Assignment 2 - suggested solution

# Assignment 2 - suggested solution - THE UNIVERSITY OF HONG...

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THE UNIVERSITY OF HONG KONG DEPARTMENT OF STATISTICS AND ACTUARIAL SCIENCE STAT0302 Business Statistics Assignment 2 Suggested Solutions 1. a. 0.90 prediction interval is σ μ 645 . 1 m So the 90% prediction interval for its diameter is 16.0 +- 1.645 (0.03) cm (15.9507cm, 16.0494cm) b. The six-sigma prediction interval is ) 2 . 179 , 8 . 170 ( 7 . 0 6 175 6 PI(x) 6 ohms ohms ohms = × = = m m The process limits are well within specification limits c. Suppose 20 1 ,..., X are the weights of those 20 persons respectively and follow normal distribution. X ) 5 . 8 , 5 . 62 ( ~ ),. .. 5 . 8 , 5 . 62 ( ~ 2 20 2 20 20 2 1 2 1 1 = = = = σμ N X N X Because are independent, 20 1 ,..., X X ) 1445 ... , 1250 ... ( ~ ... 2 20 2 1 20 1 20 1 = + = + + + σσ μμ N X X 5 . 0 ) 0 1445 1250 ... ( ) 1445 1250 1250 1445 1250 ... ( ) 1250 ... ( 20 1 20 1 20 1 = > + = > + = > + X X P X X P X X P d. ) 04 . 0 , 6 . 8 ( ~ B and ) 03 . 0 , 45 . 8 ( ~ A 2 2 2 2 = = = = B B A A N N Because A and B are independent, ) 05 . 0 03 . 0 04 . 0 , 15 . 0 45 . 8 6 . 8 ( ~ A - B 2 2 2 2 2 = + = + = = A B A B N

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1356 . 0 0.4452) - (0.5 0.4192) - 0.5 ( on) distributi normal standard a is (Z ) 6 . 1 P(Z ) 1.4 P(Z ) 05 . 0 15 . 0 23 . 0 0.05 0.15 - A - B P( ) 05 . 0 15 . 0 08 . 0 0.05 0.15 - A - B P( ) 0.23 A - P(B ) 0.08 A - P(B ) 0.23 A - B or 0.08 A - P(B = + = > + < = > + < = > + < = > < 2. a. ) 02 . 0 8.5, N( ~ X 2 2 = = σμ 0.0006 0.4994 - 0.5 .25) 3 P(Z ) 02 . 0 5 . 8 565 . 8 0.02 8.5 - X P( 8.535) P(X = = > = > = > b. ) 5 . 0 2, 5 . 11 N( ~ X 2 2 = = 0.8108 3315 . 0 4793 . 0 ) 96 . 0 Z P(-2.04 ) 5 . 0 52 . 11 12 0.5 11.52 - X 5 . 0 52 . 11 5 . 10 P( 12) X P(10.5 (ii) 0.0012 0.4988 - 0.5 ) 04 . 3 P(Z ) 5 . 0 52 . 11 10 0.5 11.52 - X P( 10) P(X (i) = + = < < = < < = < < = = < = < = < c. ) 15000 , 135000 N( ~ X 2 2 = = σ μ 85 . 0 ) 15000 135000 15000 135000 - X P( x find to 0.85 x) P(X = < = < x
0540 15 5540 1 135000 x 5540 1 1.036 15000 135000 - x B) table from 0.35 up (look 1.036 15000 135000 0.85 ) 036 . 1 P(Z = + = = × = = = < x So it keeps \$150540 in the machine just before 1:00p.m.on Saturday. d. 8 . 0 ) 2 ( = + < < σ μ k x k P 8 . 0 ) 2 ( 8 . 0 ) 2 ( = < < = + < < k Z k P k Z k P μσ k ) 2 ( k Z k P < < 0.93 ) 86 . 1 93 . 0 ( < < Z P = 0.3238+0.4686= 0.7924 0.94 ) 88 . 1 94 . 0 ( < < Z P = 0.3264+0.4699= 0.7963 0.95 ) 9 . 1 95 . 0 ( < < Z P = 0.3289+0.4713= 0.8002 0.96 ) 92 . 1 96 . 0 ( < < Z P = 0.3315+ 0.4726= 0.8041 Take = 0.95 k 3. a. X= the number of children who have blue eyes: X~Bi (n= 4, p=0.4286) i) ) X ( P 3 = + ) 3 ( = X P ) 4 ( = X P = + () ( ) 1 3 4286 . 0 1 4286 . 0 3 4 ( ) 0 4 4286 . 0 1 4286 . 0 4 4 = 0.2137 ii) Expected value = np = (4)(0.4286) = 1.7143 Variance = np(1 – p) = (4)(0.4286)(1-0.4286) = 0.9796 Standard deviation= 0.9898 b. X= the number of cans which were spoiled: X~ Bi (n= 14, p=0.20) i) ) 5 2 ( < < X P = ) 3 ( = X P + ) 4 ( = X P = + ( ) 9 3 20 . 0 1 20 . 0 3 14 ( 8 4 20 . 0 1 20 . 0 4 14 )

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= 0.4221 ii) Mean = np = (14)(0.20) = 2.8 Standard deviation = () p np 1 = ( )( ) 20 . 0 1 20 . 0 14 = 1.4967 c. P(A gets the gold medal) = P(A wins 4 games and B wins 0 game) + P(A wins 4 games and B wins 1
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## This note was uploaded on 11/06/2009 for the course STAT STAT0302 taught by Professor Prof.chiu during the Spring '09 term at HKU.

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Assignment 2 - suggested solution - THE UNIVERSITY OF HONG...

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