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Statistics II Questions

1.  The table below shows the number of defective bolts from a sample of 40

No of bolts

0

1

2

3

4

5

Frequency

20

8

6

4

1

1

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 Calculate the standard deviation of the data above   (4 mks)

2.  The table below shows the masses to the nearest kg of all the students of marigu-ini secondary. School.

 

Masses (kg)

No. of students

30-34

5

35-39

7

40-44

10

45-49

10

50-54

19

55-59

20

60-64

20

65-69

6

70-74

2

75-79

1

 

  1. Taking the assumed mean A=52kg

    Calculate:

    1. the actual mean mass of the students. (3 marks)
    2. the standard deviation of the distribution. (3 marks)
  2. Draw a cumulative frequency curve and use it to estimate the number of students whose masses lie

between 44.5kg and 59.5kg. (4 marks)

 

3.  Sixty form four students in Tahidi high sat for a mathematics examination. Their marks were  grouped into seven classes as follows: 30 – 34, 35 – 39, 40 – 44, 45 – 49, 50 – 54, 55 – 59, 60 –  64 and then named as cheetah, lion, zebra, rabbit, giraffe, elephant and buffalo respectively. The  form 4 students population was then analyzed in the form of a pie-chart as shown below.

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 Using the information above

 (a) Complete the table below.

Name

Marks

No. of students

Cheetah

Lion

Zebra

Rabbit

Giraffe

Elephant

Buffalo

30-34

35-39

40-44

45-49

50-54

55-59

60-64

 

 (2mks)

 (b) Calculate the inter quartile range. (3mks)

 (c) Using an assumed mean of 47, calculate the standard deviation of the data. (5mks)

4.  At an agricultural Research Centre, the length of a sample of 50 maize cobs were measured and recorded as shown in the frequency distribution table below.

 

Length

8 – 10

11 – 13

14 – 16

17 – 19

20 – 22

23 – 24

No. of Labs

4

7

11

15

8

5

 

 a)  State the modal class and size.  (2mks)

Calculate

 b)  the mean  (3mks)

 c)  (i)  the variance (3mks)

(ii)  the standard deviation.  (2mks)

 

5.  The table below shows the masses to the nearest kg of a number of people.

Mass (kg)

Frequency

50 – 54

19

55 – 59

23

60 – 64

40

65 – 69

28

70 – 74

17

75 – 79

9

80 – 84

4

 

a)Using an assumed mean of 67.0, calculate to one decimal place the mean mass.

(b) Calculate to one decimal place the standard deviation of the distribution.

6.  Use only a ruler and pair of compasses in this question;

 (a) construct triangle ABC in which AB = 7cm, BC = 6cm and AC = 5cm

 (b) On the same diagram construct the circumcircle of triangle ABC and measure its radius

 (c) Construct the tangent to the circle at C and the internal bisector of angle BAC. If these

lines meet at D, measure the length of AD

Image From EcoleBooks.comImage From EcoleBooks.comImage From EcoleBooks.comImage From EcoleBooks.comImage From EcoleBooks.comImage From EcoleBooks.comImage From EcoleBooks.comImage From EcoleBooks.comImage From EcoleBooks.comImage From EcoleBooks.comImage From EcoleBooks.comImage From EcoleBooks.comImage From EcoleBooks.comImage From EcoleBooks.comImage From EcoleBooks.comImage From EcoleBooks.comImage From EcoleBooks.comImage From EcoleBooks.comImage From EcoleBooks.com7.  Below is a histogram drawn by a student of Got Osimbo Girls Secondary School.

 

Image From EcoleBooks.com

 

 

 

 

 

 

Image From EcoleBooks.com

 

Image From EcoleBooks.com

 

 

 

 

 

 

 

 

 

  1. Develop a frequency distribution table from the histogram above.

b) Use the frequency distribution table above to calculate;

i) The inter-quartile range.

ii) The sixth decile.

