Find the mode of the following distribution: Marks 10 - 20 20 - 30 30 - 40 40 - 50 50 - 60 frequency 12 35 45 25 13

Marks 10 - 20 20 - 30 30 - 40 40 - 50 50 - 60 frequency 12 35 45 25 13 Here, the maximum class frequency is 45 , and the class corresponding to this frequency is 30 - 40 . So, the modal class is 30 - 40 . Now, Modal class is 30 - 40 , lower limit ( l ) of modal class = 30 , class size ( h ) = 10 , frequency ( f 1 ) of the modal class = 45 , frequency ( f 0 ) of class preceding the modal class = 35 , frequency ( f 2 ) of class succeeding the modal class = 25 . Now, let us substitute these values in the mode formula: Mode = l + ( f 1 - f 0 ) 2 f 1 - f 0 - f 2 × h = 30 + 45 - 35 90 - 35 - 25 × 10 = 30 + 10 30 × 10 = 30 + 3 . 33 = 33 . 33 Hence, the mode is 33 . 33 .

Compute the mode of the following data: Class 0 - 20 20 - 40 40 - 60 60 - 80 80 - 100 Frequency 25 16 28 20 5

Class 0 - 20 20 - 40 40 - 60 60 - 80 80 - 100 Frequency 25 16 28 20 5 Here, the maximum class frequency is 28 , and the class corresponding to this frequency is 40 - 60 . So, the modal class is 40 - 60 . Modal class = 40 - 60 , lower limit ( l ) of modal class = 40 , class size ( h ) = 20 , frequency ( f 1 ) of the modal class = 28 , frequency ( f 0 ) of class preceding the modal class = 16 , frequency ( f 2 ) of class succeeding the modal class = 20 . Now, let us substitute these values in the formula: M = l + ( f 1 - f 0 ) 2 f 1 - f 0 - f 2 × h = 40 + ( 28 - 16 ) ( 56 - 16 - 20 ) × 20 = 40 + 12 20 × 20 = 40 + 12 = 52 Hence, the mode is 52 .

Heights of students of class X are given in the following frequency distribution Height (in cm) 150 - 155 155 - 160 160 - 165 165 - 170 170 - 175 Number of students 15 8 20 12 5 Find the modal height. Also, find the mean height. Compared and interpret the two measures of central tendency.

Here, the maximum class frequency is 20 , and the class corresponding to this frequency is 160 - 165 . So, the modal class is 160 - 165 . Now, Modal class = 160 - 165 , lower limit ( l ) of modal class = 160 , class size ( h ) = 5 , frequency ( f 1 ) of the modal class = 20 , frequency ( f 0 ) of class preceding the modal class = 8 , frequency ( f 2 ) of class succeeding the modal class = 12 . Now, let us substitute these values in the formula: Mode = l + f 1 - f 0 2 f 1 - f 0 - f 2 × h . = 160 + ( 20 - 8 ) 2 × 20 - 8 - 12 × 5 = 160 + ( 12 20 ) × 5 = 160 + 3 = 163 Hence, the mode is 163 cm . It represents that the height of maximum number of students is 163 cm . Now, to find the mean let us put the data in the table below: Height ( in cm ) Number of students ( f i ) Class mark ( x i ) f i x i 150 - 155 15 152 . 5 2287 . 5 155 - 160 8 157 . 5 1260 160 - 165 20 162 . 5 3250 165 - 170 12 167 . 5 2010 170 - 175 5 172 . 5 862 . 5 Total ∑ f i = 60 ∑ f i x i = 9670 ∴ Mean = ∑ f i x i ∑ f i Mean = 9670 60 Mean = 161 . 17 . Thus, mean of the given data is 161 . 17 . It represents that on an average, the height of a student is 161 . 17 cm .

Find the mode of the following distribution: Class interval 10 - 14 14 - 18 18 - 22 22 - 26 26 - 30 30 - 34 34 - 38 38 - 42 Frequency 8 6 11 20 25 22 10 4

We know that the class 26 - 30 has the maximum frequency, thus, it is the modal class. Now, x k = 26 , h = 4 , f k = 25 , f k - 1 = 20 , f k + 1 = 22 . Now, let us substitute these values in the mode formula. ∴ M 0 = x k + { h × ( f k - f k - 1 ) 2 f k - f k - 1 - f k + 1 } = 26 + { 4 × 25 - 20 2 × 25 - 20 - 22 } = 26 + { 4 × 5 8 } = 26 + 2 . 5 = 28 . 5 Hence, the mode is 28 . 5 .

