The score on a test have a mean of 100 and standard deviation of 15. If a personnel manager wishes to select from the top 75% of applicants who take the test, find the cut off score 1:30 4A ? l l 43Areas Under Normal...团:The entries in the table are the probabilities that a randomvariable having the standard normal distribution takes on avalue between 0 and z; they are equal to the area under thecurve shaded in the figure.AREAS UNDER THE STANDARD NORMAL CURVESecond Decimal Place in z.00.01.02.03.04.05.06.07.08.09.0.0000 .0040 .0080 .0120 .0160 .0199 .0239 .0279 .0319.0359.0.1.0398 .0438 .0478 .0517 .0557 .0596 .0636 .0675 .0714 .0753.1.2.0793 .0832 .0871 .0910 .0948 .0987.1026 .1064 .1103.1141.2.3.1179 .1217.1255 .1293.1331 .1368 .1406 .1443 .1480 .1517.3.4.1554 .1591 .1628 .1664 .1700 .1736 .1772 .1808 .1844 .1879.4.1915 .1950.1985 .2019 .2054 .2088 .2123 .2157 .2190 .2224.5.2549.5.6.2257 .2291 .2324.23572389.2422 .2454 .2486 .2517.6.7.2580 .2611 .2642 .2673.2704 .2734 .2764 .2794 .2823 .2852.7.8.2881 .2910 .2939 .2967 .2995 .3023 .3051 .3078 .3106 .3133.8.9.3159 .3186 3212 .3238 .3264 .3289 .3315 .3340 .3365 .3389.91.0.3485 .3508 .3531 .3554 .3577 .3599 .3621 1.0.3749 3770 3790 .3810 3830 1.1.3849 .3869 3888 .3907 3925 .3944 .3962 .3980 .3997 .4015 1.2.4032 .4049 4066 .4082 .4090 .4115 .4131 .4147 .4162 .4177 1.3.4192 4207 4222 .4236 .4251 .4265 .4279 .4292 .4306 .4319 1.4.3413 .3438 .34611.1.3643 .3665 .3686 .3708 37291.21.31.44441 1.5.4452 .4463 .4474 4484 .4495 .4505 .4515 .4525 .4535 .4545 1.6.4554 .4564 4573 .4582 .4591 .4599 .4608 .4616 .4625 .4633 1.7.4641 .4649 4656 .4664 .4671 .4678 .4686 .4693 .4699 .4706 1.8.4738 .4744 4750 .4756 4761 .4767 1.91.5.4332 .4345 4357 4370 .4382 .4394 .4406 .4418 .44291.61.71.81.9.4713 4719 .4726 4732.4817 2.0.4854 .4857 2.1.4887 .4890 2.2.4893 .4896 4898 .4901 .4904 .4906 .4909 .4991 .4913 .4916 2.3.4918 .4920 .4922 .4925 .4927 .4929 .4931 .4932 4934 .4936 2.42.0.4772.4778.4783 .4788.4793 .4798 4803 .4808 .48122.1.4821 .4826 .4830 .4834 .4838 .4842 .4846 .48502.2.4861 .4864 4868 .4871 .4875 .4878 .4881 .48842.32.42.5.4938 .4940 .4941 .4943 .4945 .4946 .4948 4949 .4951 .4952 2.5.4960 .4961 .4962 .4963 .4964 2.6.4965 .4966 4967 .4968 .4969 .4970 .4971 .4972 .4973 .4974 2.7.4981 2.8.4986 .4986 2.92.6.4953 .4955 .4956 .4957 .49592.72.8.4974 .4975 4976.4977 .4977 4978 4979 4979.49802.9.4981 .4982 .4982 .4983 .4984 .4984 .4985 .4985.4990 3.0.4992 .4992 4992 .4992 .4993 .4993 3.1.4994 .4994 4994 .4995 4995 .4995 3.2.4997 3.3.4998 3.43.0.4987 .4987 4987 4988 .4988.4989 4989 .4989.49903.1.4990 .4991 .4991 .49913.2.4993 .4993 4994 .49943.3.4995 .4995 .4995 4996.4996 .4996 49964996 49963.4.4997 .4997 .4997 .4997 .4997 .4997 .4997 .4997.4997.4998 3.5.4998 .4998 4999 .4999 .4999 .4999 .4999 .4999 .4999 .4999 3.6.4999 3.7.4999 3.83.5.4998 .4998 .4998 .4998 .4998.4998 .4998 4998 49983.63.7.4999 .4999 4999 4999.4999 .4999 4999 .49993.8.4999 .4999 4999 .4999 .4999 .4999 .4999 .4999 .49993.9.50003.9For z 2 3.90, the areas are .5000 to four decimal places.

From the given information we find the cut off score.

Answered: The score on a test have a mean of 100…

The score on a test have a mean of 100 and standard deviation of 15. If a personnel manager wishes to select from the top 75% of applicants who take the test, find the cut off score

MATLAB: An Introduction with Applications

6th Edition

ISBN:9781119256830

Author:Amos Gilat

Publisher:Amos Gilat

Chapter1: Starting With Matlab

Section: Chapter Questions

Problem 1P

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