Draw an ogive to represent the following frequency distribution: Class interval 0 - 4 5 - 9 10 - 14 15 - 19 20 - 24 Number of students 2 6 10 5 3

The given frequency distribution is not continuous. We convert it into continuous distribution by subtracting the conversion factor from the lower limit and adding it to the upper limit of each class interval. Here, the conversion factor = 5 - 4 2 = 1 2 = 0 . 5 The cumulative frequency distribution table by less than type is as below: Class interval Number of students f i Class interval Class interval Cumulative frequency c . f . 0 - 4 2 0 - 4 . 5 Less than 4 . 5 2 5 - 9 6 4 . 5 - 9 . 5 Less than 9 . 5 6 + 2 = 8 10 - 14 10 9 . 5 - 14 . 5 Less than 14 . 5 10 + 8 = 18 15 - 19 5 14 . 5 - 19 . 5 Less than 19 . 5 5 + 18 = 23 20 - 24 3 19 . 5 - 24 . 5 Less than 24 . 5 3 + 23 = 26 Mark upper class limits on x - axis and cumulative frequency on y - axis. We plot the points 4 . 5 , 2 , 9 . 5 , 8 , 14 . 5 , 18 , 19 . 5 , 23 , and 24 . 5 , 26 . Joining the above points by line segments, the cumulative frequency polygon is as shown below:

For the following frequency distribution, draw an ogive of 'more than type' and hence find the median value. Class interval 0 - 10 10 - 20 20 - 30 30 - 40 40 - 50 50 - 60 60 - 70 Frequency 5 15 20 23 17 11 9

The cumulative frequency distribution table by more than type is as below: Class interval Frequency f i Class interval Cumulative frequency c . f . 0 - 10 5 More than 0 5 + 95 = 100 10 - 20 15 More than 10 15 + 80 = 95 20 - 30 20 More than 20 20 + 60 = 80 30 - 40 23 More than 30 23 + 37 = 60 40 - 50 17 More than 40 17 + 20 = 37 50 - 60 11 More than 50 11 + 9 = 20 60 - 70 9 More than 60 9 Mark lower class limits on x - axis and cumulative frequency on y - axis. We plot the points 0 , 100 , 10 , 95 , 20 , 80 , 30 , 60 , 40 , 37 , 50 , 20 and 60 , 9 . Joining the above points by line segments, the cumulative frequency polygon is as shown below: To locate the median point on the graph, take N 2 = 100 2 = 50 on the y - axis. From this point, draw a line parallel to the x - axis cutting the curve at a point P . From P , draw a perpendicular to the x - axis. The point of intersection of this perpendicular with the x - axis gives the median of the data. Here, the median is 35 .

The following table gives production yield per hectare of wheat of 100 farms of a village Production yield kg / hec 40 - 45 45 - 50 50 - 55 55 - 60 60 - 65 65 - 70 Number of farms 4 6 16 20 30 24 Change the distribution to a 'more than type' distribution and draw its ogive.

The cumulative frequency distribution table by more than type is as below: Production yield kg / hec Number of farms f i Production yield kg / hec Cumulative frequency c . f . 40 - 45 4 More than 40 4 + 96 = 100 45 - 50 6 More than 45 6 + 90 = 96 50 - 55 16 More than 50 16 + 74 = 90 55 - 60 20 More than 55 20 + 54 = 74 60 - 65 30 More than 60 30 + 24 = 54 65 - 70 24 More than 65 24 Mark lower class limits on x - axis and cumulative frequency on y - axis. We plot the points 40 , 100 , 45 , 96 , 50 , 90 , 55 , 74 , 60 , 54 , and 65 , 24 . Joining the above points by line segments, the cumulative frequency polygon is as shown below:

The following table gives production yield per hectare of 100 farms of a village Production yield kg / hec 50 - 55 55 - 60 60 - 65 65 - 70 70 - 75 75 - 80 Number of farms 2 8 12 24 38 16 Draw 'less than ogive' and 'more than ogive' .

The cumulative frequency distribution table by less than type is as below: Production yield kg / hec Number of farms f i Production yield kg / hec Cumulative frequency c . f . 50 - 55 2 Less than 55 2 55 - 60 8 Less than 60 8 + 2 = 10 60 - 65 12 Less than 65 12 + 10 = 22 65 - 70 24 Less than 70 24 + 22 = 46 70 - 75 38 Less than 75 38 + 46 = 84 75 - 80 16 Less than 80 16 + 84 = 100 Mark upper class limits on x - axis and cumulative frequency on y - axis. We plot the points 55 , 2 , 60 , 10 , 65 , 22 , 70 , 46 , 75 , 84 , and 80 , 100 . The cumulative frequency distribution table by more than type is as below: Production yield kg / hec Number of farms f i Production yield kg / hec Cumulative frequency c . f . 50 - 55 2 More than 50 2 + 98 = 100 55 - 60 8 More than 55 8 + 90 = 98 60 - 65 12 More than 60 12 + 78 = 90 65 - 70 24 More than 65 24 + 54 = 78 70 - 75 38 More than 70 38 + 16 = 54 75 - 80 16 More than 75 16 Mark lower class limits on x - axis and cumulative frequency on y - axis. We plot the points 50 , 100 , 55 , 98 , 60 , 90 , 65 , 78 , 70 , 54 , and 75 , 16 . Joining the above points by line segments, the less than ogive and the more than ogive are shown:

