s-assignment1 - Solution for assignment 1 1 a X = b =...

Info iconThis preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
Solution for assignment # 1 1. a) ¯ X = X i n = 5746 . 7 31 = 185 . 377 P.O.M = 0 . 5(31 + 1) = 16 ˜ X = 150 . 58 b) Since there is no outlier in this set of data, so mean is more appropriate than median. (Upper fence is 491.56). c) S 2 = X 2 i - ( X i ) 2 n n - 1 = 1341371 . 02 - (5746 . 7) 2 31 30 = 9202 . 086 S = S 2 = 9202 . 086 = 95 . 927 d) P.O.Q 1 = 0 . 25(32) = 8 Q 1 = 109 . 11 P.O.Q 3 = 0 . 75(32) = 24 Q 3 = 262 . 09 I.Q.R = 262 . 09 - 109 . 11 = 152 . 98 Lower fence=109.11-1.5(152.98)=-120.36, Upper fence=262.09+1.5(152.98)=491.56, so there is not any outlier in this data. 2. a) S = { ( HHH ) , ( HHT ) , ( HTH ) , ( THH ) , ( TTH ) , ( THT ) , ( HTT ) , ( TTT ) } b) Yes, event of a head on one flip is independent of event of a head on any other flips. c) A = { ( HHT ) , ( HTH ) , ( THH ) , ( HHH ) } P ( A ) = 3 × (0 . 6 × 0 . 6 × 0 . 4) + (0 . 6 × 0 . 6 × 0 . 6) = 0 . 648 B = { ( THH ) , ( HTH ) , ( HHT ) } P ( B ) = 3 × (0 . 6 × 0 . 6 × 0 . 4) = 0 .
Background image of page 1

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

View Full DocumentRight Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

This note was uploaded on 09/27/2010 for the course STAT 104157 taught by Professor Jhonbran during the Spring '09 term at California Coast University.

Page1 / 3

s-assignment1 - Solution for assignment 1 1 a X = b =...

This preview shows document pages 1 - 2. Sign up to view the full document.

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