1. Using the normal probability tables, calculate the areas under the standardnormal curve for the following z values:a. Between Z= 0.0 and Z = 1.2b. Between Z= 0.0 and Z= -0.9c. Between Z= 0.0 and Z= 1.45Between Z= 0.0 and Z=-1.44Between Z= 0.3 and Z= 1.56d.e.f. Between Z= -1.71 and Z= -2.03g Between Z=-1.72 and Z= 2.53Between Z= -0.02 and Z = 3.54Greater than Z = 2.50h.i.j.Greater than Z= -0.60k.Less than Z= -1.22L Less than Z= 1.66

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Answered: 1. Using the normal probability tables,…

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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If z is a standard normal variable, then P ( z > 2.17 ) is:
Construct a plot on normal probability paper Measurement results of the diameter of a shaft turned on a CNC lathe Specifications: 0.7495 – 0.7505 1 2 3 4 5 6 7 8 9 10 1 0.7501 0.7501 0.7503 0.7501 0.7502 0.7497 0.7497 0.7500 0.7502 0.7496 2 0.7501 0.7500 0.7498 0.7502 0.7496 0.7504 0.7501 0.7502 0.7502 0.7501 3 0.7500 0.7499 0.7502 0.7503 0.7500 0.7503 0.7503 0.7500 0.7500 0.7499 4 0.7500 0.7502 0.7503 0.7501 0.7498 0.7500 0.7502 0.7503 0.7498 0.7501 5 0.7501 0.7499 0.7501 0.7501 0.7499 0.7500 0.7498 0.7501 0.7501 0.7500 Draw vertical lines on the graph to represent the LSL and USL. Draw horizontal lines to represent three standard deviations. Estimate of % nonconforming above USL = Estimate of % nonconforming below LSL = Total estimated % nonconforming beyond the specification limits =
Suppose the random variable Z follows a standard normal distribution. Then the value of P(Z<-3.39) is O a. 1.000 O b. 0.0003 O c. 0.0000 O d. 0.9997 SE
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For a standard normal distribution, if we have P(z < c) = 0.9663, find c.
APEAS of a Standard Normal Distribution (b) Tuble of Areas to the Left of z When z Is Pusitive .00 .01 .02 .03 .04 .05 .06 .07 ,08 .09 (a) Table of Areas to the Lefi of z When z Is Negative 0,0 5000 .5040 5080 .5120 5160 5199 .5239 5279 .5319 .5359 .00 .01 03 -3.4 .0003 .0003 .0003 .0003 .0003 .0003 .0003 .0003 .0003 02 .04 05 06 .07 08 09 0.1 5398 .5438 5478 .5517 5557 5596 .5636 5675 .5714 .5753 .0002 0.2 .5793 .5832.5871 5910 .5948 5987 .6026 ,6064 6103 .6141 -3.3 .0005 .0005 .0005 .0004 .0004 .0004 .0004 .0004 .0004 s the areu to the .0003 0.3 .6179 ,6217 ,6255 .6293 .6331 .6368 .6406 6443 .6480 .6517 -3.2 .0007 .0007 .0006 .0006 .0006 .0006 .0006 .0005 .0005 .0005 0,4 .6554 .6591 .6628 .6664 .6700 .6736 .6772 .6808 .6844 .6879 -3.1 ,0010 ,0009 0009 .0009 .0008 .0008 .0008 ,0008 .0007 „6915 6950 6985 .7019 .7054 .7088 .7123 .7157 7190 .7224 0.5 .0007 -3.0 .0013 ,0013. .0013 .0012 .0012 0011 .0011 .0011 .0010 .0010 0.6 .7257 7291 .7324 .7357 .7389 .7422 .7454 '.7486 .7517 .7549 -2.9…
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1. Using the normal probability tables, calculate the areas under the standard normal curve for the following z values: a. Between Z= 0.0 and Z = 1.2 b. Between Z= 0.0 and Z= -0.9 c. Between Z= 0.0 and Z= 1.45