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Representation of data Questions

1.  The height of 36 students in a class was recorded to the nearest centimeters as follows.

 148  159  163  158  166  155  155  179  158  155  171  172  156  161  160  165  157  165  175  173  172  178  159  168  160  167  147  168  172  157  165  154  170  157  162  173

 (a) Make a grouped table with 145.5 as lower class limit and class width of 5. (4mks)

 

 

2.  Below is a histogram, draw.

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 Use the histogram above to complete the frequency table below:

 

Length

Frequency

11.5 ≤ x ≤13.5

13.5 ≤ x ≤15.5

15.5 ≤ x ≤ 17.5

17.5 ≤ x ≤23.5

 

 

 

 

3.  Wambui spent her salary as follows:

Food

40%

Transport

10%

Education

20%

Clothing

20%

Rent

10%

 

  Draw a pie chart to represent the above information

 

4.  The examination marks in a mathematics test for 60 students were as follows;-

 

60

54

34

83

52

74

61

27

65

22

70

71

47

60

63

59

58

46

39

35

69

42

53

74

92

27

39

41

49

54

25

51

71

59

68

73

90

88

93

85

46

82

58

85

61

69

24

40

88

34

30

26

17

15

80

90

65

55

69

89

Class

Tally

Frequency

Upper class limit

10-29

30-39

40-69

70-74

75-89

90-99

   

From the table;

 (a) State the modal class

  (b) On the grid provided , draw a histogram to represent the above information

5.  The marks scored by 200 from 4 students of a school were recorded as in the table below.

Marks

41 – 50

51 – 55

56 – 65

66 – 70

71 – 85

Frequency

21

62

55

50

12

      

 

  1. On the graph paper provided, draw a histogram to represent this information.
  2. On the same diagram, construct a frequency polygon.
  3. Use your histogram to estimate the modal mark.

 

6.  The diagram below shows a histogram representing the marks obtained in a certain test:-

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 (a) If the frequency of the first class is 20, prepare a frequency distribution table for the data

 (b) State the modal class

 (c) Estimate:   (i) The mean mark   (ii) The median mark

 

7.  The data below shows the number of sessions different subjects are taught in a week.

 Draw a pie chart to show the data:

Subject

Eng

Maths

Chemistry

C.R.E

No. of sessions

8

7

4

3

 

8.  The height of 50 athletes in Moi University team were shown below:

 

Height (cm)

150-159

160-169

170-179

180-189

190-199

200-209

Frequency

2

9

12

16

7

4

 

i) State the modal class

 ii) Calculate the median height of the athletes

9.  The table below shows the length of 40 mango tree leaves;

Length (mm)

Frequency

Cumulative frequency

118-126

127-135

136-144

145-153

154-162

163-171

172-180

3

4

10

12

5

4

2

3

7

17

29

34

38

40

 

(a) Determine the;

  (i) Modal class  

  (ii) Median class  

 (b) Calculate; (i) the mean of the leaves   (ii) the median of the leaves

 

Representation of data Answers

1.

(a)

Class

Tally mark

Freq

Freq D

145-149

//

2

2

150-154

/

1

1

155-159

//// //// /

11

11

169-164

////

5

5

165-169

//// //

7

7

170-174

//// //

7

7

175-179

///

3

3

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B1

B1

B1

B1

 

B1

B1

 

 

B1

B1

 

B1

 

B1

 

Classes

Tally mark column

Freq. column

Freq density column

(can be implied)

Freq. density (y axis

Height (x axis)

Correct spacing as per scale

Histogram drawn (bars)

Joining mid point of the bars

Joining mid point of first class to 144.5

Joining mid point of last class to 179.5

  

10

 

 

 

 

 

1.  

Length

11.5 ≤ x≤ 13.5

13.5 ≤ x≤ 15.5

15.5 ≤ x≤ 17.5

17.5 ≤ x≤ 23.5

Frequency

6

9

6

3

 

 

 

2.  Food: 40/100 x 360 = 144 °

Transport: 10/100 x 360 = 36°

Education: 20/100 x 360 = 72 °

Clothing: 20/100 x 360 = 72 °

Rent: 10/100 x 360 = 36°

 

3.  Class Tally Frequency Upper Limit

  10 – 29 ∣∣∣∣
∣∣∣ 8 29.5 B2 for

30 – 39 ∣∣∣∣
∣ 6 39.5 all tally

 40 – 69 ∣∣∣∣
∣∣∣∣
∣∣∣∣
∣∣∣∣
∣∣∣∣
∣∣∣ 28 69.5 B2 all

 70 – 74 ∣∣∣∣
∣ 6 74.5 – frequency

  75 – 89 ∣∣∣∣
∣∣∣ 8 89.5 – B1

  90 – 99 ∣∣∣∣ 4 99.5 B1

 

 Modal class 40 – 69 B1

 

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4.  See the graph paper.

For correct class boundaries

For correct class intervals.

All frequency densities

 

Correct scale

All the bars drawn.

 

Top mid pts. Of bars indicated.

For the mid pts. Joint to make a polygon.

For correctly identifying the modal mark point.

For reading correctly the modal mark ≡ 53.5  0.1

5. (a)

Marks

Frequency

5-9

10-19

20-30

40-49

20

50

40

30

(b) Modal class is 10-19

(c)(i)

Class

x

f

fx

Cf

5-9

7

20

140

20

10-19

14.5

50

725

70

20-39

29.5

40

1180

110

40-49

44.5

30

1335

140

  

F = 140

Fx =3380

 

 

x = fx = 3380 = 24.14

f 140

(ii) Median mark is at 70 + 71 = 70.5th position

Median = 119.5 + (0.5) x 20

  40

= 19.5 + 0.25

= 19.75

 

6.  Total No. of sessions

= 8 + 7 + 4 + 3 = 22

Angle for:

English = 8/22 x 360 = 130.9°

Maths = 7/22 x 360 = 114.5°

Chemistry = 4/22 x 360 = 65.5°

CRE = 3/22 x 360 = 49.01°

 

7.  180 – 189

Class limits

class

limits

f

cf

149.5

159.5

2

2

159.5

169.5

9

11

169.5

179.5

12

23

179.5

189.5

16

39

189.5

199.5

7

46

199.5

209.5

4

50

 

 

 

 

Median = 50/2 = 25

179.5 + 25 – 23 x 10

  16

= 179.5 + 20 = 180.75

16

179.5 + 26 – 23 x 10

16

179.5 + 30 = 181.38

  16

180.75 + 181.38

  2

= 181.06

 

8.  a)  i) 145 – 153

ii) Median class

(40 + 1/2)th value  median class = 145 – 153

 This is the 20.5th value

The value also in the 145 – 153 class

 

b)

Class

x

f

fx

118- 126

127- 135

136 – 144

145 – 153

154 – 162

163 – 171

172 – 180

122

131

140 B1

149

158

167

176

3

4

10 B2

12

5

4

2

366

524

1400

1788

790

668

352

  

Ef = 40

Efx = 5888

 B2 for all values of fx correct and B1 for 4 values of fx and above orrect

 Mean = Efx = 5888 = 147.2mm

  Ef 40

Median 20th = 144.5 + (11/12 x 9) = 152.75

  21st = 144.5 + (12/12 x 9) = 153.5

 Median = 152.75 + 153.5 = 153.125

2

(Alternatively one could work out the 20.5 value directly using median formula)




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