unit# 6 basic statistic exercise 6.2

7/28/2019 Unit# 6 Basic Statistic Exercise 6.2

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Mudassar Nazar Notes Published by Asghar Ali Page 1

Unit 6 Basic Statistic Exercise 6.2

Question # 3

Find Arithmetic mean by direct method for the following set of data:

(i) 12, 14, 17, 20, 24, 29, 35, 45.Solution

=

=

=

= 24.5

(ii) 200, 225, 350, 375, 270, 320, 290. Solution

=

=

=

= 290


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Question # 4

For each of the data in Q# 3 Compute arithmetic mean using indirect method

(i) 12, 14, 17, 20, 24, 29, 35, 45.Solution

X D = x 20

12 -8

14 -6

17 -3

A 20 0

24 4

29 9

35 15

45 25

= 36

= A +

= 20 +

= 20 + 4.5

= 24.5

(ii) 200, 225, 350, 375, 270, 320, 290. Solution

X D = x 290

200 -90

225 -65

270 -20A 290 0

320 30

350 60

375 85

= 0


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= A +

= 290 +

= 290 + 0

= 290

Question # 5

The marks obtained by students of class XI in mathematics are given below compute arithmetic

mean by direct and indirect methods.

Classes/ Groups Frequency

0 9 2

10 19 10

20 29 530 39 9

40 49 6

50 59 7

60 69 1

Solution

Classes/ Groups f X fX

0 9 2 4.5 9

10 19 10 14.5 145

20 29 5 24.5 122.5

30 39 9 34.5 310.5

40 49 6 44.5 267

50 59 7 54.5 381.5

60 69 1 64.5 64.5

= 40 X= 1300

Direct Method

=

=

= 32.5


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Class/ Group f X D = x 34.5 fD

0 9 2 4.5 -30 -60

10 19 10 14.5 -20 -200

20 29 5 24.5 -10 -50

30 39 9 34.5 0 0

40 49 6 44.5 10 6050 59 7 54.5 20 140

60 69 1 64.5 30 30

= 40 = -80

Short cut Method ( Indirect Method)

= A +

= 34.5 +

= 34.5 - 2

= 32.5

Class/ Group f X u = fu

0 9 2 4.5 -30 -6

10 19 10 14.5 -20 -2020 29 5 24.5 -10 -5

30 39 9 34.5 0 0

40 49 6 44.5 10 6

50 59 7 54.5 20 14

60 69 1 64.5 30 3

= 40 = -8

Coding Method ( Indirect Method)

= A + x h

= 34.5 + 10

= 34.5 2

= 32.5


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Question # 6

The following data relates to the ages of children in a school. Compute the mean age by direct

and short cut method taking any provisional mean ( Hint take A = 8)

Class limits Frequency

4 6 107 9 20

10 12 13

13 15 7

Total 50

Also compute Geometric mean and Harmonic mean.

Solution

Class limits f X fX4 6 10 5 50

7 9 20 8 160

10 12 13 11 143

13 - 15 7 14 98

= 50 X= 451

Direct Method

=

=

= 9.02

Class limits F X D = x 8 fD

4 6 10 5 -3 -30

7 9 20 8 0 0

10 12 13 11 3 39

13 - 15 7 14 6 42

= 50 = 51


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Short cut Method ( Indirect Method)

= A +

= 8 +

= 8 + 1.02

= 9.02

Class limits F X logx f logx

4 6 10 5 0.6990 6.990

7 9 20 8 0.9031 18.062

10 12 13 11 1.0414 13.538213 - 15 7 14 1.1461 8.0227

= 50 = 46.6129

G. M = Anti-log

G.M = Anti-log

G.M = Anti-log ( 0.9323)

G.M = 8.56

Class limits f X

4 6 10 5 0.2 2

7 9 20 8 0.125 2.5

10 12 13 11 0.0909 1.1817

13 - 15 7 14 0.0714 0.4998

= 50 = 6.1815

H.M =

H.M =

H.M = 8.089


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Question # 7

The following data shows the number of children in various families. Find Mode and

Median.

9, 11, 4, 5, 6, 8, 4, 3, 7 ,8, 5, 5, 8, 3, 4, 9, 12, 8, 9, 10, 6, 7, 7, 11, 4, 4, 8, 4, 3, 2, 7, 9, 10, 9,

7, 6, 9, 5.

