Find the median from the given frequency distribution. Class Interval 70 - 74 75 - 79 80 - 84 85 - 89 90 - 94 95 - 99 105 - 109 110 - 114 115 - 119 Frequency 15 25 30 4 10 8 6 4 3

Given data is Class Interval 70 - 74 75 - 79 80 - 84 85 - 89 90 - 94 95 - 99 105 - 109 110 - 114 115 - 119 Frequency 15 25 30 4 10 8 6 4 3 Now, let prepare the cumulative frequency distribution table for the given data, Class Interval Frequency Cumulative frequency 69 . 5 - 74 . 5 15 15 74 . 5 - 79 . 5 25 15 + 25 = 40 79 . 5 - 84 . 5 30 40 + 30 = 70 84 . 5 - 89 . 5 4 70 + 4 = 74 89 . 5 - 94 . 5 10 10 + 74 = 84 94 . 5 - 99 . 5 8 84 + 8 = 93 99 . 5 - 105 . 5 6 93 + 6 = 99 105 . 5 - 109 . 5 4 99 + 4 = 103 109 . 5 - 114 . 5 3 103 + 3 = 106 N = ∑ f = 106 N 2 = 106 2 = 53 ∴ Median = l + N 2 - F f m × i , where l is the boundary of the median class, F is the cumulative frequency of the class preceding the median class, f m is the frequency of the median class, i is the length of the median class, N = Total frequency Here, l = 79 . 5 , F = 40 , f m = 30 , i = 79 . 5 - 84 . 5 = 5 , N = 106 On substituting the values, we get, Median = 79 . 5 + 53 - 40 30 × 5 = 79 . 5 + 13 6 = 81 . 6 Hence, the median for the given data is 81 . 6 .

From the given data, find the median of the height of the students of a class. Height cm 161 - 167 167 - 173 173 - 179 179 - 185 185 - 191 191 - 197 Number of students 79 92 60 22 5 3

Given data is Height cm 161 - 167 167 - 173 173 - 179 179 - 185 185 - 191 191 - 197 Number of students 79 92 60 22 5 3 Now, let us prepare the frequency distribution table for the given data. Height cm Number of students f Cumulative frequency 161 - 167 79 79 167 - 173 ( Median class ) 92 79 + 92 = 171 173 - 179 60 171 + 60 = 231 179 - 185 22 231 + 22 = 253 185 - 191 5 253 + 5 = 258 191 - 197 3 258 + 3 = 261 N = ∑ f = 261 N 2 = 261 2 = 130 . 5 ∴ Median = l + N 2 - F f m × i , where l is the boundary of the median class, F is the cumulative frequency of the class preceding the median class, f m is the frequency of the median class, i is the length of the median class, N = Total frequency Here, l = 167 , F = 79 , f m = 92 , i = 173 - 167 = 6 , N 2 = 130 . 5 On substituting the values, we get, Median = 167 + 130 . 5 - 79 92 × 6 = 167 + 51 . 5 92 × 6 = 167 + 3 . 36 = 170 . 36 Hence, the median for the given data is 170 . 36 , approximately.

From the given data, if the median is 46 , then find x and y when the sum of frequencies is 230 . Class boundary 10 - 20 20 - 30 30 - 40 40 - 50 50 - 60 60 - 70 70 - 80 Frequency 12 30 x 65 y 25 18

Given data is Class boundary 10 - 20 20 - 30 30 - 40 40 - 50 50 - 60 60 - 70 70 - 80 Frequency 12 30 x 65 y 25 18 Now, let us prepare the frequency distribution table for the given data. Class boundary Frequency f Cumulative frequency 10 - 20 12 12 20 - 30 30 12 + 30 = 42 30 - 40 x 42 + x 40 - 50 ( Median class ) 65 42 + x + 65 = 107 + x 50 - 60 y 107 + x + y 60 - 70 25 132 + x + y 70 - 80 18 18 + 132 + x + y = 150 + x + y N = 150 + x + y N 2 = 150 + x + y 2 ∴ Median = l + N 2 - F f m × i , where l is the boundary of the median class, F is the cumulative frequency of the class preceding the median class, f m is the frequency of the median class, i is the length of the median class, N = Total frequency Here, l = 40 , F = 42 + x , f m = 65 , i = 50 - 40 = 10 , N 2 = 150 + x + y 2 On substituting the values, we get, Median = 40 + 150 + x + y 2 - 42 + x 65 × 10 ⇒ 46 = 40 + 150 + x + y - 84 - 2 x 130 × 10 ⇒ 46 = 40 + 66 - x + y 13 ⇒ 78 = 66 - x + y ⇒ - x + y = 78 - 66 = 12 ⇒ y - x = 12 . . . . . . . . . . . . . . i Given that the sum of frequencies is 230 . ⇒ N = x + y + 150 = 230 ⇒ x + y = 230 - 150 = 80 . . . . . . . . . . . . . . . . ii Adding equations i and ii , we get, 2 y = 92 ⇒ y = 92 2 = 46 Substituting the value of y in equation i , we get, 46 - x = 12 ⇒ - x = 12 - 46 = - 34 ⇒ x = 34 Hence, x = 34 and y = 46 .

