## Elementary Statistics (12th Edition)

a) $0.552$ b) $0.9994$ c) Part a
a. We use the z-scores to find: $z=\frac{140−165}{45.6}=−0.5482$ $z=\frac{211−165}{45.6}=1.01$ Hence, using the table of z-scores, we find that this corresponds to a probability of $0.8438−0.2912=0.552$. b) We use the z-score to find: $z=\frac{140−165}{45.6/\sqrt{36}}=−3.29$ $z=\frac{211−165}{45.6/\sqrt{36}}=6.06$ HEnce, using the table of z-scores, we find that this corresponds to a probability of $0.9999−0.0005=0.9994$ c) Part a) is the most useful. After all, there is only one pilot, so it mainly matters how likely an individual woman is this weight.