Tutorial on Propogation of Errors

8/18/2019 Tutorial on Propogation of Errors

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Question 1 : The boiling point of water is measured four times. The

results are 110.01◦C, 110.02◦C, 109.99◦C and 110.01◦C. Which of the

following statements best describes this measuring process?

i. Accurate but not precise

ii. Precise but not accurate

iii.Neither accurate nor precise

iv.Both accurate and precise

Ans : Precise but not accurate.Explanation: The measurements are close together, and thus

precise, but they are far from the true value of 100◦C, and thus

not accurate


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(2) The length of an object is given as 3.21 ± 0.02 cm. True or false:

Measurements = Measured value ± σ (uncertainty)

a. The length was measured to be 3.21 cm.Ans: True

b. The true length of the object is 3.21 cm.

Ans: False

c. The bias in the measurement is 0.02 cm.

Ans: False

d. The uncertainty in the measurement is 0.02 cm.

Ans: True


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(3) A person stands on a bathroom scale. The reading is 150 lb. After the

person gets off the scale, the reading is 2 lb.

a) Is it possible to estimate the uncertainty in this measurement? If so,

estimate it. If not, explain why not.

b) Is it possible to estimate the bias in this measurement? If so, estimateit. If not, explain why not.

a) No, it is not possible to determine the standard deviation of the

process from a single measurement.

b) Yes, As the scale reads 2 pounds when the true weight is 0, the biascan be estimated to be 2 pounds.


4/24

4: Assume that X and Y are independent measurements

with uncertainties σX =0.2 and σY =0.4. Find the

uncertainties in the following quantities:

a. 3X

b. X−Yc. 2X+3Y

2649140920494

44704020

602033

2222

32

2222

3

...

...

..

Y X Y X

Y X Y X

X X


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5) The length of a rod is to be measured by a process whose

uncertainty is 3 mm. Several independent measurements will be

taken, and the average of these measurements will be used to

estimate the length of the rod. How many measurements must be

made so that the uncertainty in the average will be 1 mm?

σ ̄ X = σ

√n

Sample mean uncertainty = σ ̄ X = 1 mm

Process measured uncertainty = σ = 3 mm

No of measurements = n = = 9σ ̄ X

σ ( )2

= 3( )

2

1


6/24

(6) In the article “The World’s Longest Continued Series of Sea Level

Observations” (M. Ekman, Paleogeography, 1988:73 –77), the mean annual

level of land uplift in Stockholm, Sweden, was estimated to be

for the years 1774 –1884 and to be for the years 1885 –1984.

Estimate the difference in the mean annual uplift between these two timeperiods, and find the uncertainty in the estimate.

4.93 0.23 mm

3.92 0.19 mm

2 2 2 20.23 0.19 0.30 mm X Y X Y


7/24

7. A force of F = 2.2 ± 0.1 N is applied to a block for a period of time,

during which the block moves a distance d = 3 m, which is

measured with negligible uncertainty. The work W is given by W = F d.

Estimate W, and Find the uncertainty in the estimate?

Answers : Estimate of the workdone :- W = F d

m d ,N .F 3103

Nm ..workdone of estimate i n Uncerta

..

Nm ..d F W

F W

3036

301033

36312

inty

X cX

c

If X is a measurement and

c is a constant, then


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8. Find the uncertainty in Y , given that X = 4.0 ± 0.4 and

(a) Y = X2

(b) Y =√X

(c) Y = 1/X

(d) Y = ln X(e) Y = eX

(f ) Y = sin X (X is in units of radians)


9/24


10/24

(9) The volume of a cone is given by V = πr2h/3, where r is the radius of

the base and h is the height. Assume the height is 6 cm, measured

with negligible uncertainty, and the radius is r = 5.00 ± 0.02 cm.

Estimate the volume of the cone, and find the uncertainty in the

estimate.

Height = h = 6 cm

Radius = r = 5 cm

Volume of cone = V = πr2h/3 = π×6×52/3 = 157.0796 cm3

σ V = dV

dr σ r

dV

dr = 2πrh/3 = 2×π×6×5/3 = 62.832

σ V = 62.832 × 0.02 = 1.3 cm3

σ V =

dV

dr σ

r

Measured uncertainty = = 0.02 cmσ r


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(10). The period T of a simple pendulum is given by

where L is the length of the pendulum and g is the acceleration due to

gravity.

a. Assume g = 9.80 m/s2 exactly, and that . Estimate T ,

and find the uncertainty in the estimate.b. Assume L = 0.742 m exactly, and that . Estimate g, and

find the uncertainty in the estimate.

