2 Measurement+and+Error PDF
2 Measurement+and+Error PDF
2 Measurement+and+Error PDF
An engineer has to make a lot of measurements, collect and analyze data, and make decisions
about the validity of his approaches and procedures. He must have a clear idea about the
results he is going to obtain. In this respect, he may develop models of his expectations and
compare the outcomes from the experiments to those from the model. He uses various
measuring instruments whose reliabilities have outmost importance in successes of his
decisions. Characteristics of measuring instruments that are used in selecting the proper ones
are reviewed in the first section. Section 2 deals with analyses of measurement data. Section 3
handles the analyses of uncertainties and establishment of engineering tolerances.
Output
B = B2
Output
B = B1
A = A1
Input A
Non-linear i/p-o/p relation
A = A2
Input B
Linear i/p-o/p relation
Input A
Input B
Measuring
Device
Output
O/p
Output (o/p) =
Sensitivity (S) x input (i/p)
Accuracy
Accuracy is defined as the degree of conformity of a measured value to the true (conventional
true value CTV) or accepted value of the variable being measured. It is a measure of the
total error in the measurement without looking into the sources of the errors. Mathematically
it is expressed as the maximum absolute deviation of the readings from the CTV. This is
called the absolute accuracy.
absolute accuracy
CTV
......................................... (2.2)
; ........................................................... (2.3)
............................................ (2.4)
Example 2.1
A voltmeter is used for reading on a standard value of 50 volts, the following readings are
obtained: 47, 52, 51, 48
Conventional true value (CTV) = 50 volts,
Maximum (VMAX) = 52 volts and minimum (VMIN) = 47 volts.
CTV VMIN = 50 47 = 3 volts; VMAX CTV = 52 50 = 2 volts.
Absolute accuracy = max of {3, 2} = 3 volts.
Relative accuracy = 3/50 = 0.06 and % accuracy = 0.06x100 = 6%
2.1.3.2.
Precision
Bias
The difference between CTV and average value (VAV) is called the bias. Ideally, the bias
should be zero. For a high quality digital voltmeter, the loading error is negligible yielding
bias very close to zero.
Bias = CTV - VAV .............................................................................................. (2.6)
In the previous example the average (VAV) = (47+48+51+52)/4 = 49.5 V
Pr = max {(49.5 47), (52 49.5)} = 2.5 volts. Thus, Bias = 50 49.5 = 0.5 volt.
A consistent bias can be due to the presence of a systematic error or instrument loading.
Hence, eliminating the causes removes the bias. However, if the bias is consistent and causes
cannot be identified and/or eliminated, the bias can be removed by re-calibrating the
instrument.
Example 2.2
A known voltage of 100 volts (CTV = 100 V) is read five times by a voltmeter and following
readings are obtained: 104, 103, 105, 103, 105
Average reading = (1/5)x(104+103+105+103+105) = 104 volts
Pr = max {(VAV VMIN), (VMAX VAV)} = max {(104 103), (105 104)} = 1 volt
Accuracy = max {(CTV VMIN), (VMAX - CTV)} = max {(100 103), (105 100)} =5 V
Bias = average CTV = 104 100 = 4 volts.
If we re-calibrated the instrument to remove the bias, then the average reading = CTV.
The new readings would be 100, 99, 101, 99, 101
Hence, after re-calibration, average = CTV = 100 volts, and accuracy = precision = 1 volt.
2.1.4.
Accuracy versus Precision
The distinction between accuracy and precision can be illustrated by an example: two
voltmeters of the same make and model may be compared. Both meters have knife-edged
pointers and mirror-backed scales to avoid parallax, and they have carefully calibrated scales.
