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Subject
Engineering
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Homework
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Question 3.6
Given:
The observed distance AB is as shown below.
Number of observation
Length L(m)
1
65.401
2
65.400
3
65.402
4
65.396
5
65.406
6
65.401
7
65.405
8
65.401
9
65.405
10
65.404
a) The line’s most probable length
4012.65
10
10
1
=
+++
=
=
=
65.404....65.40065.401
x
values the Substitute
N
x
x
i
i
Hence the line’s most probable length is 65.4012 m
b)
Calculate the standard deviation
( )
( )
00336.0
110
)4012.65404.65(...)4012.65465(
1
22
2
1
=
+++
=
=
=
65.4012-65.401
:values the Substitute
N
xx
N
i
i

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Hence the standard deviation is 0.00336
c)
Calculate the standard deviation of mean
3
1006.1
10
00336.0
=
=
=
m
:values the Substitute
N
m
Hence the standard deviation of mean is 1.06X10
-3
Question 3.9
Given:
The observed distance AB is as shown below.
Number of observation
Length L(m)
1
65.401
2
65.400
3
65.402
4
65.396
5
65.406
6
65.401
7
65.405
8
65.401
9
65.405
10
65.404
11
65.408
12
65.409
a) The line’s most probable length
4032.6512/383.784
12
12
1
==
+++++
=
=
=
65.40965.40865.404....65.40065.401
x
values the Substitute
N
x
x
i
i
Hence the line’s most probable length is 65.4032 m

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Question 3.6 Given: The observed distance AB is as shown below. Number of observation 1 2 3 4 5 6 7 8 9 10 a) The line’s most probable length Length L(m) 65.401 65.400 65.402 65.396 65.406 65.401 65.405 65.401 65.405 65.404 10 x x i 1 i N Substitute the values 65.401  65.400  ....  65.404 10  65.4012 x Hence the line’s most probable length is 65.4012 m b) Calculate the standard deviation N   x i 1 i  x N 1 Substitute the values :  65.401 - 65.40122  (65  4  65.4012) 2  ...  (65.404  65.4012) 2  0.00336 10  1 Hence the standard deviation is 0.00336 c) Calculate the standard deviation of mean m   N Substitute the values : 0.00336 m  10  1.06  10 3 Hence the standard deviation of mean is 1.06X10-3 Question 3.9 Given: The observed distance AB is as shown below. Number of observation 1 2 3 4 5 6 7 8 9 10 11 12 a) The line’s most probable length Length L(m) 65.401 65.400 65.402 65.396 65.406 65.401 65.405 65.401 65.405 65.404 65.408 65.409 12 x x i 1 i N Substitute the values 65.401  65.400  ....  65.404  65.408  65.409 x 12  784.383 / 12  65.4032 Hence the line’s most probable length is 65.4032 m b) Calculate the standard deviation N   x i 1 i  x N 1 Substitute the values :  65.401 - 65.40322  (65.400  65.4032) 2  ...  (65.409  65.4032) 2 .  ...
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