Question 1: Use a graph paper for this question. The daily pocket expenses of 200 students in a school are given below;

Pocket expenses (in Rs.) 0-5 5-10 10-15 15-20 20-25 25-30 30-35 35-40
Number of students (frequency) 10 14 28 42 50 30 14 12

Draw a histogram representing the above distribution and estimate the mode from the graph. [2014]

Answer:

icse 14 7

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Question 2: The marks obtained by 100 students in a Mathematics test are given below; 

Marks 0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80 80-90 90-100
No. of Students 3 7 12 17 23 14 9 6 5 4

Draw an ogive for the distribution on a graph sheet. (Use a scale of 2 cm = 10 units on both axis)

Use the ogive to estimate the:

(i) median

(ii) lower quartile

(iii) number of students who obtained more than 85% marks in the test.

(iv) number of students who did not pass in the if the pass percentage was 35    [2014]

Answer:

Marks No. of Students Cumulative Frequency (c.f)
0-10

10-20

20-30

30-40

40-50

50-60

60-70

70-80

80-90

90-100

8

7

12

17

23

14

9

6

5

4

3

10

22

39

62

76

85

91

96

100

On the graph paper, we plot the following points:

(10, 3), (20, 10), (30, 22), (40,39), (50, 62), (60,76), (70,85),  \\ (80, 91), (90, 96),(100, 100)

icse 14 11

(i) Medium = (\frac{n}{2})^{th} term = \frac{100}{2}=50^{th} term

From the graph 50^{th} term=44

(ii) Lower quartile = (\frac{n}{4})^{th} term = \frac{100}{4}=25^{th} term

From the graph 25^{th} term =32

(iii) The number of students who obtained more than 85\% marks in test = 100-94 = 6 students

(iv) The number of students who did not pass in the test if test if the pass percentage was 35 = 30 students.

Question 3:  The marks obtained by 30 students in a class assessment of 5 marks is given below:

Marks 0 1 2 3 4 5
No. of Students 1 3 6 10 5 5

Calculate the mean, median and mode of the above distribution.    [2015]

Answer:

Below table

x f fx cf
0 1 0 1
1 3 3 4
2 6 12 10
3 10 30 20
4 5 20 25
5 5 25 30
\Sigma f=30\ \ \ \ \ \ \  \Sigma fx=90

Mean = \frac{\Sigma fx}{\Sigma f}=\frac{90}{30} = 3

Median = size \ of \left(\frac{N}{2}\right)th\ {obs}^n

= size \ of  \left(\frac{30}{2}\right)th\ {obs}^n 

= Size \ of 15 {}^{th} \ obs{}^{n}  = 3

Mode = 3 marks (as highest frequency is 10)

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Question 4:  Calculate the mean of the following distribution.    [2015]

Class Interval 0-10 10-20 20-30 30-40 40-50 50-60
Frequency 8 5 12 35 24 16

Answer:

Class Mid Value x f  fx 
0-10 5 8 40
10-20 15 5 75
20-30 25 12 300
30-40 35 35 1225
40-50 45 24 1080
50-60 55 16 880
Total \Sigma f=100 \Sigma fx=3600

Mean = \bar{x} = \frac{\mathcal{E}fx}{ \mathcal{E}f} = \frac{3600}{100} = 36

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Question 5: The weight of 50 workers is given below:

Weight in kg 50-60 60-70 70-80 80-90 90-100 100-110 110-120
No. of workers 4 7 11 14 6 5 3

Draw on give of the given distribution using a graph sheet. Take 2 cm =10 kg on  one axis and 2 cm =5 workers along the other axis. Use a graph to estimate the following:

(i) the upper and lower quartiles

(ii) if weight 95 kg and above is considering find the number of workers who are overweight.     [2015]

Answer:

Below the table

Weight f  cf
50-60 4 4
60-70 7 11
70-80 11 22
80-90 14 36
90-100 6 42
100-110 5 47
110-120 3 50
Total 50

2015 icse 1

(i)   Lower quartile= \frac{N^{th}}{4}{ \ obs}^n

= \frac{{50}^{th}}{4}{ \ obs}^n = 12.5th { \ obs}^n = 72 \ kg

Upper quartile = \frac{3N}{4}\ th\ {obs}^n   = \frac{150}{4}th\ {obs}^n  = 37.5th { \ obs}^n    = 92 \ kg

(ii)  No. of over weight workers = 50-39 = 11 

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Question 6: The number 6,8,10,12,13 and x are arranged in an ascending order. If the mean of the observations is equal to the median, find the value of x     [2014]

Answer:

Arrange numbers in ascending order are 6,8,10,12,13, x.

