Answer
a) $P_1=50.472$
$P_{99}=104.528$
b) .999
c) Part a
Work Step by Step
a) Using the table of positive z-values, we see that the 99th percentile corresponds to a value of z of $\pm 2.33$. Thus, we find:
$P_1=77.5+(11.6)(-2.33)=50.472$
$P_{99}=77.5+(11.6)(2.33)=104.528$
b) We find the two z-scores:
$z=\frac{70-77.5}{11.6/\sqrt{25}}=-3.23$
$z=\frac{85-77.5}{11.6/\sqrt{25}}=3.23$
Thus, we find:
$p=.9994-.0004=.999$
c) The results from part a) should be used, for this is just a cutoff for further testing. Thus, the parameters from part b) are far too rare.
