Chapter 6 - The Normal Distribution - 6-2 Applications of the Normal Distribution - Exercises 6-2 - Page 339: 21

$0.006$ $821$

We have $\mu=949, \sigma=100$ What is the probability that a randomly selected PC costs more than 1200? $z1=\frac{1200-949}{100}=2.51, P(z>z1)=1-P(z1)=0.9940=0.006$ The least expensive 10% of personal computers cost less than what amount? Use Table E, $P(z2)=0.1, z2=-1.281$ $z2=\frac{X2-949}{100}=-1.281, X2=920.9\approx821$