Find the median from the following data: Class 1 - 5 6 - 10 11 - 15 16 - 20 21 - 25 26 - 30 31 - 35 36 - 40 41 - 45 Frequency 7 10 16 32 24 16 11 5 2

First we convert the given data into exclusive form, so we get: Class Frequency f Cummulative Frequency C f 0 . 5 - 5 . 5 7 7 5 . 5 - 10 . 5 10 17 10 . 5 - 15 . 5 16 33 15 . 5 - 20 . 5 32 65 20 . 5 - 25 . 5 24 89 25 . 5 - 30 . 5 16 105 30 . 5 - 35 . 5 11 116 35 . 5 - 40 . 5 5 121 40 . 5 - 45 . 5 2 123 N = ∑ f = 123 ∴ l = 15 . 5 , h = 5 , f = 32 cumulative frequency of the preceding class ( cf ) = 33 and N 2 = 61 . 5 .Now, N = 123 ⇒ N 2 = 61 . 5 . The cumulative frequency just greater than 61 . 5 is 65 and the corresponding class is 15 . 5 - 20 . 5 . Thus, the median class is 15 . 5 - 20 . 5 . Now, median is given as, M = l + ( N 2 - cf ) f × h ⇒ 15 . 5 + ( 61 . 5 - 33 ) 32 × 5 ⇒ 15 . 5 + 4 . 45 ⇒ 19 . 95 Hence, the median is 19 . 95 .

Find the median from the following data: Marks No. of students Below 10 12 Below 20 32 Below 30 57 Below 40 80 Below 50 92 Below 60 116 Below 70 164 Below 80 200

Class Cumulative frequency (cf) frequency (f) 0 - 10 12 12 10 - 20 32 20 20 - 30 57 25 30 - 40 80 23 40 - 50 92 12 50 - 60 116 24 60 - 70 164 48 70 - 80 200 36 N = ∑ f = 200 Now, N = 200 ⇒ N 2 = 100 . The cumulative frequency just greater than 100 is 116 and the corresponding class is 50 - 60 . Thus, the median class is 50 - 60 . ∴ l = 50 , h = 10 , f = 24 , cumulative frequency of preceding class ( cf ) = 92 and N 2 = 100 . Now, median is given as, M = l + ( N 2 - cf ) f × h ⇒ 50 + ( 100 - 92 ) 24 × 10 ⇒ 50 + 3 . 33 ⇒ 53 . 33 Hence, the median is 53 . 33 .

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 .

Calculate the mode from the following data: Monthly Salary (in ₹ ) No. of employees 0 - 5000 90 5000 - 10000 150 10000 - 15000 100 15000 - 20000 80 20000 - 25000 70 25000 - 30000 10

Monthly Salary (in ₹ ) No. of employees 0 - 5000 90 5000 - 10000 150 10000 - 15000 100 15000 - 20000 80 20000 - 25000 70 25000 - 30000 10 We know that class interval 5000 - 10000 has the maximum frequency, thus it is the modal class. ∴ x k = 5000 , h = 50000 , f k = 150 , f k - 1 = 90 , f k + 1 = 100 . Now, mode is given as, M 0 = x k + { h × ( f k - f k - 1 ) ( 2 f k - f k - 1 - f k + 1 ) } = 5000 + { 5000 × ( 150 - 90 ) ( 2 × 150 - 90 - 100 ) } = 5000 + { 5000 × 60 110 } = 5000 + 2727 . 27 = 7727 . 27 Hence, the mode is ₹ 7727 . 27 .

MEDIUM

10th CBSE

IMPORTANT

Earn 100

Find the median wages for the following frequency distribution:

Wages per day (in Rs) $61 - 70$ $71 - 80$ $81 - 90$ $91 - 100$ $101 - 110$ $111 - 120$

No. of women workers $5$ $15$ $20$ $30$ $20$ $8$

Important Questions on Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive

MEDIUM

10th CBSE

IMPORTANT

Secondary School Mathematics for Class 10>Chapter 18 - Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive>Exercise 18B>Q 13

Find the median from the following data:

Class	$1 - 5$	$6 - 10$	$11 - 15$	$16 - 20$	$21 - 25$	$26 - 30$	$31 - 35$	$36 - 40$	$41 - 45$
Frequency	$7$	$10$	$16$	$32$	$24$	$16$	$11$	$5$	$2$

MEDIUM

10th CBSE

IMPORTANT

Secondary School Mathematics for Class 10>Chapter 18 - Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive>Exercise 18B>Q 14

Find the median from the following data:

Marks	No. of students
Below $10$	$12$
Below $20$	$32$
Below $30$	$57$
Below $40$	$80$
Below $50$	$92$
Below $60$	$116$
Below $70$	$164$
Below $80$	$200$

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 6

Calculate the mode from the following data:

Monthly Salary (in $₹$ )	No. of employees
$0 - 5000$	$90$
$5000 - 10000$	$150$
$10000 - 15000$	$100$
$15000 - 20000$	$80$
$20000 - 25000$	$70$
$25000 - 30000$	$10$

Wages per day (in Rs)	$61 - 70$	$71 - 80$	$81 - 90$	$91 - 100$	$101 - 110$	$111 - 120$
No. of women workers	$5$	$15$	$20$	$30$	$20$	$8$

Find the median wages for the following frequency distribution: Wages per day (in Rs) 61-70 71-80 81-90 91-100 101-110 111-120 No. of women workers 5 15 20 30 20 8

Important Questions on Mean, Median, Mode of Grouped Data, Cumulative Frequency Graph and Ogive

Learn and Prepare for any exam you want !

Find the median wages for the following frequency distribution:

Wages per day (in Rs) $61 - 70$ $71 - 80$ $81 - 90$ $91 - 100$ $101 - 110$ $111 - 120$

No. of women workers $5$ $15$ $20$ $30$ $20$ $8$