0.742 0.005 m L

0.742

9.80, 0.742, 0.005, 2 2 1.72899.80 L L

g L T g

1.165204 0.005 0.0058T LdT

dL

1.7289 0.0058 sT

(a)

1.73 0.05 m L

g

LT 2

1652041897420

1

2

12 .

..Lg dL

dT


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2 2

2 2

0.7420.742, 1.73, 0.01, g 4 4 9.79

1.73T

L L T

T

2

23 32 8 0.7424 11.315

1.73dg LdT T

2

11.315 0.01 0.11

9.79 0.11 m/s

g T

dg

dT

g

1.7289 0.0058 sT

(b)


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c. Assume g = 9.80 m/s2 exactly, and that . Estimate

T , and find the relative uncertainty in the estimate.

d. Assume L = 0.855 m exactly, and that . Estimate g, and

find the relative uncertainty in the estimate.

0.8559.80, 0.742, 0.005, 2 2 1.588558

9.80 L

L g L T

g

1 12 1.08532 0.855 9.80

dT dL Lg

1.0853 0.005 0.005426T LdT

dL

1.586 .11 sT

(c)

0.855 0.005 m L

0.005426Relative uncertainty in 0.002923

1.588558

T T T

L = 0.855 1.8588

1.8588


14/24

7987985618550

22

005085618550

.g .g

.

g

LT

..T .L

L

T

dT

dg ;

L

T g

22

2

4

2

4

4

22 10498650050

85504

85612

4

2

...

.

L

T

dT

dg T T g

%..

.

g

g 005610100

79879

1049865100

4


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11. Convert the following absolute uncertainties to relative

uncertainties.

a. 37.2 ± 0.1

b. 8.040 ± 0.003

c. 936 ± 37

d. 54.8 ± 0.3

Ans:

a) The relative uncertainty = 0.1/37.2 = 0.27%

b) The relative uncertainty = 0.003/8.040 = 0.0373%

c) The relative uncertainty = 37/936 =3.953%

d) The relative uncertainty = 0.3/54.8 = 0.547%


16/24

(12) The acceleration g due to gravity is estimated by dropping an

object and measuring the time it takes to travel a certain distance.

Assume the distance s is known to be exactly 2.2 m. The time is

measured to be t = 0.67 ± 0.02 s. Estimate g, and find the relative

uncertainty in the estimate. (Note that g = 2s/t2)

Time = t = 0.67 sec Distance = s = 2.2 m

Measured uncertainty = = 0.02 secσ t

g = 2s/t2 = 2×2.2/0.672 = 9.802 m/sec2

d g

dt

-2×2×s

t

3= -2×2×2.2

0.67

3= = -29.259

σ g = d gdt

σ t σ g = -29.259 × 0.02 = 0.585 m/sec2

g =σ g

% σ g = 9.802 = 0.06 = 6.0 %0.585

g = 9.802 m/sec2 ± 6.0%

σ Y = dY

dX σ X


17/24

13

fi d i h d h di i i


18/24

14 From a fixed point on the ground, the distance to a certain tree is

measured to be s = 55.2 ± 0.1 m and the angle from the point to the

top of the tree is measured to be θ = 0.50 ± 0.02 radians. The height of

the tree is given by h = s tan θ.

a. Estimate h, and find the uncertainty in the estimate.

b. Which would provide a greater reduction in the uncertainty in h:

reducing the uncertainty in s to 0.05 m or reducing the uncertainty

in θ to 0.01 radians?

Angle = θ = 0.5 radian Distance = s = 55.2 m

= 0.02 radianσ θ = 0.1 mσ sh = s tan θ

E i h d fi d h i i h i


19/24

a. Estimate h, and find the uncertainty in the estimate.

h = s tan θ = 55.2 tan(0.5) = 55.2 × 0.546 = 30.156 m

h

θ= s × sec2 θ = 55.2 × sec2(0.5) = 55.2 × 1.1392 =71.67

h s

= tan θ = tan(0.5) = 0.546

= (0.5462 ×0.12 +71.672 ×0.022)0.5 = (0.003+2.055)0.5 = 1.434σ h

h = 30.156 ± 1.434 m

b Whi h ld id d i i h i i h


20/24

b. Which would provide a greater reduction in the uncertainty in h:

reducing the uncertainty in s to 0.05 m or reducing the uncertainty

in θ to 0.01 radians?