 

8.  ABC is a triangle drawn to scale. A point x moves inside the triangle such that

 i) AX ≤ 4 cm

 ii) BX  CX

Image From EcoleBooks.com  iii) Angle BCX ≤ Angle XCA.

  Show the locus of X.

 

 

Image From EcoleBooks.com

 

 

 

 

Image From EcoleBooks.comImage From EcoleBooks.com

 

 

 

9.  The following able shows the distribution of marks of 80 students

 

Marks

1-10

11-20

21-30

31-40

41-50

51-60

61-70

71-80

81-90

91-100

Frequency

1

6

10

20

15

5

14

5

3

1

 

(a) Calculate the mean mark

 (b) Calculate the semi-interquartile range  

 (c) Workout the standard deviation for the distribution

 

10.  The table below shows the marks of 90 students in a mathematical test

 

Marks

5-9

10-14

15-19

20-24

25-29

30-34

35-39

No. of students

2

13

31

23

14

X

1

  1. Find X  
  2. State the modal class

(c) Using a working mean of 22, calculate the; i) Mean mark ii) Standard deviation  

 

11.  (a) Using a ruler and a pair of compasses only construct triangle PQR in which PQ = 5cm,

PR = 4cm and PQR = 30o

 (b) Measure;  (i) RQ

  (ii) PQR

 (c) Construct a circle, centre O such that the circle passes through vertices P, Q, and R    (d) Calculate the area of the circle

 

12.   The ages of 100 people who attended a wedding were recorded in the distribution table below

Age

0-19

20-39

40-59

60-79

80-99

Frequency

7

21

38

27

7

 a) Draw the cumulative frequency

 b) From the curve determine:  i) Median  ii) Inter quartile range iii) 7th Decile  iv) 60th Percentile

 

13.  The marks obtained by 10 students in a maths test were:-

  25, 24, 22, 23,
x
, 26, 21, 23, 22 and 27

 The sum of the squares of the marks, x2 = 5154

(a) Calculate the:   (i) value of x       (ii) Standard deviation

 (b) If each mark is increased by 3, write down the:-

(i) New mean  

(ii) New standard deviation

 

14.  40 form four students sat for a mathematics test and their marks were distributed as follows:-

 

Marks

1 – 10

11- 20

21- 30

31 – 40

41 – 50

51 – 60

61 – 70

71 – 80

81 – 90

91 – 100

No. of students

 

1

 

3

 

4

 

7

 

12

 

9

 

2

 

1

 

0

 

1

 

a) Using 45.6 as the working mean, calculate;

  i) The actual mean.  

  ii) The standard deviation.

 b) When ranked from first to last, what mark was scored by the 30th student?

(Give your answer correct to 3 s.f.)  

 

15.  The table below shows the distribution of marks scored by pupils in a maths test at Nyabisawa

Girls.

 

Marks

11 – 20

21 – 30

31 – 40

41 – 50

51 – 60

61 – 70

71 – 80

81 – 90

Frequency

2

5

6

10

14

11

9

3

 

a)Using an Assumed mean 45.5, calculate the mean score.

b) Calculate the median mark.

c) Calculate the standard deviation.

d) State the modal class.

 

16.  The table below shows the marks scored in a mathematics test by a form four class;

 

Marks

20-29

30-39

40-49

50-59

60-69

70-79

80-89

No. of students

4

26

72

53

25

9

11

 

 (a) Using an assumed mean of 54.5, calculate:-

  (i) The mean  

  (ii) The standard deviation

 (b) Calculate the inter quartile range  

 

 

Statistics II Answers

1

 

X

0

1

2

3

4

5

f

20

8

6

4

1

1

fx

0

8

12

12

4

5

fx2

0

8

24

36

16

25

Image From EcoleBooks.com
Image From EcoleBooks.com
Image From EcoleBooks.com

Image From EcoleBooks.com

Image From EcoleBooks.com

Image From EcoleBooks.com

 

 

 

M1

 

M1

 

M1

 

 

 

A1

 

 

 

fx

 

fx2

 

 

 

 

 

Allow 1.293976429

  

4

 

 

1.  

