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Chapter 23

Graphical Representation (Histograms and Ogives)

Class - 10 Concise Mathematics Selina



Exercise 23

Question 1(a)

From the given table the values of a, b and c are :

C.I.FrequencyCumulative
Frequency
30-402424
40-50a40
50-6012b
60-70c60
  1. a = 16, b = 52 and c = 8

  2. a = 16, b = 12 and c = 8

  3. a = 40, b = 52 and c = 60

  4. a = 40, b = 12 and c = 60

Answer

From table,

⇒ 24 + a = 40

⇒ a = 40 - 24 = 16

⇒ b = 40 + 12 = 52

⇒ b + c = 60

⇒ 52 + c = 60

⇒ c = 60 - 52 = 8.

Hence, Option 1 is the correct option.

Question 1(b)

The cumulative frequency of the class 70-90 is :

  1. 80

  2. 120

  3. 180

  4. 40

The cumulative frequency of the class 70-90 is : Graphical Representation, Concise Mathematics Solutions ICSE Class 10.

Answer

From the graph, we form the cumulative frequency table :

ClassFrequencyCumulative frequency
30-505050
50-703080 (50 + 30)
70-9040120 (80 + 40)
90-11060180 (120 + 60)

Hence, Option 2 is the correct option.

Question 1(c)

For the given frequency distribution, the class mark is :

C.I.f
30-348
34-3810
38-428
  1. 38342\dfrac{38 - 34}{2}

  2. 30+342\dfrac{30 + 34}{2}

  3. 10 - 8

  4. 34 + 38

Answer

By formula,

Class mark = Lower limit of class + Upper limit of class2=30+342\dfrac{\text{Lower limit of class + Upper limit of class}}{2} = \dfrac{30 + 34}{2}.

Hence, Option 2 is the correct option.

Question 1(d)

The cumulative curve for a frequency distribution starts from :

  1. 0

  2. upper limit + lower limit2\dfrac{\text{upper limit + lower limit}}{2}

  3. lower limit of 1st class

  4. lower limit - upper limit2\dfrac{\text{lower limit - upper limit}}{2}

Answer

The cumulative curve for a frequency distribution starts from lower limit of 1st class.

Hence, Option 3 is the correct option.

Question 1(e)

The cumulative curve for a frequency distribution terminates at :

  1. 100

  2. upper limit + lower limit2\dfrac{\text{upper limit + lower limit}}{2}

  3. lower limit + upper limit2\dfrac{\text{lower limit + upper limit}}{2}

  4. upper limit of last class

Answer

The cumulative curve for a frequency distribution terminates at upper limit of last class.

Hence, Option 4 is the correct option.

Question 2(i)

Draw histograms for the following frequency distributions :

Class intervalFrequency
0 - 1012
10 - 2020
20 - 3026
30 - 4018
40 - 5010
50 - 606

Answer

Steps of construction of histogram :

  1. Take 2 cm along x-axis = 10 units.

  2. Take 1 cm along y-axis = 5 units.

  3. Construct rectangles corresponding to the above continuous frequency distribution table.

The required histogram is shown in the below figure:

Draw histograms for the following frequency distributions. Graphical Representation, Concise Mathematics Solutions ICSE Class 10.

Question 2(ii)

Draw histograms for the following frequency distributions :

Class intervalFrequency
10 - 1615
16 - 2223
22 - 2830
28 - 3420
34 - 4016

Answer

Steps of construction of histogram :

  1. Since, the scale on x-axis starts at 10, a break (kink) is shown near the origin on x-axis to indicate that the graph is drawn to scale beginning at 10.

  2. Take 2 cm along x-axis = 6 units.

  3. Take 1 cm along y-axis = 5 units.

  4. Construct rectangles corresponding to the above continuous frequency distribution table.

The required histogram is shown in the below figure:

Draw histograms for the following frequency distributions. Graphical Representation, Concise Mathematics Solutions ICSE Class 10.

