Elementary Statistics (12th Edition)

Published by Pearson
ISBN 10: 0321836960
ISBN 13: 978-0-32183-696-0

Chapter 6 - Normal Probability Distributions - 6-7 Normal as Approximation to Binomial - Beyond the Basics - Page 313: 25


a) 6; 0.46 b) 101; 0.3936 c) Neither game

Work Step by Step

a)He is betting $5$ dollars each time, so for each win, he gets $5\cdot35=175$ dollars. He bets $1000$ dollars, so he needs $\frac{1000}{175}≈6$ wins. We find that the probability that he will make a profit is: $0.5\cdot\frac{\frac{1}{38}}{\frac{1}{35}}=0.46$ b) We find: $μ=np=(200)(0.492)=98.58$ $σ=\sqrt{npq}=\sqrt{(200)(0.492)(0.507)}=7.07$ Hence, we find z: $z=\frac{100.5−98.58}{7.07}=0.6064$ Thus, using the table of z-scores, we can find that the corresponding probability is: $1−0.6064=0.3936 $. c) The roulette game has a higher chance of making a profit, so it is the better option. However, he should choose t play neither.
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