Given below is the distribution of total household expenditure of 200 manual workers in a city: Expenditure (in Rs) 1000 - 1500 1500 - 2000 2000 - 2500 2500 - 3000 3000 - 3500 3500 - 4000 4000 - 4500 4500 - 5000 Number of manual workers 24 40 31 28 32 23 17 5 Find the expenditure done by maximum number of manual workers .

Expenditure (in Rs) 1000 - 1500 1500 - 2000 2000 - 2500 2500 - 3000 3000 - 3500 3500 - 4000 4000 - 4500 4500 - 5000 Number of manual workers 24 40 31 28 32 23 17 5 We know that the class 1500 - 2000 has the maximum frequency, thus, it is the modal class. Now, x k = 1500 , h = 500 , f k = 40 , f k - 1 = 24 , f k + 1 = 31 . Now, let us substitute these values in the mode formula. ∴ M 0 = x k + { h × ( f k - f k - 1 ) 2 f k - f k - 1 - f k + 1 } = 1500 + { 500 × 40 - 24 2 × 40 - 24 - 31 } = 1500 + { 500 × 16 25 } = 1500 + 320 = 1820 Hence, the mode is ₹ 1820 .

Compute the mode from the following data: Age (in years) 0 - 5 5 - 10 10 - 15 15 - 20 20 - 25 25 - 30 30 - 35 Number of patients 6 11 18 24 17 13 5

Age (in years) 0 - 5 5 - 10 10 - 15 15 - 20 20 - 25 25 - 30 30 - 35 Number of patients 6 11 18 24 17 13 5 We know that the class 15 - 20 has the maximum frequency, thus, it is the modal class. Now, x k = 15 , h = 5 , f k = 24 , f k - 1 = 18 , f k + 1 = 17 . Now, let us substitute these values in the mode formula. ∴ M 0 = x k + { h × ( f k - f k - 1 ) 2 f k - f k - 1 - f k + 1 } = 15 + { 5 × 24 - 18 2 × 24 - 18 - 17 } = 15 + { 5 × 6 13 } = 15 + 2 . 3 = 17 . 3 Hence, the mode is 17 . 3 years.

Compute the mode from the following series: Size 45 - 55 55 - 65 65 - 75 75 - 85 85 - 95 95 - 105 105 - 115 Frequency 7 12 17 30 32 6 10

We know that the class 85 - 95 has the maximum frequency, thus, it is the modal class. Now, x k = 85 , h = 10 , f k = 32 , f k - 1 = 30 , f k + 1 = 6 . Now, let us substitute these values in the mode formula. ∴ M 0 = x k + { h × ( f k - f k - 1 ) 2 f k - f k - 1 - f k + 1 } = 85 + { 10 × 32 - 30 2 × 32 - 30 - 6 } = 85 + { 10 × 2 28 } = 85 + 0 . 71 = 85 . 71 Hence, the mode is 85 . 71 . “Note: Answer mentioned in the book is incorrect the correct answer is: 85 . 71 ”

Compute the mode from the following data: Class interval 1 - 5 6 - 10 11 - 15 16 - 20 21 - 25 26 - 30 31 - 35 6 - 40 41 - 45 46 - 50 Frequency 3 8 13 18 28 20 13 8 6 4

Clearly, we have to find the mode of the data. The given data is an inclusive series. So, we will convert it to an exclusive form as given below: Compute the mode from the following data: Class interval 0 . 5 - 5 . 5 5 . 5 - 10 . 5 10 . 5 - 15 . 5 15 . 5 - 20 . 5 20 . 5 - 25 . 5 25 . 5 - 30 . 5 30 . 5 - 35 . 5 35 . 5 - 40 . 5 40 . 5 - 45 . 5 45 . 5 - 50 . 5 Frequency 3 8 13 18 28 20 13 8 6 4 We know that the class 20 . 5 - 25 . 5 has the maximum frequency, thus, it is the modal class. Now, x k = 20 . 5 , h = 5 , f k = 28 , f k - 1 = 18 , f k + 1 = 20 . Now, let us substitute these values in the mode formula. ∴ M 0 = x k + { h × ( f k - f k - 1 ) 2 f k - f k - 1 - f k + 1 } = 20 . 5 + { 5 × 28 - 18 2 × 28 - 18 - 20 } = 20 . 5 + { 5 × 10 18 } = 20 . 5 + 2 . 78 = 23 . 28 Hence, the mode is 23 . 28 .