Question 2(iii)

Draw histograms for the following frequency distributions :

Class markFrequency
168
2412
3215
4018
4825
5619
6410

Answer

Since, the difference between the values of any two consecutive class marks is 8 (24 - 16).

Therefore, subtract 82\dfrac{8}{2} = 4, from each class mark to get the lower limit of the corresponding class interval and add 4 to each class mark to get the upper limit.

Frequency distribution table :

Class markClassFrequency
1612 - 2008
2420 - 2812
3228 - 3615
4036 - 4418
4844 - 5225
5652 - 6019
6460 - 6810

Steps of construction of histogram :

  1. Since, the scale on x-axis starts at 12, a break (kink) is shown near the origin on x-axis to indicate that the graph is drawn to scale beginning at 12.

  2. Take 1 cm along x-axis = 8 units.

  3. Take 1 cm along y-axis = 5 units.

  4. Construct rectangles corresponding to the above continuous frequency distribution table.

The required histogram is shown in the below figure:

Draw histograms for the following frequency distributions. Graphical Representation, Concise Mathematics Solutions ICSE Class 10.

Question 3(i)

Draw a cumulative frequency curve (ogive) for the following distributions :

Class IntervalFrequency
10 - 1510
15 - 2015
20 - 2517
25 - 3012
30 - 3510
35 - 408

Answer

The cumulative frequency distribution :

Class IntervalFrequencyCumulative frequency
10 - 151010
15 - 201525 (10 + 15)
20 - 251742 (25 + 17)
25 - 301254 (42 + 12)
30 - 351064 (54 + 10)
35 - 40872 (64 + 8)

Steps of construction of ogive :

  1. Take 2 cm = 5 units along x-axis.

  2. Take 1 cm = 10 units along y-axis.

  3. Ogive always starts from a point on x-axis representing the lower limit of the first class. Mark point (10, 0).

  4. Take upper class limits along x-axis and corresponding cumulative frequencies along y-axis, mark the points (15, 10), (20, 25), (25, 42), (30, 54), (35, 64) and (40, 72).

  5. Join the points marked by a free hand curve.

The required ogive is shown in the below figure:

Draw a cumulative frequency curve (ogive) for the following distributions. Graphical Representation, Concise Mathematics Solutions ICSE Class 10.

Question 3(ii)

Draw a cumulative frequency curve (ogive) for the following distributions :

Class intervalFrequency
10 - 1923
20 - 2916
30 - 3915
40 - 4920
50 - 5912

Answer

The above distribution is discontinuous converting into continuous distribution, we get :

Adjustment factor = (Lower limit of one class - Upper limit of previous class) / 2

= 20192=12\dfrac{20 - 19}{2} = \dfrac{1}{2}

= 0.5

Subtract the adjustment factor (0.5) from all the lower limits and add the adjustment factor (0.5) to all the upper limits.

Classes before adjustmentClasses after adjustmentFrequencyCumulative frequency
10 - 199.5 - 19.52323
20 - 2919.5 - 29.51639 (23 + 16)
30 - 3929.5 - 39.51554 (39 + 15)
40 - 4939.5 - 49.52074 (54 + 20)
50 - 5949.5 - 59.51286 (74 + 12)

Steps of construction of ogive :

  1. Since, the scale on x-axis starts at 9.5, a break (kink) is shown near the origin on x-axis to indicate that the graph is drawn to scale beginning at 9.5.

  2. Take 2 cm = 10 units along x-axis.

  3. Take 1 cm = 10 units along y-axis.

  4. Ogive always starts from a point on x-axis representing the lower limit of the first class. Mark point (9.5, 0).

  5. Take upper class limits along x-axis and corresponding cumulative frequencies along y-axis, mark the points (19.5, 23), (29.5, 39), (39.5, 54), (49.5, 74) and (59.5, 86).

  6. Join the points marked by a free hand curve.

The required ogive is shown in the below figure:

Draw a cumulative frequency curve (ogive) for the following distributions. Graphical Representation, Concise Mathematics Solutions ICSE Class 10.