Mean = \frac{6+8+10+12+13+x}{6}=\frac{49+x}{6}

No. of terms (n) = 6 \ (even)

Median = \frac{ (\frac{n}{2})^{th} \ term + (\frac{n}{2}+1)^{th} \ term }{2}

Median = \frac{ (\frac{6}{2})^{th} \ term + (\frac{6}{2}+1)^{th} \ term }{2} = \frac{3^{rd}+4^{th}}{2} = \frac{10+12}{2} = \frac{22}{2} = 11

According to given condition

\frac{49+x}{6}=11 \Rightarrow 49+x=66 

or x=17

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Question 8:  Calculate the mean of the distribution given below using the short cut method;     [2014]

Marks 11-20 21-30 31-40 41-50 51-60 61-70 71-80
No. of students 2 6 10 12 9 7 4

Answer:

Table as follows

Marks \\ (C.I) f Mean Value x A=45.5  d=x-A f \times d
11-20

21-30

31-40

41-50

51-60

61-70

71-80

2

6

10

12

9

7

4

15-5

25-5

35-5

45-5

55-5

65-5

75-5

-30

-20

-10

0

10

20

30

-60

-120

-100

0

90

140

120

\Sigma f=50  \Sigma fd=70

Mean = A + \frac{ \Sigma fd}{\Sigma f} = 45.5+\frac{70}{50} = 45.5 + 1.4 = 46.9

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Question 8: The median of the following observation 11,12,14 (x-2), (x+4), (x+9), 32, 38, 47 arranged in ascending order is 24 . Find the value of x and hence find the mean.     [2013]

Answer:

Given observation are 11,12,14, (x-2), (x+4), (x+9), 32, 38, 47 and mediam =24

Since n=9   which is odd, therefore

Median=\ \frac{n+1}{2}th\ term =\frac{9+1}{2}th\ term

24=5th\ term 

x+4=24 

\Rightarrow x=24-4 \Rightarrow x=20 

Therefore, 11,12,14, (20-2), (20+4), (20+9),32,38,47

=11,12,14,18,24,29,32,38,47

Now, Mean =\frac{\sum{x}}{n} 

=\ \frac{11+12+14+18+24+29+32+38+47}{9} =\frac{225}{9} =25 

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Question 9: Draw a histogram from the following frequency distribution and find the made from the graph:   [2013]

Class 0-5 5-10 10-15 15-20 20-25 25-30
Frequency 2 5 18 14 8 5

Answer:

16-1

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Question 10: Find the mean of the following distribution by step deviation method:   [2013]

Class Internal 20-30 30-40 40-50 50-60 60-70 70-80
Frequency 10 6 8 12 5 9

Answer:

C.I Frequency

(f_i)

Mid-Value

(x)

 d_i = \frac{x-a}{h} (f_i. d_i)
20-30

30-40

40-50

50-60

60-70

70-80

10

6

8

12

5

9

25

35

45

55

65

75

-2

-1

0

1

2

3

-20

-6

0

12

10

27

 \Sigma f_i = 50  \Sigma f_i.d_i = 23

Here, a=45  and h=10 

Mean = a + \frac{\Sigma f_i.d_i }{\Sigma f_i}  = 45 + \frac{23}{50} . 10 = 49.6 

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Question 11: The mark obtained by 120 students in a test are given below;

Marks 0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80 80-90 90-100
Students 5 9 16 22 26 18 11 6 4 3

Draw an ogive for the given distribution on a graph sheet; Using suitable scale for ogive to estimate the following:

(i) The mediam

(ii) The number of students who obtained more than 75% marks in the test.