They may therefore be read to the same precision. If the value of the series resistance in one
meter changes considerably, its readings may be in error by a fairly large amount. Therefore
Poor accuracy
High precision
High accuracy
High precision
Average accuracy
Poor precision
Poor accuracy
Poor precision
bullets hit the bulls eye and spaced closely enough leading to high accuracy and high
precision. The bullet hits are placed symmetrically with respect to the bulls eye in the third
case but spaced apart yielding average accuracy but poor precision. In the last example, the
bullets hit in a random manner, hence poor accuracy and poor precision.
The scatter graph in figure 2.3 shows an alternative way of presenting the accuracy and
precision. Same quantity was measured three times by 5 different analyst or methods or
measuring instruments. Distribution of readings around the true value indicates the most
accurate, most precise and least accurate and least precise readings. The last reading is too far
away from the true value and from other readings that may indicate a systematic error.
Most
accurate
and precise
6
5
4
true value
Systematic
error?
Worst
precision
2
1
0
0
2
3
analyst or different methods or measuring devices
Significant Figures
An indication of the precision of the measurement is obtained from the number of significant
figures in which the result is expressed. Significant figures convey actual information
regarding the magnitude and the measurement precision of a quantity. The more significant
figures the greater the precision of measurement.
66
68
70
67.8
68.0
68.2
Measurement errors
Human errors
(Gross errors)
Examples:
Misreading instruments
Erroneous calculations
Improper choice of
instrument
Incorrect adjustment, or
forgetting to zero
Neglect of loading effects
Methods of elimination or
reduction:
1. Careful attention to
detail when making
measurements and
calculations.
2. Awareness of
instrument limitations.
3. Use two or more
observers to take critical
data.
4. Taking at least three
readings or reduce
possible occurrences of
gross errors.
5. Be properly motivated
to the importance of
correct results.
Systematic errors
Equipment errors
Examples:
Bearing friction
Component nonlinearities
Calibration errors
Damaged equipment
Loss during transmission
How to estimate:
1. Compare with more
accurate standards
2. Determine if error is
constant or a
proportional error
Methods of reduction
or elimination:
1. Careful calibration
of instruments.
2. Inspection of
equipment to ensure
proper operation.
3. Applying
correction factors
after finding
instrument errors.
4. Use more than one
method of measuring
a parameter.
Random errors
Environmental errors
Examples:
Changes in
temperature,
humidity, stray
electric and
magnetic fields.
How to estimate:
Careful monitoring of
changes in the
variables. Calculating
expected changes.
Methods of reduction or
elimination:
2. Hermetically seal
equipment and
components under test.
3. Maintain constant
temperature and
humidity by air
conditioning.
4. Shield components
and equipment against
stray magnetic fields.
5. Use of equipment that
is not greatly effected
by the environmental
changes.
Examples:
Unknown events
that cause small
variations in
measurements.
Quite random and
unexplainable.
How to estimate:
Take many readings
and apply statistical
analysis to unexplained
variations
Methods of
reduction:
1. Careful design
of measurement
apparatus to
reduce unwanted
interference.
2. Use of statistical
evaluation to
determine best
true estimate of
measurement
readings.
x1 + x2 + x3 + + xn x
=
n
n
........................................................................ (2.7)
d1 = x1 x ; d 2 = x2 x ; ; d n = xn x ........................................................................ (2.8)
The deviation from the mean may have a positive or a negative value and that the algebraic
sum of all the deviations must be zero. The computation of deviations for the previous
example is given in Table 2.1.
2.2.2.1.
Average Deviation
The average deviation is an indication of the precision at the instruments used in making the
measurements. Highly precise instruments will yield a low average deviation between
readings. By definition average deviation is the sum of the absolute values of the deviations
divided by the number of readings. The absolute value of the deviation is the value without
respect to sign. Average deviation may be expressed as
D=
d1 + d 2 + d 3 + + d n
n
d
n
.................................................................. (2.9)
Example 2.5
The average deviation for the data given in the above example:
D=
2.2.2.2.