R S Aggarwal and Veena Aggarwal Solutions for Exercise 3: Exercise 18C

Author:R S Aggarwal & Veena Aggarwal

R S Aggarwal Mathematics Solutions for Exercise - R S Aggarwal and Veena Aggarwal Solutions for Exercise 3: Exercise 18C

Attempt the free practice questions from Exercise 3: Exercise 18C with hints and solutions to strengthen your understanding. Secondary School Mathematics for Class 10 solutions are prepared by Experienced Embibe Experts.

Questions from R S Aggarwal and Veena Aggarwal Solutions for Exercise 3: Exercise 18C with Hints & Solutions

MEDIUM

10th CBSE

IMPORTANT

Secondary School Mathematics for Class 10>Chapter 18 - Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive>Exercise 18C>Q 1

Find the mode of the following distribution:

Marks	$10 - 20$	$20 - 30$	$30 - 40$	$40 - 50$	$50 - 60$
frequency	$12$	$35$	$45$	$25$	$13$

MEDIUM

10th CBSE

IMPORTANT

Secondary School Mathematics for Class 10>Chapter 18 - Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive>Exercise 18C>Q 2

Compute the mode of the following data:

Class	$0 - 20$	$20 - 40$	$40 - 60$	$60 - 80$	$80 - 100$
Frequency	$25$	$16$	$28$	$20$	$5$

MEDIUM

10th CBSE

IMPORTANT

Secondary School Mathematics for Class 10>Chapter 18 - Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive>Exercise 18C>Q 3

Heights of students of class X are given in the following frequency distribution

Height (in cm)	$150 - 155$	$155 - 160$	$160 - 165$	$165 - 170$	$170 - 175$
Number of students	$15$	$8$	$20$	$12$	$5$

Find the modal height.
Also, find the mean height. Compared and interpret the two measures of central tendency.

MEDIUM

10th CBSE

IMPORTANT

Secondary School Mathematics for Class 10>Chapter 18 - Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive>Exercise 18C>Q 4

Find the mode of the following distribution:

Class interval	$10 - 14$	$14 - 18$	$18 - 22$	$22 - 26$	$26 - 30$	$30 - 34$	$34 - 38$	$38 - 42$
Frequency	$8$	$6$	$11$	$20$	$25$	$22$	$10$	$4$

MEDIUM

10th CBSE

IMPORTANT

Secondary School Mathematics for Class 10>Chapter 18 - Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive>Exercise 18C>Q 5

Given below is the distribution of total household expenditure of $200$ manual workers in a city:

Expenditure (in Rs)	$1000 - 1500$	$1500 - 2000$	$2000 - 2500$	$2500 - 3000$	$3000 - 3500$	$3500 - 4000$	$4000 - 4500$	$4500 - 5000$
Number of manual workers	$24$	$40$	$31$	$28$	$32$	$23$	$17$	$5$

Find the expenditure done by maximum number of manual workers

.

MEDIUM

10th CBSE

IMPORTANT

Secondary School Mathematics for Class 10>Chapter 18 - Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive>Exercise 18C>Q 7

Compute the mode from the following data:

Age (in years)	$0 - 5$	$5 - 10$	$10 - 15$	$15 - 20$	$20 - 25$	$25 - 30$	$30 - 35$
Number of patients	$6$	$11$	$18$	$24$	$17$	$13$	$5$

MEDIUM

10th CBSE

IMPORTANT

Secondary School Mathematics for Class 10>Chapter 18 - Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive>Exercise 18C>Q 8

Compute the mode from the following series:

Size	$45 - 55$	$55 - 65$	$65 - 75$	$75 - 85$	$85 - 95$	$95 - 105$	$105 - 115$
Frequency	$7$	$12$	$17$	$30$	$32$	$6$	$10$

MEDIUM

10th CBSE

IMPORTANT

Secondary School Mathematics for Class 10>Chapter 18 - Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive>Exercise 18C>Q 9

Compute the mode from the following data:

Class interval	$1 - 5$	$6 - 10$	$11 - 15$	$16 - 20$	$21 - 25$	$26 - 30$	$31 - 35$	$6 - 40$	$41 - 45$	$46 - 50$
Frequency	$3$	$8$	$13$	$18$	$28$	$20$	$13$	$8$	$6$	$4$

R S Aggarwal and Veena Aggarwal Solutions for Exercise 3: Exercise 18C

R S Aggarwal Mathematics Solutions for Exercise - R S Aggarwal and Veena Aggarwal Solutions for Exercise 3: Exercise 18C

Questions from R S Aggarwal and Veena Aggarwal Solutions for Exercise 3: Exercise 18C with Hints & Solutions

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