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Unformatted text preview: PROBLEM 4.1 KNOWN: N > 1000; T< = 9.2 units ; s)( 21.1 units
FIND: Range of x in which 50% of all measurements should fall. ASSUMPTIONS: Measurand follows a normal density function
Data set sufﬁciently large such that x z x‘ and S){ w o SOLUTION By assuming that the data is sufﬁciently large such that its population behaves as an inﬁnite
population, we can ﬁnd the interval defined by x''z.; as xin'+zlo
as follows. We can find P(x' + z. o ) from the onesided integral solution to l P(Z1): [2R1 “2 zI
je“l"'”ct[s E) This solution is given in Table 4.3 for p(z.) = 025 (one half ofthe 50% probability sought)
as 21 = 0.674. Then, we should expect that 50% ofthe xi values lie in the interval given by 9.2 — 0.?425 5 x, S 9.2 + 0.?425 (50%) COMMENT We can see from Table 4.4 that as N becomes large the value for t approaches a value given
by 24. 4—1 PROBLEM 4.2 KNOWN: N> 10 000; x=204 units;Sx= 18 units
FIND: x'—z.o$x$x'+zlo atP=90% ASSU MPTIONS Measurand follows a normal density function
Data set sufﬁciently large such that—x m x’ and S)( m o SOLUTION Using the deﬁnition 21 = (x.  x‘)/o we ﬁnd the z] value corresponding to the onesided
probability integral p(zi) = 0.45 from Table 4.3. This gives, 21 = 1.65
Then, 1.65 = (x1  x‘)/ (3
or 1.656 = x1 x'
or x1=x'+1.650' But a normal (Gaussian) distribution is symmetric about the mean value. Hence, for 90%
probability (2)(0.45), x‘1.65cr S x S x'+1.650’ (90%) 01‘ 174.3 5 x S 233.? units (90%) 4"2 PROBLEM 4.6 FIND: Histogram for Table 4.8, Column 1. SOLUTION The distribution below is not unique and is only one such solution. a:
0
U
C
d:
I.
:‘l
U
L?
O OAMWAU‘IO'JM 4.2 4.4 4.6 4.8 5 5.2 5.4
Intervals PROBLEM 4.7 FIND: Frequency distribution for Table 4.8, Column 2. SOLUTION The distribution below is not unique and is only one such solution. . .0
(.51 N) 01 00 >
u
5
=0
6'
a:
I.
LL 53
053'
01—1 0 _
4.2 4.4 4.6 4.8 5 5.2 5.4 Intervals 4—7 PROBLEM 4.8 FINDS Histogram for Table 4.8, Column 3. SOLUTION The distribution below is not unique and is only one such solution. m
a:
o
c
a:
h
:1
u
u 0 DANGPmm‘ﬂ 4.2 4.4 4.6 4.8 5 5.2 5.4
Intervals 4~8 PROBLEM 4.9 KNOWN: Table 4.8 Column 1
N = 20 FIND: 3, sp, andv SOLUTION From equation 4.143, the mean value is # N
p = ini = 4.77mm
N 1:. The standard deviation is based on the variance of (4.14b) 1 N u : f1
Sp 2: _p)w]l ‘ N _ i=1 4—9 PROBLEM 4.10 KNOWN: Data from Table 4.8 , Column 3
N = 20 FIND: E sp, and v SOLUTION From equation 4.14a, the mean value is with N = 20. The standard deviation is based on the variance of (4.14b) 1 N —
Sp =[ 120:, —p)2]”2 =0.263mm N" i—I
with v=N 1 =19. The standard deviation of the means is found (4.16) 5—: S p (NYE: =0.059mm withv=N1=19. 4—10 PROBLEM 4.11 KN OWN: Data of Table 4.8, Column 1
N = 20 FIND: E i :3, (95%) SOLUTION From Problem 4.8 (or using equations 4.14a and 4.14b), we found that p = 4.77 mm and Sp = 0.273 mm with v = N  l = 19. Then from Table 4.4, t1935 = 2.093. From (4.15) the
precision of the measured data set can be expressed by pi=p :t tsp (95%)=4.77: (2.093)(0.273)=4.77 i 0.571 (95%) Here p. denotes the value of any measurement of pressure, p. COMMENT The above probability statement reflects the precision of the data set. In effect, it provides a
range of values in which any measured value is expected to fall with 95% probability. A 95%
probability implies that 19 out of every 20 measurements are expected to fall in the range deﬁned for pi. 4—11 PROBLEM 4.12 KN OWN: Data of Table 4.8, Column 1
N = 20 FIND: 5i tSB (95%) SOLUTION For the 20 data points in Column 1, we ﬁnd using equations (4.14a and b) that?) = 4.?7
