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Unformatted text preview: Section 39 3107. e −4 4 0 = e −4 = 0.0183 0! b) P( X ≤ 2) = P( X = 0) + P( X = 1) + P( X = 2)
a) P( X = 0) = e −4 41 e −4 42 + 1! 2! = 0.2381 = e −4 + e −4 4 4 = 01954 . 4! e −4 4 8 = 0.0298 d) P( X = 8) = 8!
c) P ( X = 4 ) = 3108 a) P ( X = 0) = e −0.4 = 0.6703 e −0.4 (0.4) e −0.4 ( 0.4) 2 + = 0.9921 1! 2! e −0.4 (0.4) 4 = 0.000715 c) P ( X = 4 ) = 4! e −0.4 (0.4) 8 = 109 × 10 −8 . d) P( X = 8) = 8!
b) P( X ≤ 2) = e −0.4 + 3109.
P ( X = 0) = e − λ = 0.05 . Therefore, λ = −ln(0.05) = 2.996. Consequently, E(X) = V(X) = 2.996. 3110 a) Let X denote the number of calls in one hour. Then, X is a Poisson random variable with λ = 10. e −10 105 = 0.0378 . 5! e −10 10 e −10 102 e −10 103 + + = 0.0103 b) P( X ≤ 3) = e −10 + 1! 2! 3! c) Let Y denote the number of calls in two hours. Then, Y is a Poisson random variable with e −20 2015 = 0.0516 λ = 20. P( Y = 15) = 15! d) Let W denote the number of calls in 30 minutes. Then W is a Poisson random variable with e −5 55 = 01755 . λ = 5. P( W = 5) = 5! P( X = 5) = 3111. λ=1, Poisson distribution. f(x) =e λ λx/x! (a) P(X≥2)= 0.264 (b) In order that P(X≥1) = 1P(X=0)=1e λ exceed 0.95, we need λ=3. Therefore 3*16=48 cubic light years of space must be studied. (a) λ=14.4, P(X=0)=6*107 (b) λ=14.4/5=2.88, P(X=0)=0.056 (c) λ=14.4*7*28.35/225=12.7 P(X≥1)=0.999997 (d) P(X≥28.8) =1P(X ≤ 28) = 0.00046. Unusual. (a) λ=0.61. P(X≥1)=0.4566 (b) λ=0.61*5=3.05, P(X=0)= 0.047. 3112. 3113. 322 3114. a) Let X denote the number of flaws in one square meter of cloth. Then, X is a Poisson random variable with λ = 0.1. P( X = 2) = e −0.1 (0.1) 2 = 0.0045 2! b) Let Y denote the number of flaws in 10 square meters of cloth. Then, Y is a Poisson random variable with λ = 1. P (Y = 1) = e−111 = e−1 = 0.3679 1! c) Let W denote the number of flaws in 20 square meters of cloth. Then, W is a Poisson random variable with λ = 2. d) P (W = 0) = e −2 = 0.1353 P (Y ≥ 2) = 1 − P (Y ≤ 1) = 1 − P (Y = 0) − P (Y = 1)
= 1 − e −1 − e −1 = 0.2642 3115.a) E ( X ) = λ = 0.2 errors per test area
b) P ( X ≤ 2) = e − 0.2 e −0.2 0.2 e −0.2 (0.2) 2 + + = 0.9989 1! 2! 99.89% of test areas 3116. a) Let X denote the number of cracks in 5 miles of highway. Then, X is a Poisson random variable with λ = 10. P ( X = 0) = e = 4.54 × 10 b) Let Y denote the number of cracks in a half mile of highway. Then, Y is a Poisson random variable with λ = 1. P (Y ≥ 1) = 1 − P (Y = 0) = 1 − e = 0.6321 c) The assumptions of a Poisson process require that the probability of a count is constant for all intervals. If the probability of a count depends on traffic load and the load varies, then the assumptions of a Poisson process are not valid. Separate Poisson random variables might be appropriate for the heavy and light load sections of the highway. 