*Maths Ex 14.2 Class 10 is about Mode of Group Data. Here are the Solutions of NCERT Textbook of Maths Class 10 *

## Class 10 Maths Chapter 14

Chapter 14 of Mathematics NCERT Textbook is “STATISTICS” . In this chapter , we study about the Mean of grouped data, medium of grouped data and mode of the grouped data. This chapter also teaches to find the cumulative frequencies values, draw the cumulative frequencies curves , etc . This chapter has 4 exercise which have questions based on statistics. The solution of each exercise of Maths Class 10 NCERT Textbook are provided below.

Class 10 Mathematics Exercise 14.1 – Mean of Grouped Data

Class 10 Mathematics Exercise 14.2 – Mode of Grouped Data

Class 10 Mathematics Exercise 14.3 – Median of Grouped Data

Class 10 Mathematics Exercise 14.4 – Graphical Representation of Cumulative Frequency Distribution

## Maths Ex 14.2 Class 10 NCERT Solutions

Subject | Mathematics |

Class | 10 |

Chapter Name | Chapter 14 ( Statistics ) |

Exercise | Exercise 14.2 (Mode of Grouped Data) |

Total Questions | 6 Questions |

Text Book | NCERT Text Book (Updated Text Book ) |

**Exercise 14.2 Class 10 Maths Question 1 ****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**.

Age (in Years) | 5 -15 | 15 – 25 | 25 – 35 | 35 – 45 | 45 – 55 | 55 – 65 |

Number of Patients | 6 | 11 | 21 | 23 | 14 | 5 |

**Solution : **

First Calculating Mode

Maximum Frequency = 23

Modal Class = 35 – 45

Lower Limit of Modal Frequency (L) =35

Upper Limit of Modal Frequency (H) =

Size (h) = H – L = 45-35 = 10

Frequency of Modal Class (F1) = 23

Frequency of Class proceeding the modal class (F0) = 21

Frequency of Class succeeding the modal Class (F2) = 14

We know the Formula to calculate of Mode

Mode = \( L\ +\ \frac{F_1\ -\ F_0}{2F_1\ -\ F_0\ -\ F_2}\ \times\ h \)

Putting all the values in above formula of Mode,

we get

Mode = \( 35\ +\ \frac{23\ -\ 21}{2 \times\ 23\ -\ 21\ -\ 14}\ \times\ 10 \)

Mode = \( 35\ +\ \frac{2}{11}\ \times\ 10 \)

Mode = 35 + 1.8

Mode = 36.8 years

Now calculating Mean Using Step Deviation Method

let us take a = 40 and h = 10

Age (in years) | Number of patients (f_{i}) | Class Marks (x_{i}) | d_{i} = x_{i} – 40 | u_{i} = (x_{i} – 40) / 10 | f_{i}u_{i} |
---|---|---|---|---|---|

5-15 | 6 | 10 | -30 | -3 | -18 |

15-25 | 11 | 20 | -20 | -2 | -22 |

25-35 | 21 | 30 | -10 | -1 | – 21 |

35-45 | 23 | 40 | 0 | 0 | 0 |

45-55 | 14 | 50 | 10 | 1 | 14 |

55-65 | 5 | 60 | 20 | 2 | 10 |

TOTAL | \( \sum \) f_{i} = 80 | \( \sum \)f_{i}u_{i} = -37 |

Mean = \( a\ +\ \left(\frac{\sum{f_iu_i}}{\sum f_i}\right)\ \times\ h \)

Mean = \( 40\ +\ \left(\frac{-37}{80}\right)\ \times\ 10 \)

Mean = 35.37

**Exercise 14.2 Class 10 Maths Question 2****Q2. The following data gives the information on the observed lifetimes (in hours) of 225 electrical components **. **Determine the modal lifetimes of the components.**

Lifetimes(in hours) | 0 – 20 | 20 – 40 | 40 – 60 | 60 – 80 | 80 – 100 | 100 – 120 |

Frequency | 10 | 35 | 52 | 61 | 38 | 29 |

**Solution :**

Maximum Frequency (F1) = 61

Modal Class = 60-80

Therefore , L = 60 and H = 80

Size (h) =H – L = 80 – 60 = 20

Frequency of Modal Class (F1) = 61

Frequency of Class proceeding the modal class (F0) = 52

Frequency of Class succeeding the modal Class (F2) = 38

Mode = \( L\ +\ \frac{F_1\ -\ F_0}{2F_1\ -\ F_0\ -\ F_2}\ \times\ h \)

Mode = \( 60\ +\ \frac{61\ -\ 52}{2\ \times\ 61\ -\ 52\ -\ 38}\ \times\ 20 \)

