Suppose Appendix Table A.3 contained ?(z) only for z ?0 Explain how you could still compute a. P( –1.72? Z ?–.55) b. P( –1.72? Z ? .55) Is it necessary to tabulate ?(z) for ?z ?negative? What property of the standard normal curve justifies your answer?

Solution 48E Step1 of 2: We have (z) only for z 0. We need to compute, a).P(-1.72Z-0.55) b). (-1.72Z0.55) and We need to check Is it necessary to tabulate (z) for z negative What property of the standard normal curve justifies your answer Step2 of 2: a). Consider, P(-1.72Z-0.55) = P(0.55Z1.72) = (1.72) - (0.55) = 0.9573 - 0.7088 = 0.2485 [therefore (1.72) and (0.55) is calculated by using standard normal table] (1.72) = 0.9573 In standard normal table we have to see row 1.7 under column 0.02. Similarly, (0.55) = 0.7088 In standard...