 

Mass kg

Mid term

x

 

F

 

d = x A

 

fd

 

d2

 

fd2

50 – 54

55 – 59

60 – 64

65 – 69

70 – 74

75 – 79

80 – 84

52

57

62

67

72

77

82

19

23

40

28

17

9

4

-15

-10

-5

0

5

10

15

-285

-230

-200

0

85

90

60

225

100

25

0

25

100

225

4275

2300

1000

0

425

900

900

  

f = 140

 

fd = – 480

 

fd = 9800

Marks awarded for √ table as follows:-

  f = 140 BI

 Column for d B1

 Column for fd B1

 fd = – 480 B1

 √Column for d2 = 9800 B1

fd = 9800B1

x =A + fd

f

= 67.0 + – 480

140

= 67.0 – 3.43 = 63.57 ………… M1

= 63.6 kg …………… A1

 

Standard deviation = fd2fd

  f f

 

 

= 9800 – (3.43)2

140

 

 

= 58.24 = 7.631

 

= 7.6

 

 

2.  = 8/150 + 6/150 + 9/300 + 3/ 300

 

= 40/300 = 2/15

 

  1. Construction of AB B1

Construction of BC B1

Construction of AC B1

b) Construction of bisect of AC B1

Construction of bisect BC B1

Radius 3.6 cm B1

 

c) Construction of bisect < CAB B1 OC B1

Construction of AD B1 AD = 12.8cm B1

3.  a)

Class

f

x

d = A – x

fd  

d2

fd2

41 – 50

51 – 55

56 – 65

66 – 70

71 – 85

20

60

60

50

15

45.5

53

60.5

68

73

15

7.5

0

-7.5

-12.5

300

450

0

-375

187.5

225

56.25

0

56.25

156.25

4500

3375

0

2812.50

2343.75

    

∑fd 562.5

 

∑fd2 13031.25

 

 

 b) S = ∑fd2
∑fd

∑f ∑f

 

 

S = 13031.25 – 562.5

205 205

 

 

= 63.567 – 7.529

 

 

= 56.038

 

= 7.486

4.  15 (ax)4 (-2/x2) = 4860

 

  60a4 = 4860

 a4 = 81

 a = 3

5.  

Marks(x)

Freq.(f)

fx

d=x-x

d2

Fd2

5.5

15.5

25.5

35.5

45.5

55.5

65.6

75.5

85.5

95.5

1

6

10

20

15

5

14

5

3

1

5.5

99

255

710

682.5

277.5

917

377.5

256.5

95.5

-40.45

-30.45

-20.45

-10.45

-0.45

9.55

19.55

29.55

39.55

49.55

1636

927.2

418.2

109.2

0.2025

91.20

382.2

873.2

1564

2455

1636

5563

4182

2184

3038

456

535

4366

4692

2455

 

f=80

f99x=3676

  

fx233,923

Mean =
fx = 3676

 f  80

= 45.95

(b) Q1 = 30.5 + 3 x 10

  14

 = 62.64

S.I.R = ½ (62.64 -32)

  = 15.32

(c) Standard deciation

 

= fd2 = 33923

  f 80

= 20.59

 

6.  a) x = 90 – (2 +13 + 51 + 27 + 14 + 1)

= 90 – 84 = 6

b) 15 – 19

 

c) i)

Class

x

f

D= x-A

fd

D2

Fd2

5-9

7

2

-15

-30

225

450

10-14

12

13

-10

-130

100

1300

15-19

17

31

-5

-155

25

775

20-24

22

23

0

0

0

0

25-29

27

14

5

70

350

4900

30-34

32

6

10

60

600

3600

35-39

37

1

15

15

225

225

 

Ef = 90 Efd = 170 Efd2 = 11250

 

Mean  = E + d + A

Ef

= -170 + 22

90

= 22 – 1.888 = 20.11

 

ii) S.d = √Efd – [Efd]2

Ef Ef

= √ 122 – (-1.888)2

= √125 – 3.566 = √121.4

= 11.02

Image From EcoleBooks.com7.  