Question 4(i)

Draw an ogive for the following distribution :

Marks obtainedNo. of students
less than 108
less than 2025
less than 3038
less than 4050
less than 5067

Answer

Cumulative frequency distribution table :

Marks obtainedClass intervalNo. of students (Cumulative frequency)
less than 100 - 108
less than 2010 - 2025
less than 3020 - 3038
less than 4030 - 4050
less than 5040 - 5067

Steps of construction of ogive :

  1. Take 2 cm = 10 units along x-axis.

  2. Take 1 cm = 10 units along y-axis.

  3. Ogive always starts from a point on x-axis representing the lowest limit. Mark point (0, 0).

  4. Take upper class limits along x-axis corresponding cumulative frequencies along y-axis, mark the points (10, 8), (20, 25), (30, 38), (40, 50) and (50, 67).

  5. Join the points marked by a free hand curve.

The required ogive is shown in the below figure:

Draw an ogive for the following distribution. Graphical Representation, Concise Mathematics Solutions ICSE Class 10.

Question 4(ii)

Draw an ogive for the following distribution :

Age in years (less than)Cumulative frequency
100
2017
3032
4037
5053
6058
7065

Answer

Cumulative frequency distribution table :

Age in years (less than)Cumulative frequency
100
2017
3032
4037
5053
6058
7065

Steps of construction of ogive :

  1. Take 2 cm = 10 units along x-axis.

  2. Take 1 cm = 10 units along y-axis.

  3. Ogive always starts from a point on x-axis representing the lowest limit. Mark point (0, 0).

  4. Take upper class limits along x-axis corresponding cumulative frequencies along y-axis, mark the points (10, 0), (20, 17), (30, 32), (40, 37), (50, 53), (60, 58) and (70, 65).

  5. Join the points marked by a free hand curve.

The required ogive is shown in the adjoining figure.

Draw an ogive for the following distribution. Graphical Representation, Concise Mathematics Solutions ICSE Class 10.

Question 5

(a) Use the information given in the adjoining histogram to construct a frequency table.

(b) Use this table to construct an ogive.

(a) Use the information given in the adjoining histogram to construct a frequency table. (b) Use this table to construct an ogive. Graphical Representation, Concise Mathematics Solutions ICSE Class 10.

Answer

(a) The cumulative frequency table for above distribution is :

Class intervalFrequencyCumulative frequency
8 - 1299
12 - 161625
16 - 202247
20 - 241865
24 - 281277
28 - 32481

(b) Steps of construction of ogive :

  1. Since, the scale on x-axis starts at 8, a break (kink) is shown near the origin on x-axis to indicate that the graph is drawn to scale beginning at 8.

  2. Take 2 cm = 4 units along x-axis.

  3. Take 1 cm = 10 units along y-axis.

  4. Ogive always starts from a point on x-axis representing the lowest limit. Mark point (8, 0).

  5. Take upper class limits along x-axis corresponding cumulative frequencies along y-axis, mark the points (12, 9), (16, 25), (20, 47), (24, 65), (28, 77) and (32, 81).

  6. Join the points marked by a free hand curve.

The required ogive is shown in the adjoining figure.

(a) Use the information given in the adjoining histogram to construct a frequency table. (b) Use this table to construct an ogive. Graphical Representation, Concise Mathematics Solutions ICSE Class 10.

Question 6

Use graph paper for this question.

The table given below shows the monthly wages of some factory workers.

(i) Using the table, calculate the cumulative frequencies of workers.

(ii) Draw a cumulative frequency curve.

Wages (in ₹)No. of workers
6500 - 700010
7000 - 750018
7500 - 800022
8000 - 850025
8500 - 900017
9000 - 950010
9500 - 100008

Answer

(i) The cumulative frequency distribution table is as follows :

Wages (in ₹)No. of workersCumulative frequency
6500 - 70001010
7000 - 75001828 (10 + 18)
7500 - 80002250 (28 + 22)
8000 - 85002575 (50 + 25)
8500 - 90001792 (75 + 17)
9000 - 950010102 (92 + 10)
9500 - 100008110 (102 + 8)

(ii) Steps for construction of cumulative frequency curve :

  1. Since, the scale on x-axis starts at 6500, a break (kink) is shown near the origin on x-axis to indicate that the graph is drawn to scale beginning at 6500.

