Suppose x is a normally distributed random variable with u = 33 and 5 = 5. Find a value xO of the random variable x. a. P(x2x0)=.5 b. P(XXo)=-10 Table of Normal Curve Areas d. P(X>X0)=.95 Q Click here to view a table of areas under the standardized normal curve. A r a. x0 = (Round to the nearest hundredth as needed.) a z b. x0 = (Round to the nearest hundredth as needed) I 0 01 02 03 04 05 06 o7 08 09 _ .0 .0000 .0040 .0000 .0120 .0160 .0199 .039 .0279 .0319 .0359 0- X0 - (ROUUd '0 the \"eareSt hundredth as needed) .1 .0391; .0430 .0470 .0517 .0557 .0596 .0636 .0675 .0714 .0753 .2 .0793 .0032 .0071 .0910 .0940 .0907 .1026 .1064 .1 103 .1141 .3 .1179 .1217 .1255 .1293 .1331 .1360 .1406 .1443 .1400 .1517 d. X0 = (Round t0 the nearest hundredth as needed.) _4 st.' _]591 _1628 .1664 _17m _173(, J77: 4303 JW _1379 .5 .1915 .1950 .1906 .2019 .2054 .2008 .2123 .2157 .2190 .2224 .6 2257 .2291 .2324 2357 .2309 .2422 .2454 .2406 .2517 .2549 .7 2500 .2611 .2642 2673 .2701 .2734 .2764 .2794 2023 2052 .0 .2001 .2910 .2939 .2967 .2995 .3023 .3051 .3070 .3106 .3133 .9 .3159 .3106 .3212 .3230 .3264 .3209 .3315 .3340 .3365 .3309 1.0 .3413 .3438 .3485 .3500 .3531 .3554 .3577 .3599 .3621 1.1 3643 .3665 $708 .3729 .3749 .3770 .3790 .3810 .3830 12 3049 .3069 3907 .395 .3944 .3962 .3900 .3997 .4015 1.3 .4032 .4049 .4082 AM .4l15 .4131 .4147 .4162 .4177 1.4 .4192 .4207 .4236 .4251 .4265 .4279 .4292 .4306 .4319 1.5 .4332 .4345 .4370 .4312 .4394 .4406 .4410 .4429 .4441 1.6 .4452 .4463 .4484 .4495 .4505 .4515 .4525 .4535 .4545 1.7 .4554 .4564 .4502 .4591 .4599 .4600 .4616 .4625 .4633 1.0 .4641 .4649 .4664 .4671 .4678 .4606 .4693 .4699 .4706 1.9 .4713 .4719 .4732 .4730 .4744 .4750 .4756 .4761 .4767 2.0 .4772 .4770 .4700 .4793 .4790 .4003 .4000 .4012 .4017 v 4 1