During the medical checkup of 35 students of a class, their weight were recorded as follows: Weight (in kg ) Number of students Less than 38 0 Less than 40 3 Less than 42 5 Less than 44 9 Less than 46 14 Less than 48 28 Less than 50 32 Less than 52 35 Draw a 'less than type' ogive for the given data. Here, obtain the median weight from the graph and verify the result by using the formula.

Here, 38 , 40 , 42 , 44 , 46 , 48 , 50 , and 52 are the upper limits of the respective class-intervals. We mark the upper class limits on x - axis and cumulative frequency on y - axis. We plot the points 38 , 0 , 40 , 3 , 42 , 5 , 44 , 9 , 46 , 14 , 48 , 28 , 50 , 32 and 52 , 35 . Joining the above points by line segments, the cumulative frequency polygon is as shown below: To locate the median point on the graph, take N 2 = 35 2 = 17 . 5 on the y - axis. From this point, draw a line parallel to the x - axis cutting the curve at a point P . From P , draw a perpendicular to the x - axis. The point of intersection of this perpendicular with the x - axis gives the median of the data. Here, the median is 46 . 5 . Since we are given less than a cumulative frequency table. We prepare the frequency table: Marks Class interval Number of students c . f . Frequency f i Less than 38 36 - 38 0 0 Less than 40 38 - 40 3 3 - 0 = 3 Less than 42 40 - 42 5 5 - 3 = 2 Less than 44 42 - 44 9 9 - 5 = 4 Less than 46 44 - 46 14 14 - 9 = 5 Less than 48 46 - 48 28 28 - 14 = 14 Less than 50 48 - 50 32 32 - 28 = 4 Less than 52 50 - 52 35 35 - 32 = 3 We have, N = 35 So, N 2 = 35 2 = 17 . 5 The cumulative frequency just greater than 17 . 5 is 28 , which corresponds to the class 46 - 48 . Thus, the median class is 46 - 48 . Here, l = 46 , f = 14 , h = 2 , N = 35 and C = 14 ∴ Median = l + N 2 - C f × h = 46 + 17 . 5 - 14 14 × 2 = 46 + 3 . 5 14 × 2 = 46 + 0 . 5 = 46 . 5 Hence, the median is verified.

From the following frequency distribution, prepare the 'more than ogive': Score Number of candidates 400 - 450 20 450 - 500 35 500 - 550 40 550 - 600 32 600 - 650 24 650 - 700 27 700 - 750 18 750 - 800 34 Total 230

The cumulative frequency distribution table by more than type is as below: Score Number of candidates f i Score Cumulative frequency c . f . 400 - 450 20 More than 400 20 + 210 = 230 450 - 500 35 More than 450 35 + 175 = 210 500 - 550 40 More than 500 40 + 135 = 175 550 - 600 32 More than 550 32 + 103 = 135 600 - 650 24 More than 600 24 + 79 = 103 650 - 700 27 More than 650 27 + 52 = 79 700 - 750 18 More than 700 18 + 34 = 52 750 - 800 34 More than 750 34 Mark lower class limits on x - axis and cumulative frequency on y - axis. We plot the points 400 , 230 , 450 , 210 , 500 , 175 , 550 , 135 , 600 , 103 , 650 , 79 , 700 , 52 , and 750 , 34 . Joining the above points by line segments, the cumulative frequency polygon is as shown below:

From the following data, draw the types of cumulative frequency curves and determine the median: Height (in cm ) Frequency 140 - 144 3 144 - 148 9 148 - 152 24 152 - 156 31 156 - 160 42 160 - 164 64 164 - 168 75 168 - 172 82 172 - 176 84 176 - 180 34

