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NCERT Solutions for Class 10 Maths:Chapter 14 Statistics Exercise 14.2



Statistics Exercise 14.2

In this page we have NCERT Solutions for Class 10 Maths:Chapter 14 Statistics for Exercise 14.2 on pages 275,276. Hope you like them and do not forget to like , social_share and comment at the end of the page.
Formula The mode for grouped data can be found by using the formula
$\text{Mode} = l + \left ( \frac {f_1 - f_0}{2f_1 - f_1-f_2} \right ) \times h$
Where
l = lower limit of the modal class
h = size of the class interval (assuming all class sizes to be equal)
$f_1$ = frequency of the modal class
$f_0$ = frequency of the class preceding the modal class
$f_2$ = frequency of the class succeeding the modal class.
Question 1
The following table shows the ages of the patients admitted in a hospital during a year
NCERT Solutions for Class 10 Maths:Chapter 14 Statistics Exercise 14.2

Find the mode and the mean of the data given above. Compare and interpret the two measures of central tendency.
Solution
The mode for grouped data can be found by using the formula
$\text{Mode} = l + \left ( \frac {f_1 - f_0}{2f_1 - f_1-f_2} \right ) \times h$
Where
l = lower limit of the modal class
h = size of the class interval (assuming all class sizes to be equal)
$f_1$ = frequency of the modal class
$f_0$ = frequency of the class preceding the modal class
$f_2$ = frequency of the class succeeding the modal class.
Modal class = 35 – 45, l = 35, class width (h) = 10, f1 = 23, f0 = 21 and f2 = 14
Substituting the values, we get
Mode= 36.8 year
Now Calculation of Mean
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
  ∑ fi = 80
  ∑ fixi = 2830

$M= \frac { \sum f_i x_i}{\sum f_i}$
= 2830/80 = 35.37 yr
So, Maximum number of patients are of age 36.8 year and Mean age of the patient is 35.37 year
Question 2
The following data gives the information on the observed lifetimes (in hours) of 225
electrical components:



Determine the modal lifetimes of the components.
Solution
The mode for grouped data can be found by using the formula
$\text{Mode} = l + \left ( \frac {f_1 - f_0}{2f_1 - f_1-f_2} \right ) \times h$
Where
l = lower limit of the modal class
h = size of the class interval (assuming all class sizes to be equal)
$f_1$ = frequency of the modal class
$f_0$ = frequency of the class preceding the modal class
$f_2$ = frequency of the class succeeding the modal class.
Modal class of the given data is 60–80.
Modal class = 60-80, l = 60, f1 = 61, f0 = 52, f2 = 38 and h = 20
Substituting these values, we get
Mode = 65,625

Question 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:

 

Solution
The mode for grouped data can be found by using the formula
$\text{Mode} = l + \left ( \frac {f_1 - f_0}{2f_1 - f_1-f_2} \right ) \times h$
Where
l = lower limit of the modal class
h = size of the class interval (assuming all class sizes to be equal)
$f_1$ = frequency of the modal class
$f_0$ = frequency of the class preceding the modal class
$f_2$ = frequency of the class succeeding the modal class.
Modal class = 1500-2000, l = 1500, f1 = 40, f0 = 24, f2 = 33 and h = 500
Substituting these values, we get
Mode= Rs 1847.83
Calculation for mean
 
Class Interval
fi
xi
di = xi - a
ui = di/h
fi ui
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
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

$M= a + \frac { \sum f_i u_i}{\sum f_i} \times h$
= 2750 + (35/200) × 500
= 2750 - 87.50 = 2662.50
 

Question 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.

