Ch_3b_4th_Ed - Section 3-10 3-127 a E(X = 300(0.4 = 120 V(X...

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Section 3-10 3-127. a) E(X) = 300(0.4) = 120, V(X) = 300(0.4)(0.6) = 72 and 72 = X σ . Then, 010724 . 0 ) 30 . 2 ( 72 120 5 . 100 ) 100 ( = - = - 2245 Z P Z P X P b) 0.007760 0 - 007760 . 0 ) 42 . 2 77 . 4 ( 72 120 5 . 99 72 120 5 . 79 ) 100 80 ( = = - < - = - < - 2245 < Z P Z P X P c) 0.107488 892512 . 0 1 ) 24 . 1 ( 1 72 120 5 . 130 1 ) 130 ( = - = - = - - 2245 Z P Z P X P 3-128. 50 50 - = X Z is approximately N(0, 1) a) 217695 . 0 ) 78 . 0 ( 50 50 5 . 44 ) 45 ( = - < = - = < Z P Z P X P b) 261086 . 0 ) 64 . 0 ( 50 50 5 . 45 ) 45 ( = - < = - = Z P Z P X P c) - - - = - = < 50 50 5 . 40 50 50 5 . 60 ) 40 ( ) 60 ( ) 60 40 ( Z P Z P X P X P X P 84044 . 0 090123 . 0 930563 . 0 ) 34 . 1 ( ) 48 . 1 ( = - = - - = Z P Z P 3-129. Let X = number of defective inspected parts E(X) = 100(0.08) = 8 V(X) = 100(0.08)(0.92) = 7.36 a) P(X < 8) = P(X 7) = = - 7 0 100 ) 92 . 0 ( ) 08 . 0 ( 100 i i i i = 0.4471 b) 428576 . 0 ) 18 . 0 ( 36 . 7 8 5 . 7 ) 8 ( = - < = - < 2245 < Z P z P X P 3.130. Let X denote the number of original components that fail during the useful life of the product. Then, X is a binomial random variable with p = 0.005 and n = 2000. Also, E(X) = 2000(0.005) = 10 and V(X) = 2000(0.005)(0.995) = 9.95. 959071 . 0 040929 . 0 1 ) 74 . 1 ( 1 95 . 9 10 5 . 4 1 ) 5 ( = - = - - = - - = Z P Z P X P . 3-131. Let X denote the number of defective chips in the lot. Then, E(X) = 1000(0.02) = 20, V(X) = 1000(0.02)(0.98) = 19.6. a) 107488 . 0 892512 . 0 1 ) 24 . 1 ( 1 6 . 19 20 5 . 25 1 ) 25 ( = - = - = - - 2245 Z P Z P X P b) ) 15 . 2 11 . 0 ( 6 . 19 20 5 . 29 6 . 19 20 5 . 20 ) 30 20 ( = - - 2245 < < Z P Z P X P 440427 . 0 543795 . 0 984222 . 0 ) 11 . 0 ( ) 15 . 2 ( = - = < - = Z P Z P
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3-132. Let X denote the number of people. X ~ BIN(1000, 0.193) 4800 . 12 193 ) 193 . 0 1 ( 193 193 . 0 1000 - = - × - = X X Z is approximately N(0,1). a) 2743 . 0 ) 6 . 0 ( 1 48 . 12 193 5 . 200 1 ) 5 . 0 200 ( 1 ) 200 ( 1 ) 200 ( = Φ - = - Φ - = + - = - = X P X P X P b) - Φ - - Φ = < - < = < < 48 . 12 193 5 . 180 48 . 12 193 5 . 299 ) 180 ( ) 300 ( ) 300 180 ( X P X P X P 841345 . 0 158655 . 0 1 ) 00 . 1 ( ) 53 . 8 ( = - = - Φ - Φ = 3-133. Let X denote the number of error accounts. X ~ BIN(362000, 0.001) a) E(X) = 362 Stdev(X) =19.0168 b) 0168 . 19 362 ) 001 . 0 1 ( 362 001 . 0 362000 - = - × - = X X Z is approximately N(0,1). 2555 . 0 ) 6573 . 0 ( 0168 . 19 362 5 . 349 ) 5 . 0 349 ( ) 350 ( = - Φ = - Φ = + = < X P X P c) 7799 . 392 5 . 0 362 0168 . 19 ) 95 . 0 ( 1 = - + × Φ - 3-134. Let X denote the number of particles in 10 cm 2 of dust. Then, X is a Poisson random variable with λ = 10(1000) = 10,000. Also, E(X) = λ = 10,000 and V(X) = λ = 10 4 . 5 . 0 5 . 0 1 ) 0 ( 1 10 000 , 10 5 . 000 , 10 1 ) 000 , 10 ( 4 = - = - 2245 - - 2245 Z P Z P X P 3-135. 50 50 - = X Z is approximately N(0, 1) a) - - - = - = < < 50 50 5 . 40 50 50 5 . 59 ) 5 . 40 ( ) 5 . 59 ( ) 60 40 ( Z P Z P X P X P X P 819754 . 0 090123 . 0 909877 . 0 ) 34 . 1 ( ) 34 . 1 ( = - = - - = Z P Z P b) 930563 . 0 069437 . 0 1 ) 48 . 1 ( 1 50 50 5 . 39 1 ) 40 ( = - = - - = - - = Z P Z P X P c) 069437 . 0 ) 48 . 1 ( 50 50 5 . 39 ) 40 ( = - = - = < Z P Z P X P d) λ = 50*7 = 350 emails/seven day week 694974 . 0 305026 . 0 1 ) 51 . 0 ( 1 350 350 5 . 340 1 ) 340 ( = - = - < - = -
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