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.

$\text{Mode} = l + \left ( \frac {f_1 - f_0}{2f_1 - f_1-f_2} \right ) \times h$

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.

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.

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, f

Substituting the values, we get

Mode= 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 |

$M= \frac { \sum f_i x_i}{\sum f_i}$

= 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

The following data gives the information on the observed lifetimes (in hours) of 225

electrical components:

Determine the modal lifetimes of the components.

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, f

Substituting these values, we get

Mode = 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:

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, f

Substituting these values, we get

Mode= 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 |

$M= a + \frac { \sum f_i u_i}{\sum f_i} \times h$

= 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.

$\text{Mode} = l + \left ( \frac {f_1 - f_0}{2f_1 - f_1-f_2} \right ) \times h$

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, f

Substituting these values, we get

Mode= 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 |

$M= a + \frac { \sum f_i u_i}{\sum f_i} \times h$

= 29.2

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.

$\text{Mode} = l + \left ( \frac {f_1 - f_0}{2f_1 - f_1-f_2} \right ) \times h$

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

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 |

$\text{Mode} = l + \left ( \frac {f_1 - f_0}{2f_1 - f_1-f_2} \right ) \times h$

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

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.

Thanks for visiting our website.

**DISCLOSURE:** THIS PAGE MAY CONTAIN AFFILIATE LINKS, MEANING I GET A COMMISSION IF YOU DECIDE TO MAKE A PURCHASE THROUGH MY LINKS, AT NO COST TO YOU. PLEASE READ MY **DISCLOSURE**Â FOR MORE INFO.