Mode = 60 + 5.625 = 65.625

**Exercise 14.2 Class 10 Maths Question 3 ****Q3. 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 :**

Calculating Mode

Class with maximum frequency ( Modal Class ) = 1500 – 2000

L = 1500

H = 2000

Class Size (h) = H – L = 2000 – 1500 = 500

Frequency of Modal Class (F1) = 40

Frequency of Class proceeding the modal class (F0) = 24

Frequency of Class succeeding the modal Class (F2) = 33

Mode = \( L\ +\ \frac{F_1\ -\ F_0}{2F_1\ -\ F_0\ -\ F_2}\ \times\ h \)

Mode = \( 1500\ +\ \frac{40\ -\ 24}{2 \times\ 40\ -\ 24\ -\ 33}\ \times\ 500 \)

Mode = 1500 + 347.826

Mode = 1847.826

Now, Calculating Mean

let a = 2750

Expenditure (in Rs) | Number of Families(f_{i}) | x_{i} | d_{i} = x_{i} – a | u_{i} | 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 = 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 |

Total | \( \sum \) f_{i} =200 | \( \sum \) f_{i}d_{i} = -35 |

Mean = \( a\ +\ \left(\frac{\sum{f_iu_i}}{\sum f_i}\right)\ \times\ h \)

Mean = \( 2750\ +\ \left(\frac{-35}{ 200}\right)\ \times\ 500 \)

Mean = 2750 – 87.5

Mean = 2662.5

**Exercise 14.2 Class 10 Maths Question 4 ****Q4. 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 |

**Solution :**

Calculating Mode

Class with maximum frequency ( Modal Class ) = 30 – 35

L = 30

H = 35

Class Size (h) = H – L = 35 – 30 = 5

Frequency of Modal Class (F1) = 10

Frequency of Class proceeding the modal class (F0) = 9

Frequency of Class succeeding the modal Class (F2) = 3

Mode = \( L\ +\ \frac{F_1\ -\ F_0}{2F_1\ -\ F_0\ -\ F_2}\ \times\ h \)

Mode = \( 30\ +\ \frac{10\ -\ 9}{2 \times\ 10\ -\ 9\ -\ 3}\ \times\ 5 \)

Mode = 30 + 0.625

Mode = 30.628

Now, Calculating Mean

Number of Students per teacher | Number of states / U.T.(f_{i}) | x_{i} | d_{i} = x_{i} – a | u_{i} | 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 =a | 0 | 0 | 0 |

35-40 | 3 | 37.5 | 5 | 1 | 3 |

40-45 | 0 | 48.5 | 10 | 2 | 0 |

45-50 | 0 | 47.5 | 15 | 3 | 0 |

50-55 | 2 | 52.5 | 20 | 4 | 8 |

TOTAL | \( \sum \) f_{i} = 35 | \( \sum \) f_{i}u_{i} = -23 |

Mean = \( a\ +\ \left(\frac{\sum{f_iu_i}}{\sum f_i}\right)\ \times\ h \)

Mean = \( 32.5\ +\ \left(\frac{-23}{ 35}\right)\ \times\ 5 \)

Mean = 32.5 – 3.28

Mean = 29.22

**Exercise 14.2 Class 10 Maths Question 5 ****Q5. 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.**

Runs Scored | Number of Batsmen |
---|---|

3000 -4000 | 4 |

4000-5000 | 18 |

5000-6000 | 9 |

6000-7000 | 7 |

7000-8000 | 6 |

8000-9000 | 3 |

9000-10000 | 1 |

10000-11000 | 1 |

**Solution :**

Calculating Mode

Maximum Frequency = 18

Modal Class = 4000 – 5000

Therefore , L = 4000 and H = 5000

Size (h) =H – L = 5000 – 4000 = 1000

Frequency of Modal Class (F1) = 18

Frequency of Class proceeding the modal class (F0) = 4

Frequency of Class succeeding the modal Class (F2) = 9

Mode = \( L\ +\ \frac{F_1\ -\ F_0}{2F_1\ -\ F_0\ -\ F_2}\ \times\ h \)

Mode = \( 4000\ +\ \frac{18\ -\ 4}{2 \times\ 18\ -\ 4\ -\ 9}\ \times\ 1000 \)

Mode = 4000 + 608.7

Mode = 4608.7

**Exercise 14.2 Class 10 Maths Question 6 ****Q6. A student noted the number of cars passing through a spot on a road for 100 periods each of 3 minutes and summarised 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 :**

Calculating Mode

Maximum Frequency = 20

Modal Class = 40 – 50

Therefore , L = 40 and H = 50

Size (h) =H – L = 50 – 40 = 100

Frequency of Modal Class (F1) = 20

Frequency of Class proceeding the modal class (F0) = 12

Frequency of Class succeeding the modal Class (F2) = 11

Mode = \( L\ +\ \frac{F_1\ -\ F_0}{2F_1\ -\ F_0\ -\ F_2}\ \times\ h \)

Mode = \( 40\ +\ \frac{20\ -\ 12}{2 \times\ 20\ -\ 12\ -\ 11}\ \times\ 10 \)

Mode = 40 + 4.7

Mode = 44.7