## Elementary Statistics (12th Edition)

a)$z=\frac{value-mean}{standard \ deviation}=\frac{100.6-98.2}{0.62}=3.87$. By using the table the corresponding probability: 1-0.9999=0.0001, which is very close to 0, hence the cutoff is appropriate. b)By using the table, the z-score corresponding to 1-0.05=0.95: z=1.645. Hence the corresponding value:$mean+z⋅standard \ deviation=98.2−1.645⋅0.62=99.2.$