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110 CHAPTER 3
7. (a) Yes, the uncertainty can be estimated with the standard deviation of the five measurements, which is
21.3 µg.
10. (a) Yes, the uncertainty can be estimated with the standard deviation of the five measurements, which is
0.0109 ppm.
11. (a) No, they are in increasing order, which would be highly unusual for a simple random sample.
Page 110
SECTION 3.2 111
Section 3.2
1. (a) σ4X= 4σX= 4(0.3) = 1.2
2. Let nbe the number of measurements required. Then 1.5/√n= 0.5, so n= 9.
5. Let Xrepresent the estimated mean annual level of land uplift for the years 1774–1884, and let Y
represent the estimated mean annual level of land uplift for the years 1885–1984. Then X= 4.93,
Y= 3.92, σX= 0.23, and σY= 0.19. The difference is X−Y= 1.01 mm.
6. Let Xrepresent the measured diameter of the hole, and let Yrepresent the measured diameter of the
Page 111
10. (a) µ= 1.49, h= 10, τ= 30.0, στ= 0.1
12. τ0= 50, w= 1.0, k= 0.29, σk= 0.05
SECTION 3.2 113
14. (a) The bias is 2 g and the uncertainty is σ= 3 g.
15. (a) Let X= 87.0 denote the average of the eight measurements, let s= 2.0 denote the sample standard
deviation, and let σdenote the unknown population standard deviation. The volume is estimated
16. The average of the five measurements is X= 37.0, and the sample standard deviation is s= 1.28841.
(a) k=X±s/√5 = 37.0±1.28841/√5 = 37.0±0.6 N/m.
Page 113
17. Let Xdenote the average of the 10 yields at 65◦C, and let Ydenote the average of the 10 yields at
80◦C. Let sXdenote the sample standard deviation of the 10 yields at 65◦C, and let sYdenote the
19. (a) Let σX= 0.05 denote the uncertainty in the instrument. Then σX=σX/√10 = 0.05/√10 = 0.016.
20. (a) σX+Y=qσ2
X+σ2
Y=pσ2
1/4 + σ2
2/12 = p0.022/4 + 0.082/12 = 0.025.
Page 114
SECTION 3.3 115
Section 3.3
1. X= 2.0, σX= 0.3.
(a) dY
dX = 3X2= 12, σY=
dY
dX σX= 3.6
dY
2. X= 3.0, σX= 0.1
dY
Page 115
116 CHAPTER 3
dY
3. h= 6, r= 5, σr= 0.02, V=πr2h/3 = 157.0796.
5. (a) g= 9.80, L= 0.742, σL= 0.005, T= 2πpL/g = 2.00709√L= 1.7289
7. (a) g= 9.80, d= 0.15, l= 30.0, h= 5.33, σh= 0.02, F=pgdh/4l= 0.110680√h= 0.2555
(b) g= 9.80, l= 30.0, h= 5.33, d= 0.15, σd= 0.03, F=pgdh/4l= 0.659760√d= 0.2555
(c) g= 9.80, d= 0.15, h= 5.33, l= 30.00, σl= 0.04, F=pgdh/4l= 1.399562l−1/2= 0.2555
9. m= 750, V0= 500.0, V1= 813.2, σV0=σV1= 0.1, D=m
=750
= 2.3946.
Page 117
118 CHAPTER 3
dD
10. (a) C0= 0.1, t= 45, C= 0.0811, σC= 0.0005, k= (1/t)(1/C −1/C0) = (1/45)(1/C)−2/9 = 0.051788
(b) C0= 0.1, C= 0.0750, k= 0.0518, σk= 0.0017, t= (1/k)(1/C −1/C0) = (10/3)(1/k) = 64.350064
11. (a) The relative uncertainty is 0.4/20.9 = 1.9%
12. (a) The absolute uncertainty is 0.003(48.41) = 0.15
Page 118
13. s= 2.2, t= 0.67, σt= 0.02, g= 2s/t2= 4.4/t2= 9.802
dln g
14. T= 298.4, σT= 0.2, V= 20.04√T= 346.18
dln V
15. (a) g= 9.80, L= 0.855, σL= 0.005, T= 2πpL/g = 2.00709√L= 1.85588
dln T
(b) L= 0.855, T= 1.856, σT= 0.005, g= 4π2LT −2= 33.754047T−2= 9.799
dln g
16. c= 448, ∆Q= 1210, m= 0.75, σm= 0.01, ∆T= ∆Q/mc = 1210/[(0.75)(448)] = 3.6012
dln ∆T
17. (a) g= 9.80, d= 0.20, l= 35.0, h= 4.51, σh= 0.03, F=pgdh/4l= 0.118332√h= 0.2513
Page 119
120 CHAPTER 3
dln F