The cumulative frequency distribution table by less than type is as below: Height (in cm ) Frequency f i Height (in cm ) Cumulative frequency c . f . 140 - 144 3 Less than 144 3 144 - 148 9 Less than 148 9 + 3 = 12 148 - 152 24 Less than 152 24 + 12 = 36 152 - 156 31 Less than 156 31 + 36 = 67 156 - 160 42 Less than 160 42 + 67 = 109 160 - 164 64 Less than 164 64 + 109 = 173 164 - 168 75 Less than 168 75 + 173 = 248 168 - 172 82 Less than 172 82 + 248 = 330 172 - 176 84 Less than 176 84 + 330 = 414 176 - 180 34 Less than 180 34 + 414 = 448 Mark upper class limits on x - axis and cumulative frequency on y - axis. We plot the points 144 , 3 , 148 , 12 , 152 , 36 , 156 , 67 , 160 , 109 , 164 , 173 , 168 , 248 , 172 , 330 , 176 , 414 , and 180 , 448 . The cumulative frequency distribution table by more than type is as below: Height (in cm ) Frequency f i Height (in cm ) Cumulative frequency c . f . 140 - 144 3 More than 140 3 + 445 = 448 144 - 148 9 More than 144 9 + 436 = 445 148 - 152 24 More than 148 24 + 412 = 436 152 - 156 31 More than 152 31 + 381 = 412 156 - 160 42 More than 156 42 + 339 = 381 160 - 164 64 More than 160 64 + 275 = 339 164 - 168 75 More than 164 75 + 200 = 275 168 - 172 82 More than 168 82 + 118 = 200 172 - 176 84 More than 172 84 + 34 = 118 176 - 180 34 More than 176 34 Mark lower class limits on x - axis and cumulative frequency on y - axis. We plot the points 140 , 448 , 144 , 445 , 148 , 436 , 152 , 412 , 156 , 381 , 160 , 339 , 164 , 275 , 168 , 200 , 172 , 118 , and 176 , 34 . Joining the above points by line segments, the less than ogive and the more than ogive are shown: Here, the two ogives intersect at P . To find the median M , we draw P M perpendicular on O X . Hence, the median is 166 . 7 cm .

The marks obtained by 100 students of a class in an examination are given in the table: Marks Number of students 0 - 5 2 5 - 10 5 10 - 15 6 15 - 20 8 20 - 25 10 25 - 30 25 30 - 35 20 35 - 40 18 40 - 45 4 45 - 50 2 Draw cumulative frequency curves by using: (i) 'less than type' and (ii) 'more than type' . Find the median.

(i) The cumulative frequency distribution table by less than type is as below: Marks Number of students f i Marks Cumulative frequency c . f . 0 - 5 2 Less than 5 2 5 - 10 5 Less than 10 2 + 5 = 7 10 - 15 6 Less than 15 7 + 6 = 13 15 - 20 8 Less than 20 13 + 8 = 21 20 - 25 10 Less than 25 21 + 10 = 31 25 - 30 25 Less than 30 31 + 25 = 56 30 - 35 20 Less than 35 56 + 20 = 76 35 - 40 18 Less than 40 76 + 18 = 94 40 - 45 4 Less than 45 94 + 4 = 98 45 - 50 2 Less than 50 98 + 2 = 100 Mark upper class limits on x - axis and cumulative frequency on y - axis. We plot the points 5 , 2 , 10 , 7 , 15 , 13 , 20 , 21 , 25 , 31 , 30 , 56 , 35 , 76 , 40 , 94 , 45 , 98 , and 50 , 100 . (ii) The cumulative frequency distribution table by more than type is as below: Marks Number of students f i Marks Cumulative frequency c . f . 0 - 5 2 More than 0 2 + 98 = 100 5 - 10 5 More than 5 5 + 93 = 98 10 - 15 6 More than 10 6 + 87 = 93 15 - 20 8 More than 15 8 + 79 = 87 20 - 25 10 More than 20 10 + 69 = 79 25 - 30 25 More than 25 25 + 44 = 69 30 - 35 20 More than 30 20 + 24 = 44 35 - 40 18 More than 35 18 + 6 = 24 40 - 45 4 More than 40 4 + 2 = 6 45 - 50 2 More than 45 2 Mark lower class limits on x - axis and cumulative frequency on y - axis. We plot the points 0 , 100 , 5 , 98 , 10 , 93 , 15 , 87 , 20 , 79 , 25 , 69 , 30 , 44 , 35 , 24 , 40 , 6 , and 45 , 2 . Joining the above points by line segments, the less than ogive and the more than ogive are shown: Here, the two ogives intersect at P . Now, we draw P M perpendicular on O X . Hence, median = O M = 28 . 8 .

J P Mohindru and Bharat Mohindru Solutions for Chapter: Statistics, Exercise 4: EXERCISE

Author:J P Mohindru & Bharat Mohindru

J P Mohindru Mathematics Solutions for Exercise - J P Mohindru and Bharat Mohindru Solutions for Chapter: Statistics, Exercise 4: EXERCISE

Attempt the free practice questions on Chapter 14: Statistics, Exercise 4: EXERCISE with hints and solutions to strengthen your understanding. Modern's abc+ of Mathematics for Class 10 solutions are prepared by Experienced Embibe Experts.

