## 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 ecolebooks.com

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.

Using the information above

(a) Complete the table below.

 Name Marks No. of students CheetahLionZebraRabbitGiraffeElephantBuffalo 30-3435-3940-4445-4950-5455-5960-64

(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 – 5419 55 – 5923 60 – 6440 65 – 6928 70 – 7417 75 – 799 80 – 844

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

7.  Below is a histogram drawn by a student of Got Osimbo Girls Secondary School.

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

iii) Angle BCX ≤ Angle XCA.

Show the locus of X.

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?

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

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

M1

M1

M1

A1

fx

fx2

Allow 1.293976429

4

1.

 Mass kg Mid termx F d = x A fd d2 fd2 50 – 5455 – 5960 – 6465 – 6970 – 7475 – 7980 – 84 52576267727782 192340281794 -15-10-5051015 -285-230-2000859060 22510025025100225 4275230010000425900900 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

c) Construction of bisect < CAB B1 OC B1

3.  a)

 Class f x d = A – x fd d2 fd2 41 – 5051 – 5556 – 6566 – 7071 – 85 2060605015 45.55360.56873 157.50-7.5-12.5 3004500-375187.5 22556.25056.25156.25 4500337502812.502343.75 ∑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 5.515.525.535.545.555.565.675.585.595.5 16102015514531 5.599255710682.5277.5917377.5256.595.5 -40.45-30.45-20.45-10.45-0.459.5519.5529.5539.5549.55 1636927.2418.2109.20.202591.20382.2873.215642455 16365563418221843038456535436646922455 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

7.

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

(ii) X = 222 = 22.2

10

(X – x)2 = 2.82 + 1.82 + 0.22 + 0.82

13.22 + 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 – 1011 – 2021 – 3031 – 4041 – 5051 – 6061 – 7071 – 8081 – 9091 – 100 5.515.525.535.545.555.565.575.585.595.5 -40.1-30.1-20.1-10.1-0.19.919.929.939.949.9 13471292101 -40.1-90.3-80.4-70.7-1.289.139.829.9049.9 1608.018154.056464.164998.491.447938.811584.04894.0102410.01

i) Standard Deviation

D = e ∑fd2∑fd

∑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

(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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