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 - Basic Skills and Concepts - Page 312: 17

Answer

a) $0$ percent b) Unusually high c) Part b) d) Yes

Work Step by Step

a. We find: $μ=np=(945)(0.5)=472.5$ $σ=\sqrt{npq}=\sqrt{(945)(0.5)(0.5)}=15.37$ Hence, we find z: $z=\frac{878.5−472.5}{15.37}=26.41 $ $z=\frac{879.5−472.5}{15.37}=26.47 $ Thus, using the table of z-scores, we can find that the corresponding probability is: $0.0000$. b) We find z: $z=\frac{878.5−472.5}{15.37}=26.41$ Thus, using the table of z-scores, we can find that the corresponding probability is $0.0001$. Since the probability is so small that it is nearly $0$ percent, this number is unusually high. c) Part b) is more useful, because we do not care about getting exactly $845$ girls, rather, we care whether or not $845$ or more is a large number. d) Yes, it is, because the odds of getting this high of a number by chance is almost $0$.
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