Below is a graph of a normal distribution with mean u =-1 and standard deviation o =3. Tfrom this distribution that is greater than 2.0.40.30.2011Shade the corresponding region under the standard normal curve below.

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Answered: Below is a graph of a normal…

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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Find -z and z such that 55% of the standard normal curve lies between –z and z.
A normal distribution has a mean of μ = 100 with σ = 20. If one score is randomly selected from this distribution, what is the probability that the score will have a value between X = 100 and X = 130?
A normal distribution has a mean of µ = 100 with σ = 20. If one score is randomly selected from this distribution, what is the probability that the score will have a value between X = 90 and X = 110?
Suppose X is a non-standard normal variable, with a mean of 139, and std deviation of 15 (i.e. X~N(139,15)). You have an observation (z) on a standard normal scale (i.e. Z~N(0,1)). If z = 1.29, what is it on an X scale? How can I calculate this using excel & by hand?
Suppose X is a non-standard normal variable, with a mean of 118, and std deviation of 15 (i.e. X~N(118,15)). You have an observation (z) on a standard normal scale (i.e. Z~N(0,1)). If z = -0.03, what is it on an X scale?
Fawns between 1 and 5 months old have a body weight that is approximately normally distributed with mean ? = 27.1 kilograms and standard deviation ? = 4.0 kilograms. Let x be the weight of a fawn in kilograms.
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Find the area of the shaded region. The graph to the right depicts IQ scores of adults, and those scores are normally distributed with a mean of 100 and a standard deviation of 15. Click to view page 1 of th - × The area of the shaded r Standard Normal Table (Page 1) NEGATIVE z Scores Standard Normal (z) Distribution: Cumulative Area from the LEFT .00 .01 02 .03 .04 .05 .06 .07 .08 .09 -3.50 and lower .0001 -3.4 .0003 .0003 .0003 .0003 .0003 .0003 -3.3 .0005 .0005 .0005 .0004 .0004 .0004 .0003 0004 .0003 .0004 .0003 .0004 .0002 .0003 -3.2 .0007 .0007 .0006 0006 .0006 .0006 .0006 .0005 .0005 .0005 -3.1 .0010 .0009 .0009 .0009 .0008 .0008 .0008 .0008 0007 .0007 -3.0 .0013 .0013 0013 .0012 0012 .0011 .0011 .0011 0010 0010 -2.9 .0019 0018 0018 .0017 .0016 .0016 .0015 .0015 .0014 .0014 -2.8 .0026 .0025 .0024 .0023 .0023 .0022 .0021 .0021 .0020 .0019 -2.7 .0035 .0034 .0033 .0032 .0031 .0030 .0029 .0028 .0027 .0026 -2.6 .0047 .0045 0044 .0043 .0041 .0040 .0039 .0038 .0037 .0036 -2.5 .0062…
Suppose that the average life of a refrigerator before replacement is μ = 15 years with a (68%of data) range from 13 to 17 years. Let x = age at which a refrigerator is replaced. Assumethat x has a distribution that is approximately normal. Find a good approximation for thestandard deviation of x values.
Below is a graph of a normal distribution with mean u = -3 and standard deviation o =2. The shaded region represents theprobability of obtaining a value from this distribution that is between -4 and 0. Shade the corresponding region under the standard normal curve below.

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Below is a graph of a normal distribution with mean u=-1 and standard deviation o = 3. T from this distribution that is greater than 2. 0.4+ 0.3 0.2 0.1 -1 normal Curve below