 

 

 

 

 

 

 

 

 

 

RQ = 7.5 0.1

< PRQ 40°  1

B1 circle through P, Q and R

 

d)   r = 4.1 ° 0

A = r2

22/7 x 4.1 x 4.1 = 52.83

8.  

Class limits

f

cf

-0.5 – 19.5

7

7

19.5- 39.5

21

28

39.5 – 59.5

38

66

59.5 – 79.5

27

93

79.5 -= 99.5

7

100

 

i) from the curve   – median = 52. M1 A1

(ii) Inter quartile range = 66-38 = 28.

(iii) 7th 7/10 = 62.46marks

(iv) 60th percentile – 56.34

 

 

9.  252 + 242 + 222 + 232 + x2 + 262 + 212 + 232 + 222 + 272 = 5154

5.625 +576 + 2(484) + 2(529) + 676 + 441 + 729 + x2 = 5154

X2 = 81

X =9

Image From EcoleBooks.com

(ii) X = 222 = 22.2

Image From EcoleBooks.com   10

(X – x)2 = 2.82 + 1.82 + 0.22 + 0.82

Image From EcoleBooks.com13.22 + 3.82 + 1.22 + 0.82 + 0.22 + 4.82

(x-x)2 = 7.84 + 3.24 2(0.04) + 2(0.64)

+174.24 + 14.44 + 1.44 + 23.04

= 225.6

10

s.d 22.56

= 4.75

 

(b) (i) New mean = 22.2 + 3

  = 25.2

(ii) s.d = 4.75

 

 

 

10.  a) i) x = A + ∑fd

  ∑f

 = 45.6 + (-74)

  40

  = 43.75

 

Class

Mis-pt x

d = (x – A)

Frequency f

fd

Fd2

1 – 10

11 – 20

21 – 30

31 – 40

41 – 50

51 – 60

61 – 70

71 – 80

81 – 90

91 – 100

5.5

15.5

25.5

35.5

45.5

55.5

65.5

75.5

85.5

95.5

-40.1

-30.1

-20.1

-10.1

-0.1

9.9

19.9

29.9

39.9

49.9

1

3

4

7

12

9

2

1

0

1

-40.1

-90.3

-80.4

-70.7

-1.2

89.1

39.8

29.9

0

49.9

1608.01

8154.05

6464.16

4998.49

1.44

7938.81

1584.04

894.01

0

2410.01

 


 

 

 

 

 

 

 

 

 

i) Standard Deviation

Image From EcoleBooks.com

D = e ∑fd2∑fd

Image From EcoleBooks.com ∑f ∑f

 

= 10 34135.11 – -74

40 40

10 x 29.1531 = 29.1531

 

b) 30th student = 10th from bottom

30.5 + 10 – 8 10

7

= 30.5 + 2.9 = 33.4 marks.

 

11.  a) Mean 45. 5 + 530

60

  = 54.33

 

 (b) Median  = 50.5 + 30.5 – 23 10

14

= 55.86

Image From EcoleBooks.com

  (c) Standard deviation = 2300 – 530

60 60

 = 17.52

 (d) Modal class 51 – 60

12.  

x

f

d

d2

fd

fd2

24.5

4

-30

900

-120

3600

34.5

26

-20

400

-520

10400

44.5

72

-10

100

-720

7200

54.5

53

0

0

0

0

64.5

25

10

100

250

2500

74.5

9

20

400

180

3600

84.5

11

30

900

330

9900

 

200

  

-600

37200

(a) (i) Mean = A + fd

  f

  = 54.5 – 600

200

 = 51.5

(ii) Standard deviation

= fd2fd2

  f f

= 37200 – (-3)2

  200

= 186 – 9

= 13.30

 

(b) Q1 = 39.5 + 50 – 30 x 10

72

= 42.28

Q3 = 49.5 + 150-102 x 10

53

= 58.56

Q3 – Q1 = 58.56 – 42.28

   = 16.28

 

 

 

 

 

 

 

 


 




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