## Elementary Statistics (12th Edition)

p=0.2 $q=1-p=1-0.2=0.8$ $n⋅p=50⋅0.2=10≥5.$ $n⋅q=50⋅0.8=40≥5.$ Hence, the requirements are satisfied. mean: $\mu=n\cdotp=50\cdot0.2=10.$ standard deviation: $\sigma=\sqrt{n\cdot p\cdot q}=\sqrt{50\cdot0.2\cdot0.8}=2.83.$ 3.5 is the first value more than 3, hence: $z=\frac{value-mean}{standard \ deviation}=\frac{3.5-10}{2.83}=-2.3.$ By using the table, the probability belonging to z=-2.3: 0.0107, hence the probability: 0.0107. This probability is really close to 0, hence the evidence is strong.