Chapter 6 - The Normal Distribution - 6-2 Applications of the Normal Distribution - Exercises 6-2 - Page 338: 7

a. $0.3547$ b. $0.8391$

We have $\mu=85900,\sigma=11000$ a. The professor makes more than 90,000. $z1=\frac{90000-85900}{11000}=0.3727,P(z>z1)=1-P(z1)=1-0.6453=0.3547$ b. The professor makes more than 75,000 $z2=\frac{75000-85900}{11000}=-0.9909,P(z>z2)=1-P(z2)=1-0.1609=0.8391$