Standard Deviation
The range is an important measurement. It indicates figures at the top and bottom around the
average value. The findings farthest away from the average may be removed from the data set
without affecting generality. However, it does not give much indication of the spread of
observations about the mean. This is where the standard deviation comes in.
In statistical analysis of random errors, the root-mean-square deviation or standard
deviation is a very valuable aid. By definition, the standard deviation of a finite number of
d12 + d 22 + d32 + + d n2
=
n 1
2
i
n 1
............................................................ (2.10)
Another expression for essentially the same quantity is the variance or mean square
deviation, which is the same as the standard deviation except that the square root is not
extracted. Therefore
variance (V) = mean square deviation = 2 ........................................................ (2.11)
The variance is a convenient quantity to use in many computations because variances are
additive. The standard deviation however, has the advantage of being of the same units as the
variable making it easy to compare magnitudes. Most scientific results are now stated in terms
of standard deviation.
2.2.3.
Probability of Errors
2.2.3.1.
A practical point to note is that, whether the calculation is done on the whole population of
data or on a sample drawn from it, the population itself should at least approximately fall into
a so called normal (or Gaussian) distribution.
For example, 50 readings of voltage were taken at small time intervals and recorded to
20
16
Voltage reading
(volts)
# of reading
99.7
99.8
99.9
100.0
100.1
100.2
100.3
1
4
12
19
10
3
1
12
0
99.6
99.8
100.0
Volts
100.2
100.4
exp(
x2
)
2
Probability of Error
Pr obability of error =
2 SD
0.6745
1.0
2.0
3.0
0.5000
0.6828
0.9546
0.9972
-4
-3
-2
-1
Figure 2.8. The error distribution curve for a normal (Gaussian) distribution.
They can be positive or negative and there is equal probability of positive and negative
random errors. The error distribution curve indicates that:
Small errors are more probable than large errors.
Large errors are very improbable.
There is an equal probability of plus and minus errors so that the probability of a given
error will be symmetrical about the zero value.
Deviation
d
d2
101.
-0.1
0.01
101.7
0.4
0.16
101.3
0.0
0.00
101.0
-0.3
0.09
101.5
0.2
0.04
101.3
0.0
0.00
101.2
-0.1
0.01
101.4
0.1
0.01
101.3
0.0
0.00
101.1
-0.2
0.04
x=1013.0 |d|=1.4 d2=0.36
Reading, x
distribution, a range covered by one standard deviation above the mean and one standard
deviation below it (i.e. x 1 SD) includes about 68% of the observations. A range of 2
standard deviations above and below ( x 2 SD) covers about 95% of the observations. A
range of 3 standard deviations above and below ( x 3 SD) covers about 99.72% of the
observations.
2.2.3.2.
Range of a Variable
If we know the mean and standard deviation of a set of observations, we can obtain some
useful information by simple arithmetic. By putting 1, 2, or 3 standard deviations above and
below the mean we can estimate the ranges that would be expected to include about 68%,
95% and 99.7% of observations. Ranges for SD and 2 SD are indicated by vertical lines.
The table in the inset (next to the figure) indicates the fraction of the total area included within
a given standard deviation range.
Acceptable range of possible values is called the confidence interval. Suppose we
measure the resistance of a resistor as (2.65 0.04) k. The value indicated by the color code
is 2.7 k. Do the two values agree? Rule of thumb: if the measurements are within 2 SD, they
agree with each other. Hence, 2 SD around the mean value is called the range of the
variable.
2.2.3.3.
Probable Error
The table also shows that half of the cases are included in the deviation limits of 0.6745.
The quantity r is called the probable error and is defined as
probable error r = 0.6745
.......................................................... (2.12)
x = 1013.0
10
Standard deviation, =
= 101.3
0.36
d2
=
= 0.2
9
n 1
(a)
(b)
Gate
Clock
(b) results 4 pulses while case (a) supplies only 3 pulses. Hence, a digital read-out has an
uncertainty of 1 digit.
