Chapter 16 and 17 HW

Chapter 16 and 17 HW - a)I would expect the archer to get 3.4 bulls eyes b The standard deviation is 1.34 17.HE.C Page 448 24 a Mean 3.6 SD 1.34 b

Info iconThis preview shows page 1. Sign up to view the full content.

View Full Document Right Arrow Icon
Chapter 16: Random Variables 16.HE.A: Page 427, # 6 a.) The probability model would be Result 6 Another 6 Not a 6 Earnings $100 $50 0 b.) The expected amount is 1/6 ($100) + 1/6 ($50) + 5/6 (0) = $25 c.) I would be willing to pay up to $25 for the game 16.HE.B: Page 428, # 18 a.) 0(.05)+2(.25)+2(.35)+3(.15)+4((.15)+5(.05) =2.25 b.) 0.7 16.HE.C: Page 429, # 34 Mean: .64 SD: .18 Chapter 17: Probability Models 17.HE.A: Page 448, # 20 d, e, f d.) Getting exactly 4 bulls eyes is binompdf(6,.80,4)= .24 e.) Getting at least 4 bulls eyes is 1 - binomcdf (4,80,4) = .79 f.) Getting at most is 4 bulls eyes is binomcdf (4,80,4) = 1.0 17.HE.B: Page 448, # 22 a, b
Background image of page 1
This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: a.)I would expect the archer to get 3.4 bulls eyes b.) The standard deviation is 1.34 17.HE.C: Page 448, # 24 a.) Mean: 3.6 SD: 1.34 b.) The probability of her never missing is .832 c.) The probability of there being no more than 8 bulls eyes is .87 d.) The probability of exactly 7 bulls eyes is .653 e.) The probability of hitting more than missing is .798 17.HE.D: Page 449, # 30 a.) Mean: 68 SD: .28 b.) The Normal model is appropriate here because her serves meet all assumptions. c.) The distribution is within 2 deviations of her serves. d.) The probability of her making at least 65 serves is 76.47%...
View Full Document

This note was uploaded on 12/05/2010 for the course BUS BUS taught by Professor Fox during the Fall '10 term at University of Baltimore.

Ask a homework question - tutors are online