Some students paid a private tutor to help them improve their results on a certain mathematical test. These students had a mean change in score of + 18 points, with a standard deviation of 68 points. In a random sample of 100 students who pay a private tutor to help them improve their results, what is the likelihood that thechange in the sample mean score is less than 10 points?Click the icon to view the table of normal curve areas.Normal Curve AreasP(x< 10) = (Round to three decimal places as needed.)0..00.01.02.03.04.05.06.07.08.09.0000.0398.0793.1179.1554.1915.0.1.0040.0438.0080.0478.08711255.0160.0557.0948.1331.17002054.0120.0517.0199.059609871368.17362088.0239.06361026.1406177221230279.06751064144318082157.03190714.11031480184421900359.07531141151718792224.2.3.0832.1217.15911950.0910129316641628.5.19852019.2257.2580.2881.3159341322912611.6.7.8.91.023242642293932123461235726732967.323824222734238927042995.326424542764305133152486279430783340.3577251728232910.31863438302332893531310633653599254928523133338936213485350835543790368638884066.4222.4357137083907.4082.423643701.1.37293749.3643.3849.4032.4192.4332.36653869.4049.4207.43453770396241314279440638103997.416243064429383040154177431944411.21.3392540993944.411539804147.4292426543941.442511.543824418.4452.455446414463.45644649471947784474.457346564726.478344844582466447324788449545914671473847934505459946784744479845154608468647504803454546334706476748171.61.7453546251.81.92.045254616469347564808.469947614812.4713.4772.48304868489849224941.484648814909.49314948485448874913493448574890.48264864.4896.4920494048344871490149254943.483848754904492749454842487849064929494648212.12.22.34850488448614893.491849384911493249494916493649522.42.549514956.4967497649594969.49774984.49884960497049784984.4989496249724979.4985.498949634973.4980.4986.49904964497449814986499049612.62.72.82.93.0.4953.4965.4974.4981498749554966.4975.49824987495749684977498349884971.497949854989.49824987.00.01.02.03.04.05.06.07.08.09Clear allCheck answer

Given that. X~N( μ , ?^2 ) μ=18 , ?=68 , n=100 Z-score =( x - μ )/?

Answered: change in the sample mean score is less…

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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Use a table of cumulative areas under the normal curve to find the z-score that corresponds to the given cumulative area. If the area is not in the table, use the entry closest to the area. If the area is halfway between two entries, use the z-score halfway between the corresponding z-scores. If convenient, use technology to find the z-score. 0.053 Click to view page 1 of the table. Click to view page 2 of the table. The cumulative area corresponds to the z-score of. (Round to three decimal places as needed.) en 000 MacBook Air 2 8 DII DD B F10 F11 Next F12
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Use a table of cumulative areas under the normal curve to find the z-score that corresponds to the given cumulative area. If the area is not in the table, use the entry closest to the area. If the area is halfway between two entries, use the z-score halfway between the corresponding z-scores. If convenient, use technology to find the z-score. 0.05 Click to view page 1 of the table. Click to view page 2 of the table. The cumulative area corresponds to the z-score of (Round to three decimal places as needed.) Standard Normal Table (Page 1) Z 0.09 0.08 0.07 0.06 0.05 0.04 0.03 0.02 0.01 0.00 -3.4 0.0002 0.0003 0.0003 0.0003 0.0003 0.0003 0.0003 0.0003 0.0003 0.0003 -3.3 0.0003 0.0004 0.0004 0.0004 0.0004 0.0004 0.0004 0.0005 0.0005 0.0005 - 3.2 0.0005 0.0005 0.0005 0.0006 0.0006 0.0006 0.0006 0.0006 0.0007 0.0007 - 3.1 0.0007 0.0007 0.0008 0.0008 0.0008 0.0008 0.0009 0.0009 0.0009 0.0010 -3.0 | 0.0010 0.0010 0.0011 0.0011 0.0011 0.0012 0.0012 0.0013 0.0013 0.0013 -2.9 0.0014 0.0014 0.0015…
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change in the sample mean score is less than 10 points? Click the icon to view the table of normal curve areas. Normal Curve Areas P(x< 10)-O(Round to three decimal places as needed)