Chapter 6 - The Normal Distribution - 6-2 Applications of the Normal Distribution - Exercises 6-2 - Page 338: 8

a. $0.0747$ b. $0.2377$

We have $\mu=12837,\sigma=1500$ a. The student makes more than 15,000. $z1=\frac{15000-12837}{1500}=1.442, P(z>z1)=1-P(z1)=1-0.9253=0.0747$ b. The student makes between 13,000 and 14,000. $z2=\frac{13000-12837}{1500}=0.1087$, $z3=\frac{14000-12837}{1500}=0.7753$ $P(z2