Answer
a) .0014
b) .0062
c) Part B
d) The evidence is strong.
Work Step by Step
a. We find:
$\mu=np=(611)(.33)=201.63$
$ \sigma=\sqrt{npq}=\sqrt{(611)(.33)(.67)}=11.62$
Thus, we find z:
$z=\frac{171.5-201.63}{11.62}=-2.51$
$z=\frac{172.5-201.63}{11.62}=-2.59$
Thus, using the table of z-scores, we find that this corresponds to a probability of $.0062-.0048=.0014$
b. We find z:
$z=\frac{171.5-201.63}{11.62}=-2.59$
Thus, using the table of z-scores, we find that this corresponds to a probability of $.0062$
c. Part b. is the better choice, for we do not want an option that is any more extreme than the one that we found.
d. The evidence is strong, for the odds of getting 33 percent is extremely low.
