168 CHAPTER 4
The uncertainty is σb
+σ2
=√6.25 + 3.5 = 3.1.
18. If the mean decay rate is exactly 1 per second, then the mean number of events in 10 seconds is 10,
so X∼Poisson(10).
19. If the mean number of particles is exactly 7 per mL, then X∼Poisson(7).
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SECTION 4.3 169
20. Let λBrepresent the background rate, let λSrepresent the source rate, and let λT=λB+λSrepresent
the total rate, all in particles per second.
(a) Let Xbe the number of background events counted in 100 seconds. Then X∼Poisson(100λB).
(c) λS=λT−λB, so b
λS=b
λT−b
λB= 3.24 −0.36 = 2.88.
170 CHAPTER 4
(e) It is not possible. Assume that the background emissions are counted for 100 seconds, and the source
21. Let Xbe the number of flaws in a one-square-meter sheet of aluminum. Let λbe the mean number
Section 4.4
1. Let Xbe the number of units with broken fans among the seven chosen. Then X∼H(20,8,7).
320 −8
7−3
2. Let Xbe the number of restaurants with code violations among the 10 chosen. Then X∼H(30,4,10).
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SECTION 4.4 171
230 −4
10 −2
3. Let Xbe the number of the day on which the computer crashes. Then X∼Geom(0.1).
4. Let Xbe the number days that pass up to and including the first time the car encounters a red light.
Then X∼Geom(0.4).
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172 CHAPTER 4
6. Let X1be the number of green lights, X2be the number of yellow lights, and X3be the number of
8. Let Xbe then number of packages that will be filled before the process is stopped.
Then X∼Geom(0.01).
9. Let Xbe the number of hours that have elapsed when the fault is detected.
Then X∼Geom(0.8).
SECTION 4.4 173
10. Let Xbe the number of runs up to and including the fifth failure.
Then X∼NB(5,0.001).
11. (a) X∼H(10,3,4). P(X= 2) = 3
210 −3
4−2
10
4= 0.3
13. (a) X∼H(60,5,10), so
060 −5
10 −0
160 −5
10 −1
174 CHAPTER 4
14. (a) (X1, X2, X3, X4)∼MN(15,0.20,0.30,0.15,0.35)
15. Let X1denote the number of orders for a small drink, let X2denote the number of orders for a
medium drink, let X3denote the number of orders for a large drink.
16. (a) Let X1be the number of readings that are within 0.1◦C of the true temperature, let X2be the number
of readings that are more than 0.1◦C above the true temperature, let X3be the number of readings
that are more than 0.1◦C below the true temperature. Then (X1, X2, X3)∼MN(10,0.7,0.1,0.2).
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SECTION 4.5 175
(b) X1∼Bin(10,0.7)
1).
Section 4.5
1. (a) Using Table A.2: 1 −0.1977 = 0.8023
2. (a) Using Table A.2: 0.7123
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3. (a) Using Table A.2: c= 1
4. (a) z= (2 −2)/√9 = 0. The area to the right of z= 0 is 0.5000.
5. (a) z= (19 −16)/2 = 1.50. The area to the right of z= 1.50 is 0.0668.
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SECTION 4.5 177
7. (a) z= (700 −480)/90 = 2.44. The area to the right of z= 2.44 is 0.0073.
8. (a) For 3.7, z= (3.7−4.1)/0.6 = −0.67. For 4.4, z= (4.4−4.1)/0.6 = 0.50.
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178 CHAPTER 4
9. (a) z= (1800 −1400)/200 = 2.00. The area to the right of z= 2.00 is 0.0228.
10. No. The maximum possible score of 800 is 1.5 standard deviations above the mean of 650. If the
11. (a) z= (6 −4.9)/0.6 = 1.83. The area to the right of z= 1.83 is 1 −0.9664 = 0.0336.
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12. (a) Since 10% of the bolts have strengths less than 18.3 kN, the z-score of 18.3 is −z.10 =−1.28.
Since 5% of the bolts have strengths greater than 19.76 kN, the z-score of 19.76 is z.05 = 1.645.
13. Let Xbe the diameter of a hole, and let Ybe the diameter of a piston. Then X∼N(15,0.0252),
and Y∼N(14.88,0.0152). The clearance is given by C= 0.5X−0.5Y.
