Let z be a standard normal random variable with mean ? = 0 and standard deviation ? = 1. Use Table 3 in Appendix I to find the probability. (Round your answer to four decimal places.)P(−2.58 < z < 2.58) = Table 3 (continued)00.01.02.03.04.05.060708090.05000504050805120.516051995239.5279531953590.15398543854785517.555755965636క675571457530.25793583258715910594859876026.60646103648061410.36179621762556293.633163686406.644365170.46554659166286664.67006736.6772.6808684468790.56915695069857019.7054708871237157719072240.67257729173247357738974227454.7486751775490.775807611764276737704773477647794782378520.87881791079397967.7995802380518078810681330.98159818682128238826482898315.8340836583891.084138643843884618485850885318554.8577859986211.1866586868708872987498770.8790881088301.28849886988888907892589448962.8980899790151.39032904990669082.9099911591319147916291771.49192920792229236.9251.92659279.9292930693199382.9495940695151.5933294529345935793709394.9418942994411.694639474948495059525953595451.79554956495739582.9591959996089616962596331.89641964996569664.9671.967896869693969997061.99713971997269732.9738974497509756976197679798.98422.09772977897839788979398039812.9808.985098172.19821982698309834.98389846985498572.29861986498689871.9875987898819884988798902.39893989698989901.990499069909.9911991399162.49918992099229925.992799299931.9932993499362.599389940994199439945.994699489949995199522.69953995599569957.9959996099619962996399642.7996599669967.9968.9969997099719972997399742.89974997599769977.9977.9978.99799979998099812.99981998299829983.9984998499859985998699863.09987998799879988.998899899989.99899992999099903.19990999199919991.9992.9992.9992999399933.29993999399949994.9994.9994.9994.9995999599953.39995999599959996.9996.9996.99969996999699973.49997999799979997.9997.9997.9997.999799979998 Table 3 Areas under the Normal Curve0001.0203.04.05.06.07.08.09-3.40003.0003.00030003.0003.00030003.0003.0003.0002-33-3.2.0005.0005.00050004.0004.00040004.0004.0004.0003.0006.00080007.0007.00060006.00060006.0005.0005.0005-3.1-3.00010.0009.00090009.00080008.0008.0007.00070013.0013.0013.0012.0012.00110011.0011.00100010-2.9-2.8-2.7.0019.0018.00170017.0016.00160015.0015.001400140026.0025.0024.0023.0023.00220021.0021.0020.00190035.0034.00330032.0031.00300029.0028.00270026-2.60047.0045.00440043.0041.00400039.0038.0037.0036-250062.0060.00590057.0055.00540052.0051.00490048-240082.0080.0078.0075.0073.00710069.0068.00660064.00940122-2.30107.010401020099.00960091.0089.0087.008401100143-220139.013601320129.01250119.0116.0113.0146-2.10179.0174.01700166.0162.01580154.0150-2.00228.0222.0217.0212.0207.02020197.019201880183-1.90287.0281.02740268.0262.02560250.0244.0239.0233-1.8.0359.0351.03440336.0329.03220314.0307.0301.0294-1.7.0446.0436.04270418.0409.04010392.0384.03750367-1.60548.0537.05260516.0505.04950485.0475.04650455-1.50668.0655.06430630.0618.0606.0594.058205710559-1.40808.0793.07780764.0749.07350722.0708.06940681-1.309680918.08850869.0853.0838.1003.0951.0934.09010823-1.2.1151.1131.1112.1093.1075.1056.1038.10200985-1.1.1357.1335.1314.1292.1271.1251.1230.1210.1190.1170-1.0.1587.1562.1539.1515.1492.1469.1446.1423.1401.1379-0.9.1841.1814.1788.1762.1736.1711.1685.1660.1635.1611-0.8-0.7-0.620902061203323272119.2005.1977.1949.1922.1894.18672420238923582296226625782236220621772148274327092676264326112546251424832451-0.53085305030152981.294629122877284328102776--0.434463821340933723336.330032643228319231563121-0.3378337453707.366936323594355735203483--0.24207416841294090405240133974393638973859-0.1-0.046024562A52244834443440443644325428642475000.49604920488048404801476147214681A641

Answered: Image /qna-images/answer/3bed3ac5-55a7-4ac1-a879-206a6d7cd531.jpg

Let z be a standard normal random variable with mean ? = 0 and standard deviation ? = 1. Use Table 3 in Appendix I to find the probability. (Round your answer to four decimal places.) P(−2.58 < z < 2.58) =

Let z be a standard normal random variable with mean ? = 0 and standard deviation ? = 1. Use Table 3 in Appendix I to find the probability. (Round your answer to four decimal places.) P(−2.58 < z < 2.58) =

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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