Assignment 2 - suggested solution

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

Info iconThis preview shows pages 1–5. Sign up to view the full content.

View Full Document Right Arrow Icon
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
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
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
Background image of page 2
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 )
Background image of page 3

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
= 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
Background image of page 4
Image of page 5
This is the end of the preview. Sign up to access the rest of the document.

Page1 / 12

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

This preview shows document pages 1 - 5. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online