2.2.5.2.
The uncertainty in analog displays depends upon the organization of display screen and
capabilities of the reader. In analog multi meters it is accepted as scale divisions (the least
count). In oscilloscope displays, it depends upon the thickness of the trace and it is around
mm. For both analog and digital displays, it is recommended to take the measurement as close
to full scale as possible to minimize the effect of the reading error. The following example
illustrates the uncertainties in analog meter readings.
Example 2.7
An analog voltmeter is used to measure a voltage. It has 100
5.8
6.0
6.2
The least count of the scale (smallest division) scales are all in cm
xi = xi xi ....................................................................................................... (2.14)
i = 1, 2, , n; x i is known as the nominal value; xiis known as the uncertainty in the
variable x1; then
R = R R ........................................................................................................ (2.15)
where R =
f ( x1, x 2 ,...., x n )
The uncertainty R = wR can be computed using Taylors series expansion and statistical
analysis. All partial derivatives of R are taken. The partial derivative
f
shows the
xi
sensitivity of R to variable xi. Since the measurements have been taken, the xi values are
known and can be substituted into the expressions for the partial derivatives and partial
derivatives are evaluated at known values of x1, x2, . , xn.
2.3.1.1.
Limiting Error
Two methods are commonly used for determining the uncertainty. The first one is called the
method of equal effects and it yields the limiting (guarantied) error (maximum uncertainty
possible).
n
R = R = [
i =1
where (
f
]o xi ................................................................................... (2.16)
xi
f
) o is the partial derivative of the function with respect to xI calculated at the
xi
nominal value. The absolute value is used because some of the partial derivatives may be
negative and would have a canceling effect. If one of the partial derivative is high compared
to the others, then a small uncertainty in the corresponding variable has large effect on the
total error. Hence, the equation also illustrates which of the variable exerts strongest influence
on the accuracy of the overall results.
Example 2.8
The voltage generated by a circuit is equally dependent on the value of three resistors and is
given by the following equation: V0 = I(R1R2/R3)
Uncertainty Analysis / 19
If the tolerance of each resistor is 1 per cent, what is the maximum error of the generated
voltage?
SOLUTION:
Let us find the sensitivities first.
V0
V V
V V
V
R
R
RR
= I 2 = 0 ; 0 = I 1 = 0 ; 0 = I 1 2 2 = 0
R1
R3 R1 R2
R3 R2 R3
R3
R3
V0 =
V
V
V0
R1 + 0 R2 + 0 R3
R3
R2
R1
The total variation of the resultant voltage is 0.3 per cent, which is the algebraic sum of
the three tolerances. This is true in the first approximation. The maximum error is slightly
different from the sum of the individual tolerances. On the other hand, it is highly unlikely
that all three components of this example would have the maximum error and in such a
fashion to produce the maximum or minimum voltage. Therefore, the statistical method
outlined below is preferred.
2.3.1.2.
The second method is called the square root of sum of squares. It is based on the observations
stated before for the random errors. It yields the expected value of the uncertainty and
computed as
n
(R) = ( R ) = [(
2
i =1
f 2
) ]o (xi ) 2 ................................................................................ (2.17)
xi
This will be used throughout the course unless the question asks the limiting error, or
maximum possible uncertainty.
Example 2.9
P = VI, if V = 100 2 volt (measured) and I = 10 0.2 Amp (measured), determine the
maximum allowable uncertainty, and the expected uncertainty in power.
SOLUTION: Pm = wPm =
P
P
V + I = 10 x 2 + 100 x0.2 = 40
I
V
P
P
V ) 2 + ( I ) 2
I
V
watts.
. The
Special Case
l n k
R = Y1 Y2 Y3 ................................................................................................................... (2.18)
then
R 2 2 Y 2 2 Y 2 2 Y 2
) = l ( ) + n ( ) + k ( ) ........................................................................ (2.19)
Ro
Y1o
Y2 o
Y3o
2.3.2.
