#### Answer

a) $P_1=50.472 $
$P_{99}=104.528$
b) $0.999$
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 $z=±2.33$. Hence:
$P_1=77.5+(11.6)(−2.33)=50.472$
$P_{99}=77.5+(11.6)(2.33)=104.528$
b) We find the two z-scores:
$z=\frac{70−77.5}{11.6/\sqrt{25}}=−3.23$
$z=\frac{85−77.5}{11.6/\sqrt{25}}=3.23$
Hence:
$p=0.9994−0.0004=0.999 $
c) The results from part a) should be used, because this is just a cutoff for further testing. Hence, the parameters from part b) are too rare.