## Elementary Statistics (12th Edition)

q=1-p=1-0.4=0.6 $n\cdot p=13\cdot 0.4=5.2\geq5.$ $n\cdot q=13\cdot 0.6=7.8\geq5.$ Hence, the requirements are satisfied. mean: $\mu=n\cdotp=13\cdot0.4=5.2.$ standard deviation: $\sigma=\sqrt{n\cdot p\cdot q}=\sqrt{13 \cdot0.4\cdot0.6}=\sqrt{3.12}=1.77.$ 2.5 is the first one lower than 3, hence: $z=\frac{value-mean}{standard \ deviation}=\frac{2.5-5.2}{1.77}=-1.53.$ By using the table, the probability belonging to z=-1.53: 0.063.