Question

1. (5 points) Suppose Z is a random variable that follows the standard normal distribution. a) Find P(Z > 0.45). b) Find P(0.7 SZ 1.6). c) Find 20.09. d) Find the Z-score for having area 0.18 to its left under the standard normal curve. e) Find the value of z such that P(-2SZS2) -0.5.

Answer 1

Here, we are given Z~N(0,1). 9. P[2>,0.45] = i-p[220.45] = 1-800.45) = 1-0.6736 [obtained from normal table] = 0.3264 -0.633 b. P(0.75731.6) = Q(1.6) - $10.7) = 0.9452-0.3122 [obtained from normal table c. To find 20.oo we take the a-score table. Next we find the value 0.09 in the table and see it occurs concepending to -13 and 0.04. flence, we get the value of 7o.ug ned by combining -1.3 and.o.ou which is = 1.34 d. To find the z-score with area 0.18 to ite left under the standard normal curve we first soldragtes with giren area since, we have to locate the value with mean o. : area = 0.18-0.5=-0.32. We can split -0.32 in -0.3 and 0.02. Thus, coresponding 7-score = 0.3745.