PS1s - 1 ) ( 43 1 ) (from drawer s view) (4) (a) Ans: 0.600...

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SOLUTIONS TO ISMT111 PROBLEM SET #1 (1) Ans: (D) Statement III only. 0 is the mode number of accidents as it has the highest relative frequency of .55. Obviously, it is also the median as its cumulative relative frequency is just greater than .50. So, mode = median = 0. The mean cannot be calculated but is clearly greater than 0. (2) Ans: 3.00 Ms . Jones : 70 - x = s (1) Mr . Shaw : 28 - x = - 1 . 8 s (2) To solve for x and s, we subtract (2) from (1) : 42 = 2 . 8 s giving s = 15 and x = 55 Hence , the z score of 100 is : 100 - 55 15 = 3 (3) P (Jackpot) = n ( J ) n ( T ) = ( 6 6 )( 42 0 )( 1 0 ) ( 49 6 ) = 7 . 15112 E - 08 (from player 0 s view) also = ( 6 6 )( 43 0 ) ( 49 6 ) × ( 43 1 ) ( 43 1 ) (from drawer 0 s view) P (Second) = n ( S ) n ( T ) = ( 6 5 )( 42 0 )( 1 1 ) ( 49 6 ) = 4 . 29067 E - 07 (from player 0 s view) also = ( 6 5 )( 43 1 ) ( 49 6 ) × ( 1
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Unformatted text preview: 1 ) ( 43 1 ) (from drawer s view) (4) (a) Ans: 0.600 P (under 20 yrs | on the technical staff) = (31 + 59) (31 + 59 + 45 + 15) = 0 . 600 (b) Ans: 0.060 P (over 20 yrs ∩ on the nontechnical staff ∩ plans to retire at age 65) = 12 (31 + 59 + 5 + 25 + 45 + 15 + 12 + 8) = 0 . 060 (5) P ( A ) = ( 2 2 ) ( 6 2 ) = 1 15 P ( B ) = ( 4 2 ) ( 6 2 ) = 6 15 P ( C ) = ( 4 1 )( 2 1 ) + ( 4 2 )( 2 ) ( 6 2 ) = 14 15 also equal to P ( A c ) P ( B | C ) = P ( B ∩ C ) P ( C ) = P ( B ) P ( C ) = 6 14 because B is a subset of C. * * * 1 ATChan 2008...
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This note was uploaded on 12/22/2009 for the course ISOM ISOM111 taught by Professor Anthonychan during the Fall '09 term at HKUST.

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