The length of 30 leaves of a plant is given in millimetre below- Length of leaves ( in mm ) 111 - 120 121 - 130 131 - 140 141 - 150 151 - 160 161 - 170 Number of leaves 3 5 7 9 4 2 Make a histogram to depict the frequency distribution.

To represent the given data by a histogram, we first draw horizontal and vertical axes. Let us consider that the horizontal and vertical axes represent the class-limits and the frequencies of the class-intervals respectively. Now, the given data is converted into continuous class intervals. Length of leaves ( in mm ) 110 . 5 - 120 . 5 120 . 5 - 130 . 5 130 . 5 - 140 . 5 140 . 5 - 150 . 5 150 . 5 - 160 . 5 160 . 5 - 170 . 5 Number of leaves 3 5 7 9 4 2 The above data is a continuous grouped frequency distribution with equal class-intervals. Construct rectangles with class-intervals as bases and respective frequencies as heights. The histogram of the data is as follows:

The length of 30 leaves of a plant is given in millimetre below- Length of leaves ( in mm ) 111 - 120 121 - 130 131 - 140 141 - 150 151 - 160 161 - 170 Number of leaves 3 5 7 9 4 2 Make a frequency polygon using direct method.

First find the class marks of each class. Length of leaves ( in mm ) 110 . 5 - 120 . 5 120 . 5 - 130 . 5 130 . 5 - 140 . 5 140 . 5 - 150 . 5 150 . 5 - 160 . 5 160 . 5 - 170 . 5 Number of leaves 3 5 7 9 4 2 Mid-point 115 . 5 125 . 5 135 . 5 145 . 5 155 . 5 165 . 5 1 . We will plot the mid points on the X -axis and the frequencies on the Y -axis. Mark the mid points on the X -axis. 2 . Mark the points as A , B , C , D , E , F , H using the corresponding frequency. 3 . Mark on the X -axis the midpoint 55 . 5 ( point A ) of the class 0 - 110 . 5 , that lies before the first class and the midpoint 175 . 5 ( point H ) of the class 170 . 5 - 180 . 5 , which is after the first class. We have marked them on the X -axis because their frequencies are zero. 4 . Join all the marked mid-points sequentially. 5 . The figure A B C D E F G obtained is the frequency polygon,

The length of 30 leaves of a plant is given in millimetre below- Length of leaves ( in mm ) 111 - 120 121 - 130 131 - 140 141 - 150 151 - 160 161 - 170 Number of leaves 3 5 7 9 4 2 In which class the number of leaves is the largest?

Given, Length of leaves ( in mm ) 111 - 120 121 - 130 131 - 140 141 - 150 151 - 160 161 - 170 Number of leaves 3 5 7 9 4 2 Arranging in descending order, In the class interval 141 - 150 , the number of leaves is 9 . In the class interval 131 - 140 , the number of leaves is 7 . In the class interval 121 - 130 , the number of leaves is 5 . In the class interval 151 - 160 , the number of leaves is 4 . In the class interval 111 - 120 , the number of leaves is 3 . In the class interval 161 - 170 , the number of leaves is 2 . Hence, the largest number of leaves present in 141 - 150 class interval.

The number of runs made by two team A and B in a 10 over ( 60 ball ) match is given below: Number of balls Team A Team B 1 - 6 3 6 7 - 12 2 6 13 - 18 7 1 19 - 24 8 10 25 - 30 3 6 31 - 36 8 5 37 - 42 5 3 43 - 48 11 4 49 - 54 6 7 55 - 60 2 11 Use the above data make a frequency polygon on one graph paper.

First find the class marks of each class by making continuous class intervals. Number of balls Mid-point Team A Team B 0 . 5 - 6 . 5 3 . 5 3 6 6 . 5 - 12 . 5 7 . 5 2 6 12 . 5 - 18 . 5 15 . 5 7 1 18 . 5 - 24 . 5 21 . 5 8 10 24 . 5 - 30 . 5 27 . 5 3 6 30 . 5 - 36 . 5 33 . 5 8 5 36 . 5 - 42 . 5 39 . 5 5 3 42 . 5 - 48 . 5 45 . 5 11 4 48 . 5 - 54 . 5 51 . 5 6 7 54 . 5 - 60 . 5 57 . 5 2 11 1 . We will plot the mid points on the X -axis and the frequencies on the Y -axis. Mark the mid points on the X -axis. 2 . Mark the points B , C , D , E , F , G , H , I , J , K and I I - X I using the corresponding frequency of Team- A and Team- B . 3 . Mark on the X -axis the midpoint 3 . 5 ( point B and point I I ) of the class 1 - 6 for both the teams, that lies before the first class and the midpoint 57 . 5 ( point K and point X I ) of the class 55 - 60 for both the teams, which is after the first class. We have marked them on the X -axis because their frequencies are zero. 4 . Join all the marked mid points sequentially. 5 . The figure A B C D E F G H I J K L and I - X I I obtained is the frequency polygon.

