## Elementary Statistics (12th Edition)

a) $0.0014$ b) $0.0062$ c) Part b) d) The evidence is strong.
a. We find: $μ=np=(611)(0.33)=201.63$ $σ=\sqrt{npq}=\sqrt{(611)(0.33)(0.67)}=11.62$ Hence, we find z: $z=\frac{171.5−201.63}{11.62}=−2.59$ $z=\frac{172.5−201.63}{11.62}=−2.51$ Thus, using the table of z-scores, we can find that the corresponding probability is: $0.0062−0.0048=.0014$. b) We find z: $z=\frac{171.5−201.63}{11.62}=−2.59$ Thus, using the table of z-scores, we can find that the corresponding probability is $0.0062$. c) Part b) is the better choice, because we do not want an option that is any more extreme than the one that we found. d) The evidence is strong, because the odds of getting $33$ percent is extremely low.