stThe population mean and standard deviation are given below. Find the required probability and determine whether thegiven sample mean would be considered unusual.For a sample of n = 66, find the probability of a sample mean being less than 19.6 if μ = 20 and 1.16.Z0.00.10.20.3Standard Normal Table (Page 2)1.31.41.5Click the icon to view page 1 of the standard normal table.Click the icon to view page 2 of the standard normal table.For a sample of n = 66, the probability of a sample mean being less than 19.6 if μ = 20 and o=1.16 is.(Round to four decimal places as needed.)Arca0zz.00.01 .02.5000 5040 5080.5398 5438 5478.5793 5832 587175497794 .782378520.8.8078 810681330.97881 .7910 .7939 79678159 8186 8212 82388413 8438 8461 84858340 .8365838986211.08554 .8577 85991.11.2911592369265.922292519345 9357 .9370 93829406 9418 9429944193949505.9515 .9525 .9535.9545.03.04.05.06 .07.08.09.5120 5160 5199 5239 5279 .5319 5359.5517 5557 5596 .5636 5675 5714 .5753.5910 5948 5987 6026 6064 6103 61416179 6217 6255 .6293 6331 6368 6406 6443 6480 65170.4 6554 6591 6628 .6664 6700 6736 6772 6808 6844 68790.5 69156950 6985 7019 .7054 7088 7123.7157 719072240.6 .7257 7291 7324.7357 .7389 .7422 7454 .7486 .75170.7 .7580 7611 7642 7673 .7704 7734 7764.7995 802380518264 8289 83158508 8531.8643 8665 8686 8708 8729 8749 8770 8790 8810 88308849 8869 8888 8907 8925 .8944 8962 8980 8997 .90159032 9049 .9066 .9082 9099.9131 9147 9162 .91779192 9207.9279 .9292 .9306 .931993321.6 9452 9463 9474 .9484 .94951.7 .9554 .9564 9573 .9582 9591 .95991.8 9641 9649 .9656 .9664 9671 .9678 96861.9 9713 9719 9726 .9732 9738 .9744 .97502.0 9772 .9778 9783 9788 9793 .9798 98032.1 .9821 9826 .9830 9834 9838 .9842 .9846 .9850 .9854.9861 .9864 .9868 9871 9875 .9878 .9881 .9884 98879893 9896 .9898 .9901 9904 .9906 .9909 9911 .99139918 9920 .9922 .9925 9927 .9929 .9931 .9932 9934.9938 9940 .9941 .9943 9945 .9946 .9948 .9949 .99519953 .9955 .9956 .9957 .9959 9960 9961 .9962 .9963.9965 9966 9967 .9968 .9969 .9970 9971 .9972 .9973.9974 9975 .9976 9977 .9977 9978 9979 .9979 .9980.9981 .9982 .9982 9983 .9984.9984.9987 .9987 9987 9988 .9988 99899990 .9991 .9991 9991 .9992 9992.9993 .9993 .9994 9994 .9994.9995 .9995 .9995 9996 9996 .99969997 .9997 .9997 .9997 .9997 9997.00.02.03.04 .05.9608 .9616 .9625 .9633.9693 9699.9756 .97619706.9767.9808 .98122.22.32.42.52.62.72.82.93.03.13.23.33.4Z9985 .9985 .99869989 .9989 .99909992 .9992 .9993.9994 .9994 .9995 .9995.9996 .9996 .9996.9997 .9997 9997.06.08.01.07DPrintCIDone.9817.9857.9890991699369952996499749981.9986.9990.9993.9995.9997.9998.09- X AreaThe population mean and standard deviation are given below. Find the required probability and determine whether thegiven sample mean would be considered unusual.For a sample of n = 66, find the probability of a sample mean being less than 19.6 if μ = 20 and 1.16.zClick the icon to view page 1 of the standard normal table.Click the icon to view page 2 of the standard normal table.Standard Normal Table (Page 1)For a sample of n = 66, the probability of a sample mean being less than 19.6 if μ = 20 and o=1.16 is(Round to four decimal places as needed.)2D.09.08.07.02.01-3.4000200030003.0003-3.3 .0003 0004.0005 0005000700060009 .0009.0078 00800102 0104 01070129 0132 .0136 0139.0166 0170 .0174 01790212 0217 0222 02280281 02870268 0274.06 .05.04 .030003 .0003 0003 0003 00030004 .0004 0004 0004 0004-3.2 .0005 0005 0005 0006 0006 0006 .0006-3.1 0007 0007 0008 .0008 0008 0008 0009-3.0 0010 0010 0011 0011 0011 0012 0012 .0013 .0013-2.9 0014 0014 0015 .0015 0016 0016 0017 .0018 .0018-2.8 0019 0020 0021 0021 0022 0023 0023 0024 .0025-2.7 0026 0027 0028 .0029 0030 0031 .0032 0033 .0034-2.6 0036 0037 0038 0039 0040 0041 .0043 0044 0045-2.5 0048 0049 0051 .0052 .0054 0055.0057 .00590060-2.4 .0064 0066 00680069.00710073.0075-2.3 .0084 0087 0089 .0091 0094 0096 .0099-2.2 0110 0113 01160119 0122 0125-2.1 0143 0146 0150 0154 .0158 0162-2.0 0183 0188 0192 0197 0202 0207-1.9 0233 0239 0244 0250 0256 0262-1.8 0294 0301 0307 0314 0322 0329-1.7 .0367 .0375 0384 0392 0401 0409-1.6 0455 0465 0475 0485 0495 0505-1.5 0559 0571 0582 0594 0606-1.4 0681 0694 0708 0721 0735-1.3 0823 0838 0853 0869 0885-1.2 0985 1003 .1020 .1038 .1056 .1075 .1093 1112 1131 1151-1.1 1170 .1190 1210 12301251 1271 1292 1314 .1335 .1357-1.0 .1379 .1401 .1423 1446 .1469 .1492-0.9 .1611 .1635 .1660 .1685 1711 .1736 .1762-0.8 .1867 .1894 1922 .1949 .1977 2005 2033-0.7 2148 2177 2206 .2236 2266 2296 2327-0.6 2451 2483 2514 .2546 2578 2611 2643-0.5 2776 2810 28432877 2912 2946 2981-0.4 3121 3156 3192 .3228 3264 3300 3336-0.3 3483 3520 3557 3594 3632 3669 3707-0.2 3859 3897 3936 .3974-0.1 4247 4286 4325 4364-0.0 4641 4681 4721 4761z0427 .04360336 03440418.0516.06.30 0643 .0655.0351 0359044605480526 .0537.0668061807490901.0764 .0778 0793 08080918 0934 .0951 09681515 .1539 .1562 1587.1788 .1814 18412061 2090 21192358 2389 24202676 270927433015 .3050 3085344638214207460250003372 .34093745 .37834013 4052 4090 4129 41684404 4443 44834801 4840 488004 .034522 45624920 4960.02 .01.09.08.07.06.05.00PrintDone.00000300050007001000130019002600350047.00620082- X

GivenMean(μ)=20standard deviation(σ)=1.16sample size(n)=66

Answered: The population mean and standard…

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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The population mean and standard deviation are given below. Find the required probability and determine whether the given sample mean would be considered unusual. For a sample of n = 66, find the probability of a sample mean being less than 19.6 if μ=20 and a = 1.16.