The frequency distribution of marks obtained by 100 students is as follows: Marks obtained 0 - 10 10 - 20 20 - 30 30 - 40 40 - 50 50 - 60 Total Number of students 7 10 23 51 6 3 100 Make a cumulative frequency curve using the above data.

The cumulative frequency of a class is the sum of the frequency of that class and frequencies of all the classes before it. Marks obtained 0 - 10 10 - 20 20 - 30 30 - 40 40 - 50 50 - 60 Total Number of students 7 10 23 51 6 3 100 Based on following table let us try to plot the less than cumulative curve:- Class Frequency Cumulative frequency Less than 0 0 0 Less than 10 7 7 Less than 20 10 17 = 10 + 7 Less than 30 23 40 = 23 + 10 + 7 Less than 40 51 91 = 51 + 23 + 10 + 7 Less than 50 6 97 = 51 + 23 + 10 + 7 + 6 Less than 60 3 100 = 51 + 23 + 10 + 7 + 6 + 3 The cumulative frequency curve is:

Make a frequency table and the cumulative frequency curve for the following data: Marks Obtained Number of students less than 10 3 less than 20 8 less than 30 12 less than 40 19 less than 50 31 less than 60 42 less than 70 60

The cumulative frequency of a class is the sum of the frequency of that class and frequencies of all the classes before it. Based on following table let us try to plot the less than cumulative curve:- Class Frequency Cumulative frequency Less than 0 0 0 Less than 10 3 3 Less than 20 8 11 = 8 + 3 Less than 30 12 23 = 12 + 8 + 3 Less than 40 19 42 = 19 + 12 + 8 + 3 Less than 50 31 73 = 31 + 19 + 12 + 8 + 3 Less than 60 42 115 = 42 + 31 + 19 + 12 + 8 + 3 Less than 70 60 175 = 60 + 42 + 31 + 19 + 12 + 8 + 3 The cumulative frequency curve is:

The marks obtained by two groups of students in a test are given below:- Class Interval Group A Group B 50 - 52 4 8 47 - 49 10 9 44 - 46 15 10 41 - 43 18 14 38 - 40 20 12 35 - 37 12 17 32 - 34 13 22 Total 92 92 Make frequency polygons for each of these groups on one graph paper.

First find the class marks of each class by making continuous class intervals. Class Interval Mid-point Group A Group B 31 . 5 - 34 . 5 33 13 22 34 . 5 - 37 . 5 36 12 17 37 . 5 - 40 . 5 39 20 12 40 . 5 - 43 . 5 42 18 14 43 . 5 - 46 . 5 45 15 10 46 . 5 - 49 . 5 48 10 9 49 . 5 - 52 . 5 51 4 8 Total 92 92 1 . We will plot the mid points on the X -axis and the frequencies on the Y -axis. Mark the mid points on the X -axis. 2 . Mark the points B , C , D , E , F , G , H and I I - V I I I using the corresponding frequency of Group- A and Group- B . 3 . Mark on the X -axis the midpoint 33 ( point B and point I I ) of the class 31 . 5 - 34 . 5 for both the teams, that lies before the first class and the midpoint 51 ( point H and point V I I I ) of the class 49 . 5 - 52 . 5 for both the teams, which is after the first class. We have marked them on the X -axis because their frequencies are zero. 4 . Join all the marked mid points sequentially. 5 . The figure A B C D E F G H I and I - I X obtained is the frequency polygon.

Chhattisgarh Board Solutions for Exercise 4: Exercise

Author:Chhattisgarh Board

Chhattisgarh Board Mathematics Solutions for Exercise - Chhattisgarh Board Solutions for Exercise 4: Exercise

Attempt the practice questions from Exercise 4: Exercise with hints and solutions to strengthen your understanding. Mathematics Class - 9 solutions are prepared by Experienced Embibe Experts.

