## Elementary Statistics (12th Edition)

p=0.2 $q=1-p=1-0.2=0.8$ $n⋅p=1000⋅0.2=200≥5.$ $n⋅q=1000⋅0.8=800≥5.$ Hence, the requirements are satisfied. mean: $\mu=n\cdotp=1000\cdot0.2=200.$ standard deviation: $\sigma=\sqrt{n\cdot p\cdot q}=\sqrt{1000\cdot0.2\cdot0.8}=12.65.$ 170.5 is the first value more than 17%, hence: $z=\frac{value-mean}{standard \ deviation}=\frac{170.5-200}{12.65}=-2.33.$ By using the table, the probability belonging to z=-2.33: 0.0099, hence the probability: 0.0099. This probability is really close to 0, hence the evidence is strong.