A survey regarding the heights (in cm) of 51 girls of Class X of a school was conducted and the following data was obtained: Height (in cm) Number of girls Less than 140 4 Less than 145 11 Less than 150 29 Less than 155 40 Less than 160 46 Less than 165 51 Find the median height.

Here, we are given less than cumulative frequency table. First, we prepare the frequency table: Height (in cm ) Class interval Number of Girls c . f . Frequency f i Less than 140 135 - 140 4 4 Less than 145 140 - 145 11 11 - 4 = 7 Less than 150 145 - 150 29 29 - 11 = 18 Less than 155 150 - 155 40 40 - 29 = 11 Less than 160 155 - 160 46 46 - 40 = 6 Less than 165 160 - 165 51 51 - 46 = 5 We have, N = 51 So, N 2 = 51 2 = 25 . 5 The cumulative frequency just greater than 25 . 5 is 29 , which corresponds to the class 145 - 150 . Thus, the median class is 145 - 150 . Here, l = 145 (the lower limit of the median class), f = 18 (frequency of the median class), h = 5 (width of each class), N = 51 (number of observations), and C = 11 (cumulative frequency of class preceding the median class) ∴ Median = l + N 2 - C f × h = 145 + 25 . 5 - 11 18 × 5 = 145 + 14 . 5 18 × 5 = 145 + 4 . 03 = 149 . 03 approx Hence, the median is 149 . 03 .

The median of the following data is 525 . Find the values of x and y , if the total frequency is 100 . Class Interval Frequency 0 - 100 2 100 - 200 5 200 - 300 x 300 - 400 12 400 - 500 17 500 - 600 20 600 - 700 y 700 - 800 9 800 - 900 7 900 - 1000 4

Cumulative frequency for the given data; Class interval Frequency f Cumulative frequency CF 0 - 100 2 2 100 - 200 5 7 200 - 300 x x + 7 300 - 400 12 x + 19 400 - 500 17 x + 36 500 - 600 20 x + 56 600 - 700 y x + y + 56 700 - 800 9 x + y + 65 800 - 900 7 x + y + 72 900 - 1000 4 x + y + 76 total Σ f = 76 + x + y = 100 Given that the median is 525 . So median class is 500 - 600 Now median = l + N 2 - C F f × h where l = 500 , h = 100 , N = 100 , C F = 36 + x , f = 20 so, 525 = 500 + 100 2 - ( 36 + x ) 20 × 100 ⇒ 525 - 500 = ( 50 - ( 36 + x ) ) × 5 ⇒ 25 = ( 14 - x ) × 5 ⇒ x = 9 Also, Σ f = 76 + x + y = 100 ⇒ 9 + y = 100 - 76 ⇒ y = 15 Hence, x = 9 and y = 15

The annual profits earned by 30 shops of a shopping complex in a locality give rise to the following distribution: Profit (in lakhs in ₹) Number of shops More than or equal to 5 30 More than or equal to 10 28 More than or equal to 15 16 More than or equal to 20 14 More than or equal to 25 10 More than or equal to 30 7 More than or equal to 35 3 Draw both ogives for the data above. Hence obtain the median profit.

More than method: Profit (in lakhs in ₹) Number of shops More than equal to 5 30 More than equal to 10 28 More than equal to 15 16 More than equal to 20 14 More than equal to 25 10 More than equal to 30 7 More than equal to 35 3 Now, we mark the lower class limits on x -axis and the more than cumulative frequencies along y -axis. Thus, we plot the points ( 5 , 30 ) , ( 10 , 28 ) , ( 15 , 16 ) , ( 20 , 14 ) , ( 25 , 10 ) , ( 30 , 7 ) & ( 35 , 3 ) . Less than method: Profit (in lakhs in ₹) Number of shops (frequency) Profit less than Cumulative frequency 0 - 10 2 10 2 10 - 15 12 15 14 15 - 20 2 20 16 20 - 25 4 25 20 25 - 30 3 30 23 30 - 35 4 35 27 35 - 40 3 40 30 Now, we mark the upper class limits on x -axis and the less than cumulative frequencies along y -axis. Thus, we plot the points ( 10 , 2 ) , ( 15 , 14 ) , ( 20 , 16 ) , ( 25 , 20 ) , ( 30 , 23 ) , ( 35 , 27 ) & ( 40 , 30 ) . We find that the two types of cumulative frequency curves intersect at point P . From point P , we draw perpendicular PM to x -axis. The value of profit corresponding to M is ₹ 17 . 5 lakh . Hence, the median is ₹ 17 . 5 lakh .

Karnataka Board Solutions for Chapter: Solved Examples, Exercise 14: Exercise 14

Author:Karnataka Board

Karnataka Board Mathematics Solutions for Exercise - Karnataka Board Solutions for Chapter: Solved Examples, Exercise 14: Exercise 14

Attempt the free practice questions on Chapter NA: Solved Examples, Exercise 14: Exercise 14 with hints and solutions to strengthen your understanding. Mathematics 10 : Part II solutions are prepared by Experienced Embibe Experts.