 
Solution
$\text{Mode} = l + \left ( \frac {f_1 - f_0}{2f_1 - f_1-f_2} \right ) \times h$
Where
l = lower limit of the modal class
h = size of the class interval (assuming all class sizes to be equal)
$f_1$ = frequency of the modal class
$f_0$ = frequency of the class preceding the modal class
$f_2$ = frequency of the class succeeding the modal class.
Modal class = 30-35, l = 30, f1 = 10, f0 = 9, f2 = 3 and h = 500
Substituting these values, we get
Mode= 30.6
Calculation for mean  by step deviation method ( Assumed mean 32.5)
 
Class Interval
fi
xi
di = xi - a
ui = di/h
fi ui
15 − 20
3
17.5
-15
-3
-9
20 -  25
8
22.5
-10
-2
-16
25 - 30
9
27.5
-5
-1
-9
30 - 35
10
32.5
0
0
0
35 - 40
3
37.7
5
1
3
40 – 45
0
42.5
10
2
0
45 – 50
0
47.5
15
3
0
50 - 55
2
52.5
20
4
8
  ∑ fi  = 35
      ∑fiui = -23

$M= a + \frac { \sum f_i u_i}{\sum f_i} \times h$
= 29.2
 
Question 5
The given distribution shows the number of runs scored by some top batsmen of the
world in one-day international cricket matches.

Find the mode of the data.
Solution
$\text{Mode} = l + \left ( \frac {f_1 - f_0}{2f_1 - f_1-f_2} \right ) \times h$
Where
l = lower limit of the modal class
h = size of the class interval (assuming all class sizes to be equal)
$f_1$ = frequency of the modal class
$f_0$ = frequency of the class preceding the modal class
$f_2$ = frequency of the class succeeding the modal class.
From the given data, it can be observed that the maximum class frequency is 18,
belonging to class interval 4000 − 5000. Therefore, modal class = 4000 − 5000
Lower limit (l) of modal class = 4000
Frequency (f1) of modal class = 18
Frequency (f0) of class preceding modal class = 4
Frequency (f2) of class succeeding modal class = 9
Class size (h) = 1000
Substituting these values, we get
Mode =4608.7 runs
 
Question 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
0-10
10-20
20-30
30-40
40-50
50-60
60-70
70-80
Frequency
7
14
13
12
20
11
15
8
 
Solution
$\text{Mode} = l + \left ( \frac {f_1 - f_0}{2f_1 - f_1-f_2} \right ) \times h$
Where
l = lower limit of the modal class
h = size of the class interval (assuming all class sizes to be equal)
$f_1$ = frequency of the modal class
$f_0$ = frequency of the class preceding the modal class
$f_2$ = frequency of the class succeeding the modal class.
Here maximum class frequency is 20, belonging to 40 − 50 class intervals. Therefore, modal class = 40 − 50
Lower limit (l) of modal class = 40
Frequency (f1) of modal class = 20
Frequency (f0) of class preceding modal class = 12
Frequency (f2) of class succeeding modal class = 11
Substituting these values, we get
ode =44.7
 
 
Download Class 10 Statistics Exercise 14.2 as pdf


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Reference Books for class 10

Given below are the links of some of the reference books for class 10 math.

  1. Oswaal CBSE Question Bank Class 10 Hindi B, English Communication Science, Social Science & Maths (Set of 5 Books)
  2. Mathematics for Class 10 by R D Sharma
  3. Pearson IIT Foundation Maths Class 10
  4. Secondary School Mathematics for Class 10
  5. Xam Idea Complete Course Mathematics Class 10

You can use above books for extra knowledge and practicing different questions.


Class 10 Maths Class 10 Science

Practice Question

Question 1 What is $1 - \sqrt {3}$ ?
A) Non terminating repeating
B) Non terminating non repeating
C) Terminating
D) None of the above
Question 2 The volume of the largest right circular cone that can be cut out from a cube of edge 4.2 cm is?
A) 19.4 cm3
B) 12 cm3
C) 78.6 cm3
D) 58.2 cm3
Question 3 The sum of the first three terms of an AP is 33. If the product of the first and the third term exceeds the second term by 29, the AP is ?
A) 2 ,21,11
B) 1,10,19
C) -1 ,8,17
D) 2 ,11,20






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