3117. a) Let X denote the number of flaws in 10 square feet of plastic panel. Then, X is a Poisson random variable with λ = 0.5. P ( X = 0) = e = b) Let Y denote the number of cars with no flaws,
−0.5 −1 −10 −5 0.6065 ⎛10 ⎞ P (Y = 10) = ⎜ ⎟(0.6065)10 (0.3935) 0 = 0.0067 ⎜10 ⎟ ⎝⎠
c) Let W denote the number of cars with surface flaws. Because the number of flaws has a Poisson distribution, the occurrences of surface flaws in cars are independent events with constant probability. From part a., the probability a car contains surface flaws is 1−0.6065 = 0.3935. Consequently, W is binomial with n = 10 and p = 0.3935. ⎛10 ⎞ P (W = 0) = ⎜ ⎟ (0.3935)0 (0.6065)10 = 0.0067 ⎝0⎠ ⎛10 ⎞ P (W = 1) = ⎜ ⎟ (0.3935)1 (0.6065)9 = 0.0437 ⎝1⎠ P (W ≤ 1) = 0.0067 + 0.0437 = 0.0504 323 3118. a) Let X denote the failures in 8 hours. Then, X has a Poisson distribution with λ = 0.16. P( X = 0) = e −0.16 = 0.8521
b) Let Y denote the number of failure in 24 hours. Then, Y has a Poisson distribution with λ = 0.48. P(Y ≥ 1) = 1 − P(Y = 0) = 1 − e −48 = 0.3812 Supplemental Exercises 3119. E( X ) = 1 ⎛1⎞ 1 ⎛1⎞ 3 ⎛1⎞ 1 ⎜ ⎟+ ⎜ ⎟+ ⎜ ⎟ = , 8 ⎝3⎠ 4 ⎝ 3⎠ 8 ⎝ 3⎠ 4
2 2 2 2 3120. ⎛1⎞ ⎛1⎞ ⎛ 1⎞ ⎛1⎞ ⎛3⎞ ⎛1⎞ ⎛ 1 ⎞ V ( X ) = ⎜ ⎟ ⎜ ⎟ + ⎜ ⎟ ⎜ ⎟ + ⎜ ⎟ ⎜ ⎟ − ⎜ ⎟ = 0.0104 ⎝8⎠ ⎝3⎠ ⎝ 4⎠ ⎝3⎠ ⎝8⎠ ⎝3⎠ ⎝ 4⎠ ⎛1000 ⎞ 1 999 a) P ( X = 1) = ⎜ ⎟ ⎜ 1 ⎟0.001 (0.999) = 0.3681 ⎠ ⎝ ⎛1000 ⎞ 999 0 b) P( X ≥ 1) = 1 − P( X = 0) = 1 − ⎜ ⎟ ⎜ 0 ⎟0.001 (0.999) = 0.6319 ⎠ ⎝ ⎛1000 ⎞ ⎛1000 ⎞ ⎛1000 ⎞ 1000 999 2 998 ⎟0.0011 (0.999) + ⎜ ⎟0.0010 (0.999) + ⎜ c) P( X ≤ 2) = ⎜ ⎟ ⎜ 2 ⎟0.001 0.999 ⎜1⎟ ⎜0⎟ ⎠ ⎝ ⎠ ⎝ ⎠ ⎝ = 0.9198 d) E ( X ) = 1000(0.001) = 1 V ( X ) = 1000(0.001)(0.999) = 0.999
3121. a) n = 50, p = 5/50 = 0.1, since E(X) = 5 = np. ⎛ 50 ⎞ ⎛ 50 ⎞ ⎛ 50 ⎞ 50 49 48 ≤ 2) = ⎜ ⎟0.10 (0.9) + ⎜ ⎟0.11 (0.9) + ⎜ ⎟0.12 (0.9) = 0.112 ⎜2⎟ ⎜1⎟ ⎜0⎟ ⎝⎠ ⎝⎠ ⎝⎠ ⎛ 50 ⎞ ⎛ 50 ⎞ 49 1 0 ⎟0.1 (0.9) + ⎜ ⎟0.150 (0.9 ) = 4.51 × 10− 48 c) P ( X ≥ 49) = ⎜ ⎜ 50 ⎟ ⎜ 49 ⎟ ⎝⎠ ⎝⎠
b) P( X 3122. (a)Binomial distribution, p=0.01, n=12. (b) P(X>1)=1P(X≤1)= 1 ⎜ ⎜ V(X)=np(1p) = 0.1188 ⎛12 ⎞ 0 ⎛12 ⎞ ⎟ p (1 − p)12  ⎜ ⎟ p 1 (1 − p )14 =0.0062 ⎟ ⎜1 ⎟ ⎝0 ⎠ ⎝⎠
σ= (c) µ =E(X)= np =12*0.01 = 0.12 V ( X ) = 0.3447 3123. (a) (b) (c) (0.5)12 = 0.000244 6 C12 (0.5)6 (0.5)6 = 0.2256
12 12 C5 (0.5) 5 (0.5) 7 + C6 (0.5) 6 (0.5) 6 = 0.4189 3124. (a) Binomial distribution, n=100, p=0.01. (b) P(X≥1) = 0.634 (c) P(X≥2)= 0.264 324 ...
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