03exS_4

# 03exS_4 - x rooms were booked be Y x i y i(\$ P Y = y i x 6...

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4. Let the number of customers that will show up be X (a) X is a binomial random variable because: There are 2 possible outcomes in each trial: a customer will show up or not show up. n and p are known. The number of rooms being booked: n =13 . The probability of each customer will show up: p =0 . 80 . The additional assumption is whether each customer will show up or not is independent. X is a binomial random variable with n =13 , p =0 . 80 P ( X = x )= C 13 x 0 . 80 x (1 0 . 80) 13 x (b) The expected number of customers that will show up is: E ( X )= np =13 × 0 . 80 = 10 . 4 (c) P ( x 6 12) = P 12 x =0 C 15 x p x (1 p ) n x =1 C 15 13 0 . 8 13 (1 0 . 8) 2 C 15 14 0 . 8 14 (1 0 . 8) 1 C 15 15 0 . 8 15 (1 0 . 8) 0 =0 . 601977 (d) Let the revenue of
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Unformatted text preview: x rooms were booked be Y x i y i (\$) P ( Y = y i ) x 6 12 375 × 15 = 3386 . 11945 P 12 x =0 C 15 x p x (1 − p ) n − x = 0 . 601977 x = 13 375 × 15 − 500 = 1183 . 34939 C 15 13 . 8 13 (1 − . 8) 2 = 0 . 230897 x = 14 375 × 15 − 1000 = 610 . 22895 C 15 14 . 8 14 (1 − . 8) 1 = 0 . 131941 x = 15 375 × 15 − 1500 = 145 . 13553 C 15 15 . 8 15 (1 − . 8) = 0 . 035184 The expected net revenue if 15 room bookings were accepted is: E ( Y ) = P y i P ( Y = y i ) = \$5324 . 8333 3...
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## This note was uploaded on 01/06/2011 for the course ISOM 111 taught by Professor Hu,inchi during the Fall '10 term at HKUST.

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