Table 22.3 cumulative Probablltles of the Standard Normal Distrilbution Function
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Table 22.3 cumulative Probablltles of the Standard Normal Distrilbution Function .00 01 0203 04.05.0« .07 0809 0000 0040 .0080 0120 .0160 .0199 .0239 .0279 .0319 .0359 .0398 .0438 .0478 ,0517 .0557 .0596 .0636 .0675 .0714 .0753 0793 0832 .0871 .0910 .0948 .0987 .1026 .1064 .1103 .114 1179 .1217 .1255 .1293 .133 1368 .1406 .1443 .1480 .1517 4 1554 .1591 .1628 .1664 .1700 .1736 .1772 .1808 .1844 .1879 5.1915 .1950 .1985 .2019 2054 .2088 2123 .2157 .2190 .2224 6 2257 .2291 .2324 2357 .2389 2422 .2454 2486 .2517 .2549 .7 .2s80 ·2611 .2642.2673 .2704 ,2734 2764 .2794 .2823 .2852 8 28 2910 2939 2967 .2995 .3023 .3051 .3078 3106 .3133 9 3159 .3186 .3212 .3238 .3264 3289 .3315 .3340 .3365 .3389 1.0 3413 3438 3461 .3485 3508 .3531 .3554 3577 .3599 3621 I.1.3643 .3665 .3686 .3708 .3729 3749 .3770 .3790 .3810 .3830 3849 3869 .3888 3907 .3925 .3944 .3962 .3980 .3997 4015 4032 4049 4066 4082 4099 4115 413 4147 4162 4177 1.4 .4192 .4207 .4222 .4236 .4251 .4265 .4279 .4292 .4306 .43 19 1,5 .4332 .4345 .4357 .4370 ,4382 .4394 .4406 .4418 .4429 .4441 16 4452 4463 4474 4484 4495 4505 4515 4525 4535 4545 1.7.4554 4564 4573 4582 459 4599 4608 .4616 4625 4633 1.8464 4649 4656 4664 467 4678 4686 .4693 4699 4706 1.94713 4719 4726 4732 4738 4744 4750 4756 4761 4767 2.0 4773 4778 4783 4788 .4793 .4798 4803 4808 4812 4817 2.14821 4826 4830 4834 48384842 4846 4850 4854 4857 2.24861 4866 4868 4871 4875 4878 4881 4884 4887 4890 2.3 4893 4896 4898 .4901 4904 4906 4909 4911 4913 4916 2.4 .4918 .4920 .4922 .4925 .4927 .4929 .4931 .4932 .4934 .4936 2.54938 4940 4941 4943 4945 4946 4948 4949 4951 4952 4953 4955 4956 4957 .4959 4960 49614962 4963 4964 2.74965 4966 4967 .4968 4969 A970 4971 4972 4973 4974 2.8 4974 4975 4976 4977 4977 4978 4979 4979 4980 4981 4981 4982 4982 4982 4984 4984 4985 4985 4986 .4986 3.04987 A987 4987 4988 4988 4989 49894989 4990 4990 .2 .3 1.3 2.6 2.9 N(c) ropresants aroas under ths standard normal dlstribution function. Suppose that di 24.The table Implies a cmulative probablity of 5000 + .0948594BI s equal to 2452, we must estimate the probablity by interpolating becween N(.25) and N(2)Explanation / Answer
WE need to find value of N(d) where d1= 0.2452
as per table at 0.24 d1 value is 0.0948
and at 0.25 d1 value is 0.0987
so with increase in d1 = 0.25 - 0.24 = 0.01 increase in value of d1 = 0.0987 - 0.0948 = 0.0039
but we need to find value d1 at increase of = 0.2452-0.24 = 0.0052
so we will do some maths here
increase in value / increase in d1 = 0.0039 / 0.01
=0.39 this show the increase in value with incraese of 1.0 in d1
but we need to find the incraese in value with increase of 0.0052, so we multiply this with 0.39
=0.39 * 0.0052
=0.002028
so value of d1 at 0.2452 = 0.0948 + 0.002028
=0.096828
value of N(d) = 0.5 + 0.096828
=0.596828.
so what we done here is
( Value at greater d1 – value at lower d1 / Greater d1 – lower d1) x d1 we need to find - lower d1
0.0987- 0.0948 / 0.25-0.24) * 0.2452-0.24
=0.002028
add this in value of 0.24 i.e =0.0948 + 0.002028 = 0.096828
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