Solution

Arranged Data

(n = 38)

2, 3, 3, 3, 4, 4, 4, 4, 4, 4, 5, 5, 5, 5, 6, 6, 6, 7, 7, 7, 7, 7, 8, 8, 8, 8, 8, 9, 9, 9, 9, 9, 9, 10, 10, 11, 11,

12.

Mode = the most frequent number

Mode = 4 , 9

Median = [th

observation +th

observation ]

Median = [th

observation +th

observation ]

Median = [ 19th observation + 20th observation ]

Median = [ 7 + 7]

Median = [14]

Median = 7


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Question # 8

Find modal number of heads for the following distribution showing the number of heads

when 5 coins are tossed. Also determine Median.

X ( Number of heads) Frequency ( number of times)

1 3

2 8

3 5

4 3

5 1

Solution

X f C.F

1 3 3

2 8 11

3 5 16

4 3 19

5 1 20

= 20

n =

Mode = the most frequent observation

Mode = 2

Median = the class containing ( )th

observation

Median = the class containing ( )th

observation

Median = the class containing 10th

observation

Median = 2

Question # 9


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The following frequency distribution is the weights of boys in kilograms. Compute mean,

median and mode.

Class Intervals Frequency

1 3 2

4 6 37 9 5

10 12 4

13 15 6

16 18 2

19 21 1

Solution

Class Intervals f X fX

1 3 2 2 4

4 6 3 5 15

7 9 5 8 40

10 12 4 11 44

13 15 6 14 84

16 18 2 17 34

19 21 1 20 30

= 23 =241

Mean

=

=

= 10.48


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Class Intervals f C.F C. Boundaries

1 3 2 2 0.5 3.5

4 6 3 5 3.5 6.5

7 9 5 8 6.5 9.5

10 12 4 11 l 9.5 12.5

13 15 6 14 12.5 15.516 18 2 17 15.5 18.5

19 21 1 20 18.5 21.5

= 23

=

=

= 11.5

Median = l + [ - c]

Median = 9.5 + ( 11.5 10 )

Median = 9.5 + ( 1.5)

Median = 9.5 + 1.125

Median = 10.625

Class Limits f C. Boundaries

1 3 2 0.5 3.5

4 6 3 3.5 6.5

7 9 5 6.5 9.5

10 12 4 f1 9.5 12.5

13

15 6 fm 12.5

15.516 18 2 f2 15.5 18.5

19 21 1 18.5 21.5

= 23

Mode = l +


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Mode = 12.5 +

Mode = 12.5 +

Mode = 12.5 +

Mode = 12.5 + 1

Mode = 13.5

Question # 10

A student obtained the following marks at a certain examination. English 73, Urdu 82, Maths

80, History 67 and Science 62.

(i) If the weights accorded these marks are 4, 3, 3, 2 and 2 respectively, what is anappropriate average marks?

(ii) What is the average mark if equal weights are used?Solution

X (marks) w ( weight) Xw

73 4 292

82 3 246

80 3 240

67 2 134

62 2 124

= 364 = 14 w = 1036

(i) Weighted Meanw =

w =

w = 74

(ii) Arithmetic Mean=

=

= 72.8


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Question # 11

On a routine trip a family bought 21.3 liters of petrol at 39.90 rupees per liter, 18.7

liters at 42.90 rupees per liter and 23.5 liters at 40.90 rupees per liter. Find the mean

price paid per liter.

Solution

Mean price =

Mean price =

Mean price =

Mean price = 41.15 rupees per liter.

Question # 12

Calculate simple moving average of 3 years from the following data:

Year 2001 2002 2003 2004 2005 2006 2007 2008 2009 2010

Value 102 108 130 140 158 180 196 210 220 230

Solution

Year Value3 Year Moving

Total Average

2001 102 . .

2002 108 340 113.33

2003 130 378 126

2004 140 428 142.67

2005 158 478 159.33

2006 180 534 178

2007 196 586 195.332008 210 626 208.67

2009 220 660 220

2010 230 ------- ------

unit# 6 basic statistic exercise 6.2

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