(iii) The number of students who did not pass the test if minimum marks required to pass is 40.   [2013]

Answer:

Marks No. of students Cumulative Frequency
0-10

10-20

20-30

30-40

40-50

50-60

60-70

70-80

80-90

90-100

5

9

16

22

26

18

11

6

4

3

5

14

30

52

78

96

107

113

117

120

N=120

On the graph paper we plot the following points:

(10,5), (20,14), (30,30), (40,52), (50,78), (60,96), (70,107), (80,113), (90,117), (100,120)

16-11

(i) Mediam = (\frac{n}{2})^{th} \ term = \frac{120}{2} =60^{th} \ term 

From the graph 60th term =42

(ii) The number of students who obtained more than 75\% marks in test = 120-110 = 10

(iii) The number of students who did not pass the test if the minimum pass marks 40=52

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Question 12: Marks obtained by 40 students in a short assessment is given below, where a \ and \  b are two missing data:

Marks 5 6 7 8 9
No. of Students 6 A 16 13 B

If the mean of the distribution is 7.2 find a \ and \  b .    [2012]

Answer:

Given, the total number of students = 40

Therefore 6 + a+ 16+ 13+ b  = 40

\Rightarrow a + b = 5 … … … (i)

Given mean (\overline{x}) = \frac {\Sigma fx}{\Sigma f} = 7.2

Therefore 7.2 = \frac{5 \times 6 + 6 \times a +7 \times 16 +8 \times 13 +9 \times b}{40}

246 + 6a + 9 b = 288

6a + 9 b = 42

2a + 3b = 14 … … … (ii)

Solving (i) and (ii) we get a = 1 and b = 4 

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Question 13: The following distribution represents the height of 160 students of a school

Height 140-145 145-150 150-155 155-160 160-165 165-170 170-175 175-180
Students 12 20 30 38 24 16 12 8

 Draw a given for the given distribution taking 2 cm =5 cm of height on one axis and 2 cm=20 students on the other axis. Using the graph, determine:

(i) The medium height

(ii) The interquartile range

(iii) The number of students whose height is above 172 cm.     [2012]

Answer:

Following table:

Height F c.f.
140-145

145-150

150-155

155-160

160-165

165-170

170-175

175-180

12

20

30

38

24

16

12

8

12

32

62

100

124

140

152

160

icse 2012 5

(i) Mean = 157.3

(ii) Interquartile range = Q_3 - Q_1 = 163.5 - 152 = 11.5

(iii) No. of students above 172 \ cm = 160-144 = 16

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Question 14: Find the mode and medium of the following frequency distribution:    [2012]  

X 10 11 12 13 14 15
f 1 4 7 5 9 3

Answer:

Mode is the value of the highest frequency.

Therefore Mode = 14

For Median, first write the data in ascending order as follows:

10, 11, 11, 11, 11, 12, 12, 12, 12, 12, 12, 12,  \\ 13, 13, 13, 13,13, 14, 14, 14,14, 14, 14, 14, 14, 14,15, 15, 15 .

Since Median is the middle most value.

Median = (\frac{N+1}{2})^{th} observation = (\frac{29+1}{2})^{th} observation = 15^{th}  \ observation = 13

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Question 15: A mathematics aptitude test of 50 students was recorded as follows:

Marks 50-60 60-70 70-80 80-90 90-100
No. of students 4 8 14 19 5

Draw a histogram for the above data using a graph paper and locate the mode.   [2011]

Answer:

2011-3

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Question 16: Using step division method, calculate the mean marks of the following distribution: State the modal class:     [2011]

Class Interval 50-55 55-60 60-65 65-70 70-75 75-80 80-85 85-90
Frequency 5 20 10 10 9 6 12 8

Answer:

C.I f  x   d=x-67.5   u   f.u.
50-55 5 52.5 -15 -3 -15
55-60 20 57.5 -10 -2 -40
60-65 10 62.5 -5 -1 -10
65-70 10 67.5 0 0 0
70-75 9 72.5 5 1 9
75-80 6 77.5 10 2 12
80-85 12 82.5 15 3 36
85-90 8 87.5 20 4 32
\Sigma f = 80   \Sigma fu = 24 

A.M = 67.5 

(i) \bar{x} = A.M. + \frac{\Sigma fu}{\Sigma f} \times i = 67.5 + \frac{24}{80} \times 5 = 69 

(ii) Modal class is 55-60  (class with highest freq.)

