- Flash back from IX Class
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- Mean for Ungroup Frequency table
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- Mean for group Frequency table
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- Various ways to calculate mean
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- Mode for grouped frequency table
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- Cumulative Frequency chart
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- Median of a grouped data frequency table
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- Empirical Formula between Mode, Mean and Median

In this page we have *NCERT Solutions for Class 10 Maths:Chapter 14 Statistics* for
Exercise 14.2 . Hope you like them and do not forget to like , social_share
and comment at the end of the page.

The following table shows the ages of the patients admitted in 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.

odal class = 35 – 45, l = 35, class width (h) = 10, f

Substituting the values, we get

ode= 36.8 year

Now Calculation of Mean

Class Interval |
Frequency (f_{i}) |
Mid-point (x_{i}) |
f_{i}x_{i} |

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 |

∑ f_{i} = 80 |
∑ f_{i}x_{i} = 2830 |

= 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

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.

odal class of the given data is 60–80.

odal class = 60-80, l = 60, f

Substituting these values, we get

ode = 65,625

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 |
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 |

odal class = 1500-2000, l = 1500, f

Substituting these values, we get

ode= Rs 1847.83

Calculation for mean

Class Interval |
f_{i} |
x_{i} |
d_{i} = x_{i} - a |
u_{i} = d_{i}/h |
f_{i} u_{i} |

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 |

∑ f_{i} = 200 |
∑f_{i}u_{i} = -35 |

= 2750 + (35/200) ? 500

= 2750 - 87.50 = 2662.50

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.

Number 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 |

odal class = 30-35, l = 30, f

Substituting these values, we get

ode= 30.6

Calculation for mean by step deviation method ( Assumed mean 32.5)

Class Interval |
f_{i} |
x_{i} |
d_{i} = x_{i} - a |
u_{i} = d_{i}/h |
f_{i} u_{i} |

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 |

∑ f_{i} = 35 |
∑f_{i}u_{i} = -23 |

= 29.2

The given distribution shows the number of runs scored by some top batsmen of the

world in one-day international cricket matches.

Runs scored |
Number of batsmen |

3000-4000 |
4 |

4000-5000 |
18 |

5000-6000 |
9 |

6000-7000 |
7 |

7000-8000 |
6 |

8000-9000 |
3 |

9000-1000 |
1 |

1000-1100 |
1 |

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

ode =4608.7 runs

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 ofcars |
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 |

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 Statistics Exercise 14.2 as pdf

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

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

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