Questions from Chhattisgarh Board Solutions for Exercise 4: Exercise with Hints & Solutions

MEDIUM

9th Chhattisgarh Board

IMPORTANT

Mathematics Class - 9>Chapter 16 - Data Handling and Analysis>Exercise>Q 2

The time taken (in second) by $25$ students in doing a question is as follows:

$16, 20, 26, 27, 28, 30, 33, 37, 38, 40, 42, 46, 46, 46, 48, 49, 50, 53, 58, 59, 60, 64, 52, 20$

Use the histogram to make a frequency polygon.

MEDIUM

9th Chhattisgarh Board

IMPORTANT

Mathematics Class - 9>Chapter 16 - Data Handling and Analysis>Exercise>Q 3

The length of $30$ leaves of a plant is given in millimetre below-

Length of leaves ( in $mm$ )	$111 - 120$	$121 - 130$	$131 - 140$	$141 - 150$	$151 - 160$	$161 - 170$
Number of leaves	$3$	$5$	$7$	$9$	$4$	$2$

Make a histogram to depict the frequency distribution.

HARD

9th Chhattisgarh Board

IMPORTANT

Mathematics Class - 9>Chapter 16 - Data Handling and Analysis>Exercise>Q 3

The length of $30$ leaves of a plant is given in millimetre below-

Length of leaves ( in $mm$ )	$111 - 120$	$121 - 130$	$131 - 140$	$141 - 150$	$151 - 160$	$161 - 170$
Number of leaves	$3$	$5$	$7$	$9$	$4$	$2$

Make a frequency polygon using direct method.

MEDIUM

9th Chhattisgarh Board

IMPORTANT

Mathematics Class - 9>Chapter 16 - Data Handling and Analysis>Exercise>Q 3

The length of $30$ leaves of a plant is given in millimetre below-

Length of leaves ( in $mm$ )	$111 - 120$	$121 - 130$	$131 - 140$	$141 - 150$	$151 - 160$	$161 - 170$
Number of leaves	$3$	$5$	$7$	$9$	$4$	$2$

In which class the number of leaves is the largest?

HARD

9th Chhattisgarh Board

IMPORTANT

Mathematics Class - 9>Chapter 16 - Data Handling and Analysis>Exercise>Q 4

The number of runs made by two team $A$ and $B$ in a $10$ over ( $60$ ball ) match is given below:

Number of balls	Team $A$	Team $B$
$1 - 6$	$3$	$6$
$7 - 12$	$2$	$6$
$13 - 18$	$7$	$1$
$19 - 24$	$8$	$10$
$25 - 30$	$3$	$6$
$31 - 36$	$8$	$5$
$37 - 42$	$5$	$3$
$43 - 48$	$11$	$4$
$49 - 54$	$6$	$7$
$55 - 60$	$2$	$11$

Use the above data make a frequency polygon on one graph paper.

EASY

9th Chhattisgarh Board

IMPORTANT

Mathematics Class - 9>Chapter 16 - Data Handling and Analysis>Exercise>Q 5

The frequency distribution of marks obtained by $100$ students is as follows:

Marks obtained	$0 - 10$	$10 - 20$	$20 - 30$	$30 - 40$	$40 - 50$	$50 - 60$	Total
Number of students	$7$	$10$	$23$	$51$	$6$	$3$	$100$

Make a cumulative frequency curve using the above data.

EASY

9th Chhattisgarh Board

IMPORTANT

Mathematics Class - 9>Chapter 16 - Data Handling and Analysis>Exercise>Q 6

Make a frequency table and the cumulative frequency curve for the following data:

Marks Obtained	Number of students
less than $10$	$3$
less than $20$	$8$
less than $30$	$12$
less than $40$	$19$
less than $50$	$31$
less than $60$	$42$
less than $70$	$60$

HARD

9th Chhattisgarh Board

IMPORTANT

Mathematics Class - 9>Chapter 16 - Data Handling and Analysis>Exercise>Q 7

The marks obtained by two groups of students in a test are given below:-

Class Interval	Group $A$	Group $B$
$50 - 52$	$4$	$8$
$47 - 49$	$10$	$9$
$44 - 46$	$15$	$10$
$41 - 43$	$18$	$14$
$38 - 40$	$20$	$12$
$35 - 37$	$12$	$17$
$32 - 34$	$13$	$22$
Total	$92$	$92$

Make frequency polygons for each of these groups on one graph paper.

Chhattisgarh Board Solutions for Exercise 4: Exercise

Chhattisgarh Board Mathematics Solutions for Exercise - Chhattisgarh Board Solutions for Exercise 4: Exercise

Questions from Chhattisgarh Board Solutions for Exercise 4: Exercise with Hints & Solutions

Learn and Prepare for any exam you want !