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Question 17: Marks obtained by 200 students in an examination are given below:

Marks 0-10 10-20 20-30 30-40 40-50 50-60 60-70 70-80 80-90 90-100
No. of students 5 11 10 20 28 37 40 29 14 6

Draw an ogive for the given distribution taking 2 \ cm=10 marks on one axis and 2 \ cm=20 students on the other axis. Using the graph, determine.

(i) The median marks

(ii) The number of students who failed if minimum marks required to pass is 40 .

(iii) If scoring 85 and more marks is considered as grade one, find the number of students who secured grade on in the examination;     [2011]

Answer:

Class Interval Frequency Cumulative Frequency
0-10 5 5
10-20 11 16
20-30 10 20
30-40 20 46
40-50 28 74
50-60 37 111
60-70 40 151
70-80 29 180
80-90 14 194
90-100 6 200

2011-8

(i) n = 200 (even)

Median = (\frac{n}{2})^{th} observation = (\frac{200}{2})^{th}  observation = 100^{th} observation = 57

(ii) Number of student who failed = 46 

(iii) Number of students who secured grade one  200 - 1888 = 12 

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Question 18:  The distribution given below shows the marks obtained by 25 students in an aptitude test. Find the mean, median and mode of the distribution.     [2010]

Marks obtained 5 6 7 8 9 10
No. of students 3 9 6 4 2 1

Answer:

x_i f_i x_if_i cf
5 3 15 3
6 9 54 12
7 6 42 18
8 4 32 22
9 2 18 24
10 1 10 25
\Sigma = 25 \Sigma x_if_i = 171

Mean = \frac{\Sigma x_if_i}{N}  = \frac{171}{25} = 6.84

Since N = 25 is odd,

Median = (\frac{n+1}{2})^2 \ term = 13^{th} \ term = 7

Mode = 6 (maximum frequency)

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Question 19: The mean of the following distribution is 52 and the frequency of class interval 30-40 is f . Find f      [2010]

Class Interval 10-20 20-30 30-40 40-50 50-60 60-70 70-80
Frequency 5 3 f 7 2 6 13

Answer:

Interval Frequency (f_i) x-i d_i = x_i - A f_id_i
10-20 5 15 -30 -150
20-30 3 25 -20 -60
30-40 F 35 -10 -10f
40-50 7 45 (A) 0 0
50-60 2 55 10 20
60-70 6 65 20 120
70-80 13 75 30 390
36+f \Sigma f_id_i= 320 - 10f

Mean = A + \frac{\Sigma f_id_i}{N}

\Rightarrow 52 = 45 + \frac{320-10f}{36+f}

\Rightarrow 7 = \frac{320-10f}{36+f}

\Rightarrow 252 + 7f = 320 - 10f

\Rightarrow 17f = 68

\Rightarrow f = 4

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Question 20: The monthly income of a group of 320 employees in a company is given below:

Monthly Income No. of Employees
6000-7000 20
7000-8000 45
8000-9000 65
9000-10000 95
10000-11000 60
11000-12000 30
12000-13000 5

Draw an ogive of the given. distribution on a graph sheet taking 2 \ cm = Rs. \ 1000 on one axis and 2 \ cm = 50 \ employees on the other axis. From the graph determine:

(i) the median wage

(ii) the number of employees whose income is below Rs.  \ 8500

(iii) If the salary of a senior employee is above Rs.  \ 11500 find the number of senior employees in the company.

(iv) The upper quartile     [2010]

Answer:

Monthly Income No. of Employees Cumulative Frequency
6000-7000 20 20
7000-8000 45 65
8000-9000 65 130
9000-10000 95 225
10000-11000 60 285
11000-12000 30 315
12000-13000 5 320

2010-6

Here n (no. of employees) = 320 (even)

(i) Median = \frac{1}{2} \{ \frac{n}{2} + (\frac{n}{2} + 1) \} = \frac{1}{2} (160+161) = 160.5

Required median = Rs.  \ 9800 (from graph)

(ii) Number of employees whose income is below Rs. \ 8500 = 95 approx

(iii) Number of senior employees in the company = 320 - 305 = 15

(iv) Upper Quartile = (\frac{3n}{4})^{th} = \frac{3 \times 320}{4} = 240^{th} term

Upper Quartile = Rs. \ 10200

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