Questions from J P Mohindru and Bharat Mohindru Solutions for Chapter: Statistics, Exercise 4: EXERCISE with Hints & Solutions

HARD

10th CBSE

IMPORTANT

Modern's abc+ of Mathematics for Class 10>Chapter 14 - Statistics>EXERCISE>Q 2

Draw an ogive to represent the following frequency distribution:

Class interval	$0 - 4$	$5 - 9$	$10 - 14$	$15 - 19$	$20 - 24$
Number of students	$2$	$20 - 24$	$10$	$5$	$3$

HARD

10th CBSE

IMPORTANT

Modern's abc+ of Mathematics for Class 10>Chapter 14 - Statistics>EXERCISE>Q 3

For the following frequency distribution, draw an ogive of 'more than type' and hence find the median value.

Class interval	$0 - 10$	$10 - 20$	$20 - 30$	$30 - 40$	$40 - 50$	$50 - 60$	$60 - 70$
Frequency	$5$	$15$	$(24.5, 26)$	$23$	$17$	$11$	$9$

HARD

10th CBSE

IMPORTANT

Modern's abc+ of Mathematics for Class 10>Chapter 14 - Statistics>EXERCISE>Q 4

The following table gives production yield per hectare of wheat of $100$ farms of a village

Production yield $(kg / hec)$	$40 - 45$	$45 - 50$	$50 - 55$	$55 - 60$	$60 - 65$	$65 - 70$
Number of farms	$4$	$6$	$16$	$20$	$30$	$24$

Change the distribution to a 'more than type' distribution and draw its ogive.

HARD

10th CBSE

IMPORTANT

Modern's abc+ of Mathematics for Class 10>Chapter 14 - Statistics>EXERCISE>Q 5

The following table gives production yield per hectare of $100$ farms of a village

Production yield $(kg / hec)$	$50 - 55$	$55 - 60$	$60 - 65$	$65 - 70$	$70 - 75$	$75 - 80$
Number of farms	$2$	$8$	$12$	$24$	$38$	$16$

Draw 'less than ogive' and 'more than ogive'.

HARD

10th CBSE

IMPORTANT

Modern's abc+ of Mathematics for Class 10>Chapter 14 - Statistics>EXERCISE>Q 6

During the medical checkup of $35$ students of a class, their weight were recorded as follows:

Weight (in $kg$ )	Number of students
Less than $38$	$0$
Less than $40$	$3$
Less than $42$	$5$
Less than $44$	$9$
Less than $46$	$14$
Less than $48$	$28$
Less than $50$	$32$
Less than $52$	$35$

Draw a 'less than type' ogive for the given data. Here, obtain the median weight from the graph and verify the result by using the formula.

HARD

10th CBSE

IMPORTANT

Modern's abc+ of Mathematics for Class 10>Chapter 14 - Statistics>EXERCISE>Q 7

From the following frequency distribution, prepare the 'more than ogive':

Score	Number of candidates
$400 - 450$	$20$
$450 - 500$	$35$
$500 - 550$	$40$
$550 - 600$	$32$
$600 - 650$	$24$
$650 - 700$	$27$
$700 - 750$	$18$
$750 - 800$	$34$
Total	$230$

HARD

10th CBSE

IMPORTANT

Modern's abc+ of Mathematics for Class 10>Chapter 14 - Statistics>EXERCISE>Q 8

From the following data, draw the types of cumulative frequency curves and determine the median:

Height (in $cm$ )	Frequency
$140 - 144$	$3$
$144 - 148$	$9$
$148 - 152$	$24$
$152 - 156$	$31$
$156 - 160$	$42$
$160 - 164$	$64$
$164 - 168$	$75$
$168 - 172$	$82$
$172 - 176$	$84$
$176 - 180$	$34$

HARD

10th CBSE

IMPORTANT

Modern's abc+ of Mathematics for Class 10>Chapter 14 - Statistics>EXERCISE>Q 9

The marks obtained by $100$ students of a class in an examination are given in the table:

Marks	Number of students
$0 - 5$	$2$
$5 - 10$	$5$
$10 - 15$	$6$
$15 - 20$	$8$
$20 - 25$	$10$
$25 - 30$	$25$
$30 - 35$	$20$
$35 - 40$	$18$
$40 - 45$	$4$
$45 - 50$	$2$

Draw cumulative frequency curves by using: (i) 'less than type' and (ii) 'more than type'.

Find the median.

J P Mohindru and Bharat Mohindru Solutions for Chapter: Statistics, Exercise 4: EXERCISE

J P Mohindru Mathematics Solutions for Exercise - J P Mohindru and Bharat Mohindru Solutions for Chapter: Statistics, Exercise 4: EXERCISE

Questions from J P Mohindru and Bharat Mohindru Solutions for Chapter: Statistics, Exercise 4: EXERCISE with Hints & Solutions

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