Bidyut Guha Thakurta and Amitava Mitra Solutions for Exercise 2: EXERCISE 1B

Author: Bidyut Guha Thakurta & Amitava Mitra

Bidyut Guha Thakurta Mathematics Solutions for Exercise - Bidyut Guha Thakurta and Amitava Mitra Solutions for Exercise 2: EXERCISE 1B

Attempt the practice questions from Exercise 2: EXERCISE 1B with hints and solutions to strengthen your understanding. SCHOOL MATHEMATICS Class 10 solutions are prepared by Experienced Embibe Experts.

Questions from Bidyut Guha Thakurta and Amitava Mitra Solutions for Exercise 2: EXERCISE 1B with Hints & Solutions

HARD

10th West Bengal Board

IMPORTANT

SCHOOL MATHEMATICS Class 10>Chapter 9 - Measures of Central Tendency>EXERCISE 1B>Q 4

Find the median from the given frequency distribution.

$x$	$1$	$2$	$3$	$4$	$5$	$6$
$f$	$7$	$12$	$17$	$19$	$21$	$24$

HARD

10th West Bengal Board

IMPORTANT

SCHOOL MATHEMATICS Class 10>Chapter 9 - Measures of Central Tendency>EXERCISE 1B>Q 5

Find the median from the given data.

Income $(₹)$	$100$	$150$	$80$	$200$	$250$	$180$
Number of persons	$24$	$26$	$16$	$20$	$6$	$30$

HARD

10th West Bengal Board

IMPORTANT

SCHOOL MATHEMATICS Class 10>Chapter 9 - Measures of Central Tendency>EXERCISE 1B>Q 6

Find the median from the given frequency distribution.

Marks obtained	$0 - 10$	$10 - 30$	$30 - 60$	$60 - 70$	$70 - 90$
Number of female students	$15$	$25$	$30$	$4$	$10$

HARD

10th West Bengal Board

IMPORTANT

SCHOOL MATHEMATICS Class 10>Chapter 9 - Measures of Central Tendency>EXERCISE 1B>Q 7

Find the median from the given frequency distribution.

Class Interval	$70 - 74$	$75 - 79$	$80 - 84$	$85 - 89$	$90 - 94$	$95 - 99$	$105 - 109$	$110 - 114$	$115 - 119$
Frequency	$15$	$25$	$30$	$4$	$10$	$8$	$6$	$4$	$3$

HARD

10th West Bengal Board

IMPORTANT

SCHOOL MATHEMATICS Class 10>Chapter 9 - Measures of Central Tendency>EXERCISE 1B>Q 8

Find the median from the given data.

Mid-value	$15$	$25$	$35$	$45$	$55$	$65$	$75$	$85$	$95$
Frequency	$6$	$8$	$10$	$2$	$4$	$7$	$9$	$4$	$6$

HARD

10th West Bengal Board

IMPORTANT

SCHOOL MATHEMATICS Class 10>Chapter 9 - Measures of Central Tendency>EXERCISE 1B>Q 9

From the given data, find the median of the height of the students of a class.

Height $(cm)$	$161 - 167$	$167 - 173$	$173 - 179$	$179 - 185$	$185 - 191$	$191 - 197$
Number of students	$79$	$92$	$60$	$22$	$5$	$3$

HARD

10th West Bengal Board

IMPORTANT

SCHOOL MATHEMATICS Class 10>Chapter 9 - Measures of Central Tendency>EXERCISE 1B>Q 10

An incomplete frequency distribution table is given. If from the complete distribution the median obtained is $53.1 kg$ , then find the missing frequency $x$ .

Weight $(kg)$	$46 - 48$	$49 - 51$	$52 - 54$	$55 - 57$	$58 - 60$
Number of girls	$5$	$8$	$15$	$x$	$4$

HARD

10th West Bengal Board

IMPORTANT

SCHOOL MATHEMATICS Class 10>Chapter 9 - Measures of Central Tendency>EXERCISE 1B>Q 11

From the given data, if the median is $46$ , then find $x$ and $y$ when the sum of frequencies is $230$ .

Class boundary	$10 - 20$	$20 - 30$	$30 - 40$	$40 - 50$	$50 - 60$	$60 - 70$	$70 - 80$
Frequency	$12$	$30$	$x$	$65$	$y$	$25$	$18$

Bidyut Guha Thakurta and Amitava Mitra Solutions for Exercise 2: EXERCISE 1B

Bidyut Guha Thakurta Mathematics Solutions for Exercise - Bidyut Guha Thakurta and Amitava Mitra Solutions for Exercise 2: EXERCISE 1B

Questions from Bidyut Guha Thakurta and Amitava Mitra Solutions for Exercise 2: EXERCISE 1B with Hints & Solutions

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