Unit# 6 Basic Statistic Exercise 6.2
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Transcript of Unit# 6 Basic Statistic Exercise 6.2
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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 ------- ------