## Elementary Statistics (12th Edition)

a) $P_1=50.472$ $P_{99}=104.528$ b) $0.999$ c) Part a
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.