Questions from Karnataka Board Solutions for Chapter: Solved Examples, Exercise 14: Exercise 14 with Hints & Solutions

HARD

10th Karnataka Board

IMPORTANT

Mathematics 10 : Part II>Chapter NA - Solved Examples>Exercise 14>Q 1

The marks obtained by $30$ students of Class $X$ of a certain school in Mathematics paper consisting of $100$ marks are presented in the table below. Find the mean of the marks obtained by the students.

Marks obtained $(x_{i})$	$10$	$20$	$36$	$40$	$50$	$56$	$60$	$70$	$72$	$80$	$88$	$92$	$95$
Number of students $(f_{i})$	$1$	$1$	$3$	$4$	$3$	$2$	$4$	$4$	$1$	$1$	$2$	$3$	$1$

HARD

10th Karnataka Board

IMPORTANT

Mathematics 10 : Part II>Chapter NA - Solved Examples>Exercise 14>Q 3

The distribution below shows the number of wickets taken by bowlers in one-day cricket matches. Find the mean number of wickets by choosing a suitable method. What does the mean signify?

Number of wickets	$20 - 60$	$60 - 100$	$100 - 150$	$150 - 250$	$250 - 350$	$350 - 450$
Number of bowlers	$7$	$5$	$16$	$12$	$2$	$3$

MEDIUM

10th Karnataka Board

IMPORTANT

Mathematics 10 : Part II>Chapter NA - Solved Examples>Exercise 14>Q 4

The wickets taken by a bowler in $10$ cricket matches are as follows:

2

6

4

5

0

2

1

3

2

3

Find the mode of the data.

HARD

10th Karnataka Board

IMPORTANT

Mathematics 10 : Part II>Chapter NA - Solved Examples>Exercise 14>Q 5

A survey conducted on $20$ households in a locality by a group of students resulted in the following frequency table for the number of family members in a household:

Family size	$1 - 3$	$3 - 5$	$5 - 7$	$7 - 9$	$9 - 11$
Number of families	$7$	$8$	$2$	$2$	$1$

Find the mode of this data.

HARD

10th Karnataka Board

IMPORTANT

Mathematics 10 : Part II>Chapter NA - Solved Examples>Exercise 14>Q 6

The marks distribution of $30$ students in a mathematics examination is given in the table. Find the mode of this data. Also, compare and interpret the mode and the mean.

Class interval	$10 - 25$	$25 - 40$	$40 - 55$	$55 - 70$	$70 - 85$	$85 - 100$
Number of students	$2$	$3$	$7$	$6$	$6$	$6$

MEDIUM

10th Karnataka Board

IMPORTANT

Mathematics 10 : Part II>Chapter NA - Solved Examples>Exercise 14>Q 7

A survey regarding the heights (in cm) of $51$ girls of Class $X$ of a school was conducted and the following data was obtained:

Height (in cm)	Number of girls
Less than $140$	$4$
Less than $145$	$11$
Less than $150$	$29$
Less than $155$	$40$
Less than $160$	$46$
Less than $165$	$51$

Find the median height.

HARD

10th Karnataka Board

IMPORTANT

Mathematics 10 : Part II>Chapter NA - Solved Examples>Exercise 14>Q 8

The median of the following data is $525$ . Find the values of $x$ and $y$ , if the total frequency is $100$ .

Class Interval	Frequency
$0 - 100$	$2$
$100 - 200$	$5$
$200 - 300$	$x$
$300 - 400$	$12$
$400 - 500$	$17$
$500 - 600$	$20$
$600 - 700$	$y$
$700 - 800$	$9$
$800 - 900$	$7$
$900 - 1000$	$4$

HARD

10th Karnataka Board

IMPORTANT

Mathematics 10 : Part II>Chapter NA - Solved Examples>Exercise 14>Q 9

The annual profits earned by $30$ shops of a shopping complex in a locality give rise to the following distribution:

Profit (in lakhs in ₹)	Number of shops
More than or equal to $5$	$30$
More than or equal to $10$	$28$
More than or equal to $15$	$16$
More than or equal to $20$	$14$
More than or equal to $25$	$10$
More than or equal to $30$	$7$
More than or equal to $35$	$3$

Draw both ogives for the data above. Hence obtain the median profit.

Karnataka Board Solutions for Chapter: Solved Examples, Exercise 14: Exercise 14

Karnataka Board Mathematics Solutions for Exercise - Karnataka Board Solutions for Chapter: Solved Examples, Exercise 14: Exercise 14

Questions from Karnataka Board Solutions for Chapter: Solved Examples, Exercise 14: Exercise 14 with Hints & Solutions

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