  24 
7174  3  72.5  3  9 
7477  8  75.5  0  0 
7780  7  78.5  3  21 
8083  4  81.5  6  24 
8386  2  84.5  9  18 
 =30 

 =12 
Mean,
Therefore, the mean heartbeats per minute of these women are 75.9
NCERT solutions for class 10 maths chapter 14 Statistics Excercise: 14.2
Q1 The following table shows the ages of the patients admitted in a hospital during a year:
Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency.
Answer:
The class having maximum frequency is the modal class.
The maximum frequency is 23 and hence the modal class = 3545
Lower limit (l) of modal class = 35, class size (h) = 10
Frequency ( ) of the modal class = 23, frequency ( ) of class preceding the modal class = 21, frequency ( ) of class succeeding the modal class = 14.
Now,
Age
 Number of patients  Classmark  
515  6  10  60 
1525  11  20  220 
2535  21  30  630 
3545  23  40  920 
4555  14  50  700 
5565  5  60  300 
 =80 
 =2830 
Mean,
The maximum number of patients are in the age group of 36.8, whereas the average age of all the patients is 35.37.
Statistics Class 10 Maths Chapter 14 Solution Excercise: 14.3
Q1 The following frequency distribution gives the monthly consumption of electricity of 68 consumers of a locality. Find the median, mean and mode of the data and compare them.
Answer:
Let the assumed mean be a = 130 and h = 20
Class  Number of consumers  Cumulative Frequency  Classmark    
6585  4  4  70  60  3  12 
85105  5  9  90  40  2  10 
105125  13  22  110  20  1  13 
125145  20  42  130  0  0  0 
145165  14  56  150  20  1  14 
165185  8  64  170  40  2  16 
185205  4  68  190  60  3  12 

 = 68 


 = 7 
MEDIAN:
Median class = 125145; Cumulative Frequency = 42; Lower limit, l = 125;
c.f. = 22; f = 20; h = 20
Thus, the median of the data is 137
MODE:
The class having maximum frequency is the modal class.
The maximum frequency is 20 and hence the modal class = 125145
Lower limit (l) of modal class = 125, class size (h) = 20
Frequency ( ) of the modal class = 20; frequency ( ) of class preceding the modal class = 13, frequency ( ) of class succeeding the modal class = 14.
Thus, Mode of the data is 135.76
MEAN:
Mean,
Thus, the Mean of the data is 137.05
Q2 If the median of the distribution given below is 28.5, find the values of x and y.
Answer:
Class  Number of consumers  Cumulative Frequency 
010  5  5 
1020  x  5+x 
2030  20  25+x 
3040  15  40+x 
4050  y  40+x+y 
5060  5  45+x+y 
 = 60 

Now,
Given median = 28.5 which lies in the class 2030
Therefore, Median class = 2030
Frequency corresponding to median class, f = 20
Cumulative frequency of the class preceding the median class, c.f. = 5 + x
Lower limit, l = 20; Class height, h = 10
Also,
Therefore, the required values are: x=8 and y=7
Q4 The lengths of 40 leaves of a plant are measured correct to the nearest millimeter, and the data obtained is represented in the following table :
Find the median length of the leaves.
(Hint: The data needs to be converted to continuous classes for finding the median since the formula assumes continuous classes. The classes then change to
117.5  126.5, 126.5  135.5, . . ., 171.5  180.5.)
Answer:
The data needs to be converted to continuous classes for finding the median since the formula assumes continuous classes.
Class  Frequency  Cumulative Frequency 
117.5126.5  3  3 
126.5135.5  5  8 
135.5144.5  9  17 
144.5153.5  12  29 
153.5162.5  5  34 
162.5171.5  4  38 
171.5180.5  2  40 
Therefore, Median class = 144.5153.5
Lower limit, l = 144.5; Class height, h = 9
Frequency corresponding to median class, f = 12
Cumulative frequency of the class preceding the median class, c.f. = 17
Thus, the median length of the leaves is 146.75 mm
Q5 The following table gives the distribution of the lifetime of 400 neon lamps :
Find the median lifetime of a lamp.
Answer:
Class  Frequency  Cumulative Frequency 
15002000  14  14 
20002500  56  70 
25003000  60  130 
30003500  86  216 
35004000  74  290 
40004500  62  352 
45005000  48  400 
Therefore, Median class = 30003500
Lower limit, l = 3000; Class height, h = 500
Frequency corresponding to median class, f = 86
Cumulative frequency of the class preceding the median class, c.f. = 130
Thus, the median lifetime of a lamp is 3406.97 hours
Thus, the median length of the leaves is 146.75 mm
Statistics Class 10 Maths  Important Topics
A direct method of finding mean
Assumed mean method of finding mean
Step deviation method of finding mean
Median
Mode
NCERT Solutions for Class 10 subject wise
NCERT solutions for Class 10 maths  Chapter Wise
Key Features of NCERT Statistics Class 10 Maths Solutions Chapter 14
Valuable Student Resources: These materials offer invaluable guidance to students studying statistics in Class 10.
Enhanced Academic Performance: By utilizing class 10 math chapter 14 resources, students can significantly improve their performance on questions related to chapter 14 class 10 maths, ensuring they achieve high marks.
Versatile Study Aid: Whether for exam preparation or chapter review, this resource proves to be a versatile tool for students.
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Also, check  NCERT Solutions for Class 10
How to use NCERT solutions for class 10 maths chapter 14 Statistics?
First of all, go through NCERT Textbook to have an understanding of Statistics Class 10 chapter.
Learn the techniques to calculate the mean, median, mode.
Once you learn the techniques and formulae, then move to the practice exercises available in the Statistics Class 10 chapter.
While practising, if you get stuck anywhere then you can take the help of NCERT Solutions for chapter 14 class 10 maths PDF Download, which cab be used offline.
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