(b) g= 9.80, l= 35.0, h= 4.51, d= 0.20, σd= 0.008, F=pgdh/4l= 0.561872√d= 0.2513
dln F
(c) g= 9.80, d= 0.20, h= 4.51, l= 35.00, σl= 0.4, F=pgdh/4l= 1.486573l−1/2= 0.2513
dln F
18. θ= 0.90, σθ= 0.01, n= 1/sin θ= 1.276606
dln n
19. m= 288.2, V0= 400.0, V1= 516.0, σV0= 0.1, σV1= 0.2, D=m
=288.2
= 2.484
20. (a) C0= 0.04, t= 30, C= 0.0038, σC= 0.0002, k= (1/t)(1/C −1/C0) = 1/(30C)−5/6 = 7.94
dln k
k= 7.94 L/mol·s±5.8%
(b) C0= 0.04, t= 50, C= 0.0024, σC= 0.0002, k= (1/t)(1/C −1/C0) = 1/(50C)−1/2 = 7.83
dln k
Section 3.4
1. (a) X= 10.0, σX= 0.5, Y= 5.0, σY= 0.1, U=XY 2= 250
(b) X= 10.0, σX= 0.5, Y= 5.0, σY= 0.1, U=X2+Y2= 125
(c) X= 10.0, σX= 0.5, Y= 5.0, σY= 0.1, U= (X+Y2)/2 = 17.50
2. (a) r= 5.00, h= 6.00, σr= 0.02, σh= 0.01, V=πr2h/3 = 157.0796
∂r = 2πrh/3 = 62.8319, ∂V
∂h =πr2/3 = 26.1799, σV=s∂V
∂r 2
∂h 2
Page 121
(b) σV=s∂V
σ2
r+∂V
σ2
h,∂V
3. (a) s= 55.2, θ= 0.50, σs= 0.1, σθ= 0.02, h=stan θ= 30.1559
∂h
∂s 2
∂θ 2
h= 30.2±1.4
(b) σh=s∂h
∂s 2
σ2
s+∂h
∂θ 2
σ2
θ,∂h
∂s = 0.5463, ∂h
∂θ = 71.6742
4. (a) µ= 1.49, τ= 30.0, στ= 0.1, h= 10.0, σh= 0.2, V=τh/µ =τh/1.49 = 201.3
(b) σV=s∂V
∂τ 2
σ2
τ+∂V
∂h 2
σ2
h,∂V
∂τ = 6.71141, ∂V
∂h = 20.1342
Page 122
5. (a) P1= 10.1, σP1= 0.3, P2= 20.1, σP2= 0.4, P3=√P1P2= 14.25
(b) σP3=s∂P3
∂P12
σ2
P1+∂P3
∂P22
σ2
P2,∂P3
∂P1
= 0.705354, ∂P3
∂P2
= 0.354432
6. (a) M1= 1.32, σM1= 0.01, M2= 1.04, σM2= 0.01, W= (M1−M2)/M1= 0.2121
(b) σW=s∂W
∂M12
σ2
M1+∂W
∂M22
σ2
M2,∂W
∂M1
= 0.596878, ∂W
∂M2
=−0.757576
7. (a) p= 2.3, σp= 0.2, q= 3.1, σq= 0.2, f=pq/(p+q) = 1.320
Page 123
124 CHAPTER 3
(b) σf=s∂f
∂p 2
σ2
p+∂f
∂q 2
σ2
q,∂f
∂p = 0.329561, ∂f
∂q = 0.181413
8. (a) P= 242.52, σP= 0.03, V= 10.103, σV= 0.002, T=P V/8.31 = 294.847
(b) P= 242.52, σP= 0.03, T= 290.11, σT= 0.02, V= 8.31T/P = 9.9407
(c) V= 10.103, σV= 0.002, T= 290.11, σT= 0.02, P= 8.31T/V = 238.624
∂V 2
∂T 2
P= 238.624 ±0.050 kPa
9. (a) C= 1.25, σC= 0.03, L= 1.2, σL= 0.1, A= 1.30, σA= 0.05,
(b) σM=s∂M
∂C 2
σ2
C+∂M
∂L 2
σ2
L+∂M
∂A 2
σ2
A,
10. Let I1be the estimated index at a temperature of 166 K, let I2be the estimated index at a temperature
of 196 K, and let R=I1/I2be the ratio. Then I1= 0.00116, σI1= 0.0001, I2= 0.00129, σI2= 0.0001.
∂I12
∂I22
11. (a) τ0= 50, στ0= 1, w= 1.2, σw= 0.1, k= 0.29, σk= 0.05,
126 CHAPTER 3
(b) στ=s∂τ
∂τ02
σ2
τ0+∂τ
∂k 2
σ2
k+∂τ
∂w 2
σ2
w,
(c) If στ0= 0, σk= 0.05, and σw= 0, στ= 3.0. Thus implementing the procedure would reduce the
uncertainty in τonly to 3.0 MPa. It is probably not worthwhile to implement the new procedure for
a reduction this small.
12. θ1= 0.3672, σθ1= 0.005, θ2= 0.2943, σθ2= 0.004, n=sin θ1
sin θ2
= 1.238
13. (a) g= 3867.4, σg= 0.3, b= 1037.0, σb= 0.2, m= 2650.4, σm= 0.1, y=mb/g = 710.68.
SECTION 3.4 127
(b) σy=s∂y
∂g 2
σ2
g+∂y
∂m 2
σ2
m+∂y
∂b 2
σ2
b,
14. (a) l= 14.0, σl= 0.1, d= 4.4, σd= 0.1, R=kl/d2= 0.723k
(b) σR=s∂R
∂l 2
σ2
l+∂R
∂d 2
σ2
d,∂R
∂l =k/d2= 0.0516529k,∂R
∂d =−2kl/d3=−0.3287k
15. (a) F= 800, σF= 1, R= 0.75, σR= 0.1, L0= 25.0, σL0= 0.1, L1= 30.0, σL1= 0.1,
Page 127
128 CHAPTER 3
∂L1
πR2(L1−L0)2=−452.707
σY=s∂Y
∂F 2
σ2
F+∂Y
∂R 2
σ2
R+∂Y
∂L02
σ2
L0+∂Y
∂L12
σ2
L1= 608
(b) σY=s∂Y
∂F 2
σ2
F+∂Y
∂R 2
σ2
R+∂Y
∂L02
σ2
L0+∂Y
∂L12
σ2
L1,
16. T= 50, k= 0.025, T0= 70.1, σT0= 0.2, Ta= 35.7, σTa= 0.1,
Page 128