Example 2.11
Two resistors R1 and R2 are connected first in series, then in parallel. Let R1 = 10 0.5
and R2 = 10 0.5. Find the maximum and expected values for the uncertainty in the
combination.
Series analysis
Rs = R1 + R2 ; Rs/R1 = Rs/R2 = 1; Rs
= R1 + R2 = 10 + 10 = 20
Uncertainty Analysis / 21
The limiting error (maximum uncertainty) = Rsm =
R
Rs
1 1
R1 + s R2 = + = 1
R2
R1
2 2
R
Rs 2
) (R1 ) 2 + ( s ) 2 (R2 ) 2 = (1) 2 ( 12 ) 2 + (1) 2 ( 12 ) 2 = 14 + 14 =
R2
R1
R1
R2
1
2
R1
R2
yielding Rs 0.7 . The relative uncertainty = 0.7/20 = 0.035, and the percentage
uncertainty = 3.5%. Therefore, Rs = 20 0.7 = 20 3.5%
Parallel analysis
10x10
RR
= 5
R p = 1 2 = R p R p ; R p =
10 + 10
R1 + R2
R p
R1
R p
R1
Hence,
( R1 + R 2 ) ) R 2 R1 R 2
( R1 + R 2 ) 2
( R1 + R 2 ) ) R 2 R1 R 2
( R1 + R 2 ) 2
R p
R1
(R p ) 2 = (
R p
R1
R2
( R1 + R 2 ) 2
R2
R p
( R1 + R 2 ) 2 R2
( R1 + R2) ) R1 R1R2
( R1 + R2 ) 2
R1
=
( R1 + R2 ) 2
102
100 1 R p
=
= =
2
(10 + 10)
400 4 R2
) 2 (R1 ) 2 + (
R p
1
1 1
1 1
1 1
) 2 (R2 ) 2 = ( ) 2 ( ) 2 + ( ) 2 ( ) 2 = ( )( )(1 + 1) =
32
R2
16 4
4 2
4 2
1
= 0.175
32
1 1 1
Limiting error in Rp = ( + ) = 0.25
4 2 2
2.3.3.
Summary of how to propagate the errors
Addition and subtraction ( x+y; x-y): add absolute errors;
2.3.4.
Population
Sample
called the population. Instead, we decide to take a small fraction of the adults, say 1 out of
every 1000, and average these weights. This small fraction is called our sample population.
Now we have an average weight for our sample population. We want to know if our sample
population average weight is a good estimation of the population average weight. In addition,
measurement is a costly process. Hence, we also want to know the minimum sample size that
yields uncertainties within the tolerance range.
Figure 2.11 illustrates the distribution for the population and the sample. For the
Estimated mean x-s
standard deviation s-x
Population standard
deviation
Frequency
Mean
normal distribution, 68% of the data lies within 1 standard deviation. By measuring samples
and averaging, we obtain the estimated mean x s, which has a smaller standard deviation sx.
is the tail probability that xs does not differ from by more than .
Uncertainty Analysis / 23
The population standard deviation is
population =
(deviations )2
n
( x x)
.................................................................. (2.20)
sample = s = s =
(deviations )2 = ( xi x) 2
n 1
......................................................... (2.21)
n 1
The sample standard deviation allows for more variation in the sample compared to the
population, since sample is only part of population. Dividing by (n-1) increases the estimate
of the population variation. This attempts to eliminate the possibility of bias.
The estimated sample standard deviation is a measure of the spread of data about the
mean. The standard deviation of the mean x is
x =
s
n 1
.................................................................................................................... (2.23)
Equation 2.23 illustrates an important fact. The standard deviation doesnt change much,
but the error on the mean improves dramatically! It goes as
measurements. As a rule of thumb, the range R of the random variable x can be roughly taken
as R 4. If is the error that can be tolerated in the measurement, then the number of
samples required to achieve the desired: n
. Then = x
2.4. PROBLEMS
2.4.1.