mm and SD = 0.273 mm. From equation 4.16, the standard deviation of the means is found to
be S 3. gm =0.273r20" =0.06l mm P ll with v = N  1 = 19. Then from Table 4.4, t19_95 = 2.093. Then, we can expect that the true
mean value would lie within the precision interval defined by (4.17) as p'= E i is; = 4.77 i (2.093)(0.061) = 4.7? i 0.13 (95%) This gives the range of mean values we would expect to find from a given data set. A 95%
probability indicates that 19 out of every 20 complete data sets would show a sample mean
value within the range deﬁned for p'. Compare the meaning of this statement to that found in
Problem 4.1]. They are very different! COMMENT The reasoning behind this precision interval for the mean value lies within the limitations of
ﬁnite statistics. While the sample mean value deﬁnes the mean value of the 20 data points
exactly, it is not necessarily the true mean value of the measured variable (compare the
results for these three real data sets in columns 1,2, and 3 — each has a different sample
mean). Different data sets of the same variable will give somewhat different mean values. As
N becomes large, the sample mean will approach the true mean and the precision interval
will go to zero. Remember this assumes that there is no bias error acting on the measurement. 4—12 PROBLEM 4.13
KN OWN: Data ofTable 4.8, Column 2
N = 20
FIND: E i is. (90%)
SOLUHON From equation 4.1421, the mean value is
5 2
with N = 20. The standard deviation is based on the variance of (4.14b) N _
8,, ail—120:. —p)2]“2 =0.299mm
_ i=1 with v=N1=l9. From Table 4.4, tlg‘gﬂ = 1.729. From (4.15) the precision of the measured data set can be
expressed by pi=p i tSp (90%)=4.7l i (1.729)(O.299)=4.71 i 0.51? (90%)
Here pi denotes the value of any measurement of pressure, p. COMMENT The above probability statement reﬂects the precision of the data set. In effect, it provides a
range of values in which any measured value is expected to fall with 90% probability. A 90%
probability implies that 18 out of every 20 measurements are expected to fall in the range
deﬁned for p;. 4—13 PROBLEM 4.14 KNOWN: Data of Table 4.8, Column 2
N : 20 FIND: 5i tsE (90%) SOLUTION For the 20 data points in Column 2, we ﬁnd using equations (4.14a and b) that p = 4.71 mm
and SI:I = 0.299 mm. From equation 4.16, the standard deviation of the means is found to be s
s = ﬂu, =0.295v20”2 =0.06? mm
1) with v = N  1 = 19. Then from Table 4.4, tlgm = 1.729. Then, we can expect that the true
mean value would lie within the precision interval deﬁned by (4.17) as p’=pi:tSE =4.71:(1.729)(0.067)=4.71i012 (90%) This gives the range of mean values we would expect to ﬁnd from a given data set. A 90%
probability indicates that 18 out of every 20 complete data sets would show a sample mean
value within the range deﬁned for p'. Compare the meaning of this statement to that found in
Problem 4.13. They are very different! COMMENT The reasoning behind this precision interval for the mean value lies within the limitations of
ﬁnite statistics. While the sample mean value defines the mean value of the 20 data points
exactly, it is not necessarily the true mean value of the measured variable (compare the
results for these three real data sets in columns 1,2, and 3 ~ each has a different sample
mean). Different data sets of the same variable will give somewhat different mean values. As
N becomes large, the sample mean will approach the true mean and the precision interval
will go to zero. Remember this assumes that there is no bias error acting on the measurement. 4—14 ...
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 Spring '08
 Singh,V

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