(a) µC=µ0.5X−0.5Y= 0.5µX−0.5µY= 0.5(15) −0.5(14.88) = 0.06 cm
(b) σC=p0.52σ2
X+ (−0.5)2σ2
Y=p0.52(0.0252) + 0.52(0.0152) = 0.01458 cm
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180 CHAPTER 4
14. (a) The specification interval is 0.645 to 0.655 cm.
The z-score of 0.645 is (0.645 −0.652)/0.003 = −2.33.
The z-score of 0.655 is (0.655 −0.652)/0.003 = 1.00.
The area between z=−2.33 and z= 1.00 is 0.8413 −0.0099 = 0.8314.
15. (a) The z-score of 12 is (12 −12.05)/0.03 = −1.67. The area to the left of z=−1.67 is 0.0475. The
proportion is 0.0475.
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16. (a) The z-score of 6.2 is (6.2−6)/0.3 = 0.67. The area to the left of z= 0.67 is 0.7486. The probability
is 0.7486.
17. (a) The proportion of strengths that are less than 65 is 0.10. Therefore 65 is the 10th percentile of
strength. The z-score of the 10th percentile is approximately z=−1.28. Let σbe the required
18. (a) The z-score of 10.3 is (10.3−10)/0.2 = 1.5. The area to the right of z= 1.5 is 1 −0.9332 = 0.0668.
19. Let a= 1/σ and let b=−µ/σ. Then Z=aX +b. Now aµ +b= (1/σ)µ−µ/σ = 0, and
a2σ2= (1/σ)2σ2= 1. Equation (4.25) now shows that Z∼N(0,1).
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20. (a) The z-score of 160 is (160 −200)/10 = −4.00. The area to the left of z=−4.00 is negligible.
Therefore P(X≤160) ≈0.
21. (a) Let D=R2−R1. The event that R2> R1is the event that D > 0.
µD=µR2−µR1= 120 −100 = 20.
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SECTION 4.5 183
22. (a) Since Mis a linear combination of independent normal random variables, Mis normally distributed.
µM= 0.5µX+µY= 0.5(0.450) + 0.250 = 0.475.
23. (a) If m= 0, then X=E, so X∼N(0,0.25).
P(error) = P(X > 0.5). The z-score of 0.5 is (0.5−0)/√0.25 = 1.00.
The area to the right of z= 1.00 is 1 −0.8413 = 0.1587.
24. (a) If m= 0, then X=−1.5 + E, so X∼N(−1.5,0.64).
P(error) = P(X > 0.5). The z-score of 0.5 is [0.5−(−1.5)])/√0.64 = 2.50.
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184 CHAPTER 4
The area to the left of z=−1.25 is 0.1056.
Therefore P(error) = 0.1056.
(c) Let Sbe the value from the sent string (0 or 1). Let Cbe the event that the value is received correctly.
From part (a) we know that P(C|S= 0) = 1 −0.0062 = 0.9938.
25. (a) The sample mean is 114.8 J and the sample standard deviation is 5.006 J.
(b) The z-score of 100 is (100 −114.8)/5.006 = −2.96. The area to the left of z=−2.96 is 0.0015.
Therefore only 0.15% of bolts would have breaking torques less than 100 J, so the shipment would be
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SECTION 4.6 185
(e) The method is certainly not valid for the bolts in part (c). This sample contains an outlier (140), so
26.
k P (|X−µX| ≥ kσX) 1/k2
1 0.3174 1
The actual probabilities are much smaller
Section 4.6
1. Let Ybe the lifetime of the component.
(a) E(Y) = eµ+σ2/2=e1.2+(0.4)2/2= 3.5966
(b) P(3 < Y < 6) = P(ln 3 <ln Y < ln 6) = P(1.0986 <ln Y < 1.7918). ln Y∼N(1.2,0.42).
The z-score of 1.0986 is (1.0986 −1.2)/0.4 = −0.25.
2. Let Ybe the amount of active ingredient in the skin.
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186 CHAPTER 4
(b) Let mbe the median of Y. Then P(Y≤m) = P(ln Y≤ln m) = 0.5.
3. Let Yrepresent the BMI for a randomly chosen man aged 25–34.
(a) E(Y) = eµ+σ2/2=e3.215+(0.157)2/2= 25.212
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SECTION 4.6 187
4. Let Ybe the increase in the risk.
(a) E(Y) = eµ+σ2/2=e−15.65+(0.79)2/2= 2.1818 ×10−7
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