Solved Examples
1. A digital thermometer is used to measure the boiling point of water (100.0C). The
measurement is repeated 5 times and following readings are obtained: 99.9, 101.2, 100.5,
100.8, 100.1. Determine the accuracy, the precision and the bias of the thermometer.
A digital voltmeter uses 4 digit display (it can display up to 19999). It is used to
measure a voltage across a standard cell whose value is 1.2341 volt 4 times and following
readings are obtained: 1.2202, 1.2115, 1.2456, 1.2218. Determine the accuracy, the precision
and the bias of the voltmeter.
3.
A recently calibrated digital voltmeter is used to read a voltage and it consistently yields
75 volts. Another meter in the lab is also used five times to measure the same voltage and
following readings are obtained: 77, 75, 74, 76, 77. For the second meter,
a.
CTV = 75 V, yielding absolute accuracy = max {(77 - 75), (75 74)} = 2 volts,
b.
c.
4.
V2
V1
V 2 V 2
) +(
) ]
V1
V2
1
V
G = 20 log10 ( 2 ) = (20 log10 e)[ln(V2 ) ln(V1 )
V1
G
1
and
= 20 log10 e
V1
V1
1
G
yielding the uncertainty as defined above.
= 20 log10 e
V2
V2
Five resistors are available, one of 20 and four of 10 each. The uncertainty of the 20
5.
resistor is 5% and that of each 10 resistor is 10%. 3 possible connections using these
resistors are shown below. Which one would you use to obtain a 30 resistance with the
least uncertainty? What is the uncertainty of this best connection?
Problems / 25
10 10
10
10
20
10
20
10
10 = 1 ; 20 = 1 ;
10 10
B =1.414 ; in (C), RP =
10 RP = (20 1.414)//
6.
Find the power dissipated in this resistance with its uncertainty and limiting error.
P = I2R; P/I = 2IR; P/R = I2, P = 10 x103 x(10 x103 ) 2 = 1 W
7.
R = R0 [1 + 0 (T T0 )]
R = R0 + 0 R0T 0 R0T0
b.
T =
1 R
R R0
+ T0 = ( 1) + T0
0 R0
0 R0
(T ) 2 = (T0 ) 2 +
1 R 2 R0 2 R 2
( ) [(
) +( ) ]
02 R0
R0
R
1
1
R
R
T
T
T
T
,
,
=
=
=1
= 2 (1 ) , and
2
0 R0 0 0
R0
R 0 R0 R0
T0
(T ) 2 = (T0 ) 2 + (
c.
1 2
R 2
) (R) 2 + (
) (R0 ) 2
0 R0
0 R02
T=
30
1
( 1) + 280 = 407.6 K
0.00392 20
(T )2 = 10 4 + (
d.
1.5 2 6 6
) (10 + 10 ) = 0.29295
0.00392
;
yielding T = 0.54K
R
of the thermometer.
T
R
= 0 R0 = 0.00392 x 20 x103 = 78.4
K
T
e. Calculate the maximum error in T.
Tm = T0 +
1
R
1.5
R + 2 R0 = 0.01 +
(0.001 + 0.001) = 0.7753K
0 R0
0 R0
0.00392
2.4.2.
Questions
2.4.2.1.
True-False
True
False
1.
2.4.2.2.
Multiple-Choice Questions
Please choose and CICRLE the most appropriate statement in the following questions
1. Gross (human) errors
a. Are due to equipment failures
b. Can be minimized by making multiple measurements
c. Can not be treated mathematically
d. Do not effect the accuracy of the measurement
2. Resolution is
a. An indicator of how close the reading to the true value
b. The smallest incremental quantity that we can identify
c. The difference between the minimum and maximum values of the measurement
d. The total error in the measurement
3. Systematic errors
a. Can not be treated mathematically
b. Can be eliminated by making multiple measurements
c. Indicate the accuracy of the measurement
Problems / 27
d.
