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Measures of central tendency Questions

1. The results of a mathematics test that a hundred students took are as shown below:-

Marks

No. of students

30-34

35-39

40-44

45-49

50-54

55-59

60-64

65-69

4

6

10

14

ecolebooks.com

X

24

14

6

 

 

 

 

 

 

 

 

 

 

 

(a) Determine (i) the value of X  

(ii) The modal class  

 (b) Calculate the mean

(c) The median

 

2.  Without using logarithms or calculator evaluate:

2log105 – 3log102 + log1032

 

3.  The table below shows heights of 50 students :-

Height (cm)

Frequency

140-144

145-149

150-154

155-159

160-164

3

15

19

11

2

 

(a) State the modal class

 (b) Calculate the median height

4.  In an experiment, the height of 100 seedlings were measured to the nearest centimeter

and the results were recorded as shown below;

Height (cm)

20-24

25-29

30-34

35-39

40-44

45-49

Frequency

3

19

25

20

18

15

  Calculate the median height

5.  Given that x = -4 is a root of the equation 2x2 + 6x – 2k = 0; Find;

 (a) the value of k  

 (b) the second root  

Marks

60 – 62

63 – 68

69 – 73

74 – 80

Frequency

10

20

40

15

 

7.  The table below shows the distribution of marks obtained by some candidates in a mathematics

test

Marks

30-39

40-49

50-59

60-69

70-79

80-89

90-99

No. of candidates

2

3

10

12

8

3

2

c.f

       

 

a) state the total number of candidates who sat the test

 b) state the modal class

 c) calculate the mean mark using an assumed mean of 64.5 marks  

 d) calculate the median mark

 

8.  Find these statistics of the following data 4, 2, 2, 6, 1, 3, 4, 1, 4

 a) Mode

 b) Median

 c) Mean

 

9.  (a) The marks scored by a group of form two students in a mathematical test were as recorded

in the table below:-

Marks

0-9

10-19

20-29

30-39

40-49

50-59

60-69

70-79

80-89

90-99

Frequency

1

2

4

7

10

16

20

6

3

1

 

 

 

(a) (i) State the modal class

  (ii) Determine the class in which the median mark lies

  (iii) Using an assumed mean of 54.5, calculate the mean mark  

 

10.  Six weeks after planting, the height of maize plants were measured correct to the nearest

centimeter. The frequency distribution is given in the table below:

Height (x)

0 ≤ x < 4

4 ≤ x < 8

8 ≤ x < 12

12 ≤ x < 16

16 ≤ x < 20

Frequency

3

8

19

14

6

Estimate the median height of the plants

 

11.  Below are marks scored by student in maths talk in science congress.

 Marks

1 – 5

6 – 15

16 – 20

21 – 35

36 – 40

41 – 50

No. of students

1

3

6

12

5

3

 

Draw a histogram from the table above.

 

Measures of central tendency Answers

 

1. 4 + 6 + 10 + 14 + x + 24 + 14 + 6 = 100

78 + x = 100

(i) x = 22

(ii) Modal class = 55 -59

 

Marks

x

x

c

30-34

35-39

40-44

45-49

50-54

55-59

60-64

65-69

32

37

42

47

52

57

62

67

4

6

10

14

22

24

14

6

128

222

420

659

1144

1368

868

462

4

10

20

34

56

80

94

100

B1

 = 100 B1

x = 5210

B1

 

x = 5210

  1. Mean = 5210

100

= 52.10

 

(ii) Median = 49.5 + 50-34 x 5

22

= 53.14

2.  Log10 52 – log10 23 + log 25

Image From EcoleBooks.comImage From EcoleBooks.com

  Log10
25 x 32

Image From EcoleBooks.com 8

Image From EcoleBooks.com

  Log10 100 = log1010

= 2 log 1010

  But log1010 = 1

    ∴ = 2

3.  Modal class 150-154

Height

Frequency

c.f

140- 144

145 – 149

150 – 154

155 – 159

160 -164

3

15

19

11

2

3

18

37

48

50

Height Frequency c.f

= 149.5 + (25-18) x 5

  19

= 149.5 + 7 x 5

  19

= 149.5 + 1.842

= 15.34

 

4.

H

20-24

25-29

30-34

35-39

40-44

45-49

F

3

19

25

20

18

15

CF

3

22

47

67

85

100

 

 

 

 

Md = 34.5 + (50 – 47) x 4

 20

 = 34.5 + 12/20 = 35.1

 

5.  a)  2x2 + 6x – 2x = 0

32 – 24 – 2x = 0

-2x = -8

x = 4

 

b) 2x2 + 6x – 8 = 0

x2 + 3x – 4= 0

x2 + 4x – x – 4 = 0

x(x – 4) – (x + 4) = 0

(x – 1) (x + 4) = 0

 the other root is 1

 

6.  ∑xf = 61 x 10 + 65.5 x 20 + 71 x 40 + 77 x 15

= 610 + 1310 + 2840 + 1155

= 5915


∑xf = 5915

 ∑f 85

 X Mean = 69.59

7.

Marks

30-39

40-49

50-59

60-69

70-79

80-89

90-99

No. of candidates

2

3

10

12

8

3

2

C.F

2

5

15

27

35

38

40

 

  1. Number who sat = 40
  2. The modal class = 60 – 69

Marks

x

f

X – 64.5= d

fd

30-39

40-49

50-59

60-69

70-79

80-89

90-99

34.5

44.5

54.5

64.5

74.5

84.5

94.5

2

3

10

12

8

3

2

£f = 40

-30

-20

-10

0

10

20

30

-60

-60

-100

0

80

60

60

£ fd = -20

Mean = 64.5 + -20

40

  = 64.0

d) The median mark

  = ½ (20th and 21st ) marks

  = ½ (59.5 + 5 x 10 + 59.5 + 6 x 10)

12 12

  = ½ (59.5 + 4.16666 + 59.5 + 5)

  = ½ (128.16666667)   = 64.083

 

8.  1, 1, 2, 2, 3, 4, 4, 6

a) Mode = 4

b) Median = 3

c) Mean = 1 x2 + 2 x 2 + 3 x 1 + 4 x 3 + 6 x 1

9

= 3

9.  a)  i) Modal class = 60 – 69

 

 ii) class where medium lies

 median class 50- 59

Class

0 – 9

10 – 19

20 – 29

30 – 39

40 – 49

50 – 59

60 – 69

70 – 79

80 – 89

90 – 99

Centre X

4.5

14.5

24.5

34.5

44.5

54.5

64.5

74.5

84.5

94.5

Fd

-50

-80

-120

-140

-100

0

200

120

90

40

εfd -40

D= x – A

-50

-40

-30

-20

-10

0

10

20

30

40

 

Mean = 54.5 – 40

70

= 53.93

 

10.   Cumulative frequency

 3,11, 30, 44, 50

  Median = L1t (n/2 – cfa)

  Fn

  = 8 + (25 – 11) X 4

  19

 

 = 10.947

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