  2. Take 2 cm along x-axis = ₹ 500.

  3. Take 1 cm along y-axis = 10 workers.

  4. Ogive always starts from a point on x-axis representing the lower limit of first class. Mark point (6500, 0).

  5. Take upper class limits along x-axis, corresponding cumulative frequencies along y-axis, mark the points (7000, 10), (7500, 28), (8000, 50), (8500, 75), (9000, 92), (9500, 102) and (10000, 110).

  6. Join the points marked by a free hand curve.

The required ogive is shown in the below figure:

Use graph paper for this question. The table given below shows the monthly wages of some factory workers. (i) Using the table, calculate the cumulative frequencies of workers. (ii) Draw a cumulative frequency curve. Graphical Representation, Concise Mathematics Solutions ICSE Class 10.

Question 7

The following table shows the distribution of the heights of a group of factory workers :

Ht. (cm)No. of workers
150 - 1556
155 - 16012
160 - 16518
165 - 17020
170 - 17513
175 - 1808
180 - 1856

(i) Determine the cumulative frequencies.

(ii) Draw the 'less than' cumulative frequency curve on graph paper.

Answer

(i) The cumulative frequency distribution table is as follows :

Ht. (cm)No. of workersCumulative frequency
150 - 15566
155 - 1601218 (6 + 12)
160 - 1651836 (18 + 18)
165 - 1702056 (36 + 20)
170 - 1751369 (56 + 13)
175 - 180877 (69 + 8)
180 - 185683 (77 + 6)

(ii) Steps for construction of 'less than' cumulative frequency curve :

  1. Since, the scale on x-axis starts at 150, a break (kink) is shown near the origin on x-axis to indicate that the graph is drawn to scale beginning at 150.

  2. Take 2 cm along x-axis = 5 cm.

  3. Take 1 cm along y-axis = 10 workers.

  4. Ogive always starts from a point on x-axis representing the lower limit of first class. Mark point (150, 0).

  5. Take upper class limits along x-axis, corresponding cumulative frequencies along y-axis, mark the points (155, 6), (160, 18), (165, 36), (170, 56), (175, 69), (180, 77) and (185, 83).

  6. Join the points marked by a free hand curve.

The required ogive is shown in the below figure:

The following table shows the distribution of the heights of a group of factory workers. (i)Determine the cumulative frequencies. (ii) Draw the 'less than' cumulative frequency curve on graph paper. Graphical Representation, Concise Mathematics Solutions ICSE Class 10.

Question 8(i)

Construct a frequency distribution table for the following distribution :

Marks (less than)Cumulative frequency
00
107
2028
3054
4071
5084
60105
70147
80180
90196
100200

Answer

The frequency distribution table for above distribution is :

MarksCumulative frequencyFrequency
0 - 1077
10 - 202821 (28 - 7)
20 - 305426 (54 - 28)
30 - 407117 (71 - 54)
40 - 508413 (84 - 71)
50 - 6010521 (105 - 84)
60 - 7014742 (147 - 105)
70 - 8018033 (180 - 147)
80 - 9019616 (196 - 180)
90 - 1002004 (200 - 196)

Question 8(ii)

Construct a frequency distribution table for the following distribution :

Marks (more than)Cumulative frequency
0100
1087
2065
3055
4042
5036
6031
7021
8018
907
1000

Answer

The frequency distribution table for above distribution is :

MarksCumulative frequencyFrequency
0 - 1010013 (100 - 87)
10 - 208722 (87 - 65)
20 - 306510 (65 - 55)
30 - 405513 (55 - 42)
40 - 50426 (42 - 36)
50 - 60365 (36 - 31)
60 - 703110 (31 - 21)
70 - 80213 (21 - 18)
80 - 901811 (18 - 7)
90 - 10077
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