4.
a.
b.
c.
d.
5.
a.
b.
c.
d.
6.
a.
b.
c.
d.
7.
a.
b.
c.
d.
Are due to environmental factors upsetting the user and the equipment
Accuracy of a measurement is an indication of
How far the reading is away from the average value
How many digits we use to display the data
How close the reading is to the conventional true value
The smallest incremental quantity that we can identify
Precision is
An indicator of how close the reading is to the true value
The total error in the measurement
An indicator of how close the reading is to the average value
The smallest incremental quantity that we can identify
What is the result of 1.264+ 10.5 (use significant figures only)
12
11.8
11.7
11.764
Mathematical treatment of errors is possible for
Systematic and random errors
Human and systematic errors
Human and random errors
Errors that are small
2.4.2.3.
General Questions
1.
2.
3.
4.
Four capacitors are placed in parallel. The values are (in F) 47.23, 2.35, 18.026 and
0.428, with an uncertainty of one digit in the last place. Find the total capacitor and
express the result using significant figures only. Also prove your result using uncertainty
analysis.
5.
The value of the equivalent resistor and expected error in percent when the resistors
are connected in parallel.
6.
7.
The following values were obtained from the measurements for a resistor in ohms: 220.2,
119.5, 221.1, 119.9, 220.0, 220.5, 119.8, 220.1, 220.4, and 119.8. Calculate
a. The arithmetic mean
b. The average deviation
c. The standard deviation
d. The probable error of the average of the ten readings.
8.
T=
1 R
( 1) + T0 . The temperature at R0 = 5 k 1% is T0 =25C 0.1C, while at a T
0 R0
9.
A digital thermometer is used to measure the boiling water whose temperature is 96.2C.
The measurement is repeated 5 times and following readings are obtained: 95.9, 96.2, 96.5,
95.8, 96.1. Determine the percentile accuracy, the precision and the bias of the thermometer.
10.
The following values were obtained from the measurements for the line voltage in
Jeddah: 125.2, 125.5, 126.1, 126.2, 126.0, 125.8, 125.7, 126.1, 126.3, and 125.6. Write down
the formulas and calculate
a. The arithmetic mean
b. The standard deviation and the probable error of the average of the ten readings.
11. The boiling temperature
of water is measured 15
times using two
thermometers A and B, and
the readings presented in the 95.9
96.2
96.5
96.0
96.3
96.6
graph are obtained.
Thermometer A
Thermometer B
Conventional value for the
boiling temperature of water
Figure problem 11.
is 96.2C.
a. Which thermometer
(A or B) is more precise, why?
b. Calculate the percentage accuracy and bias of thermometer A.
Problems / 29
12.
13.
What is the addition of 12.5 and 1.364 with each having the last digit doubtful?
For the electronic counter show that the uncertainty in the period measurement can be
reduced by a factor of
1
if the average of N time periods is taken. Hint:
N
1
(T1 + T2 + + TN ) The TIs are statistically independent, Ti = T T ,i
N
What is the systematic error, from where it comes and how it can be eliminated?
TAV =
14.
15.
Three resistors are in series. The values are (in k) 47.23, 2.205, and 180.2, with an
uncertainty of one digit in the last place. Find the total resistor and express the result using
significant figures only.
16.
17.
a.
b.
The limiting error in ohms and in percent when the resistors are connected in series.
c. The value of the equivalent resistor and expected error in percent when the resistors
are connected in parallel.
18.
There are 1500 chickens in a poultry farm. 15 chickens are randomly selected and
weighted. The average value is 950 grams and the standard deviation is 60 grams.
a.
b.
How many chickens we will have weighing between 890 grams and 1010 grams?
c. How many chickens must be weighted to reduce the error in the average value down
to 5 grams?