= 0.02 radianσ θ

= 0.05 mσ s

h

θ =71.67

h

s= 0.546

= (0.5462 ×0.052 +71.672 ×0.022)0.5σ h

= (0.0007+2.055)0.5 = 1.434 = (0.5462 ×0.12 +71.672 ×0.012)0.5σ h

= (0.003+0.514)0.5 = 0.719

= 0.01 radianσ θ

= 0.1 mσ s

h

θ=71.67

h

s= 0.546

Reducing the uncertainty in to 0.01 radians provides the greater reduction.


21/24

(15) When air enters a compressor at pressure P1 and leaves at pressure

P2, the intermediate pressure is given by . Assume that

1 210.1 0.3 and 20.1 0.4 P MPa P MPa

3 1 2 P P P

a. Estimate P3,andfindthe uncertainty in the estimate.b. Which would provide a greater reduction in the uncertainty in P3:

reducing the uncertainty in P1 to 0.2 MPa or reducing the

uncertainty in P2 to 0.2 MPa?

(a)1 11 2 3 1 210.1, 0.3, 20.1, 0.3, 14.25 P P P P P PP

3 1

3

2 1

1

3

1 22

2 2

2 23 3

2

1 2

3

0.5 / 0.705354

0.5 / 0.354432

0.25

14.25 0.25

P P P

P P P

P

P

P P P

P P

P P

P MPa


22/24

(b)1 2 3 1 210.1, 20.1, 14.25 P P P PP

3 1

1 2

3

1 2

3

3

2 1

1

3

1 2

2

2 2

2 23 3

21 2

2 2 2 2

0.5 / 0.705354

0.5 / 0.354432

0.25

When 0.2 and 0.4, then

0.705354 0.2 0.354432 0.4 0.20

When 0.3 and 0.2, then

0.70

P P P

P P

P

P P

P

P P P

P

P P P

P

P P

P P

2 2 2 25354 0.3 0.354432 0.2 0.22

Reducing the uncertainty in P 1 to 0.2 MPa provides the greater reduction.

16 The lens equation says that if an object is placed at a distance p from a lens and an image


23/24

16. The lens equation says that if an object is placed at a distance p from a lens, and an image

is

formed at a distance q from the lens, then the focal length f satisfies the equation

1/f = 1/p + 1/q. Assume that p = 2.3 ± 0.2 cm and q = 3.1 ± 0.2 cm.

a. Estimate f and find the uncertainty in the estimate.

b. Which would provide a greater reduction in the uncertainty in f : reducing theuncertainty in p to 0.1 cm or reducing the uncertainty in q to 0.1 cm?

Answers : a.) Estimate f and its uncertainty :

q

1

p

1

f

1

7573013

1

32

1

q

1

p

1

f

1.

.. 3201f .

2

2

2

2

2

2

2

1

.......21 n X

n

X X U X

U

X

U

X

U

q p

pqf

2

q

2

2

p

2

f q

f

p

f

2

2

2

22

2

2

22

q

2

2

22

p

2

2

2

f 201332

32201332

13

q p

p

q p

q ...

....

.

075803e316413e43444f ...

075803201f .. b reduction in the uncertainty in f : reducing the uncertainty in p to 0 1 cm or reducing


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b. reduction in the uncertainty in f : reducing the uncertainty in p to 0.1 cm or reducing

the uncertainty in q to 0.1 cm

If p = +/- 0.1 and q = +/- 0.2

2

2

2

22

2

2

22

q

2

2

22

p

2

2

2

f 201332

32

101332

13

q p

p

q p

q

...

.

...

.

0490103e316413e0861f ...

0490103201f ..

If p = +/- 0.2 and q = +/- 0.1

2

2

2

22

2

2

22

q

2

2

22

p

2

2

2

f 101332

3220

1332

13

q p

p

q p

q.

..

..

..

.

0690103201f ..

0690103e291033e43444f ...

Reduction in p uncertainty will increase accuracy

Tutorial on Propogation of Errors

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