Tutorial ExerciseGiven that z is a standard normal random variable, compute the following probabilities.(a) P(-1.95 sz s 0.46)(b) P(0.55 sz s 1.23)(c) P(-1.85 ≤zs -1.05)Step 1(a) P(-1.95 sz≤ 0.46)For z to be a standard normal variable means that the random variable z follows a normal distribution with mean 0 and standard deviation 1. The normal distribution is a bell-shaped curve that is symmetric about the mean. The graph of a standard normal curve is below.-3 -2 -10mean, "z-2.0-1.91This distribution is continuous, so probabilities will be for a range of values and represented by the area under the graph between the values.The desired probability is for the region between z = -1.95 and z = 0.46. Since z = -1.95 is to the left of z = 0.46, we can find this probability by subtracting the area under the curve to the left of z = -1.95230.000.01 0.02 0.03 0.04 0.05 0.06 0.07 0.080.090.0228 0.0222 0.0217 0.0212 0.0207 0.0202 0.0197 0.0192 0.0188 0.01830.0287 0.0281 0.0274 0.0268 0.0262 0.0256 0.0250 0.0244 0.0239 0.0233-1.8 0.0359 0.0351 0.0344 0.0336 0.0329 0.0322 0.0314 0.0307 0.0301 0.02940.000.6179Step 2To find P(-1.95 sz≤ 0.46), subtract the area to the left of z = -1.95 from the area to the left of z = 0.46. Tables can be used to find areas to the left of z values. Along the leftmost column are values of z precise to one decimal place. Trace along the necessary row until you get to the column for the needed hundredths place. The value where the row and column intersect is the area under the curve tothe left of that z value.Use the table excerpt above to find the area under the standard normal curve to the left of z = -1.95, P(z ≤ -1.95).P(z -1.95) = 0.65280.020.62550.040.050.060.070.080.030.62930.6331 0.6368 0.6406 0.6443 0.64800.30.010.090.62170.65170.6554 0.6591 0.6628 0.6664 0.6700 0.6736 0.6772 0.6808 0.6844 0.68790.5 0.6915 0.6950 0.6985 0.7019 0.7054 0.7088 0.7123 0.7157 0.7190 0.72240.4-1.95 from the area under the curve to the left of z= 0.46✔Use the table excerpt above to find the area under the standard normal curve to the left of z = 0.46, P(z ≤ 0.46).P(Z < 0.46) =0.46

Let Z be standard normal distribution

Tutorial Exercise Given that is a standard normal random variable, compute the following probabilities. (a) P-1.95 $260.44) (b) P(0.55 s2s 1.23) (c) P(-1.85 s2s-1.05) Step 1 (a) P(-1.95 20.46) For 2 to be a standard normal variable means that the random variable z follows a normal distribution with mean 0 and standard deviation 1. The normal distribution is a bell-shaped curve that is symmetric about the mean. The graph of a standard normal curve is below.

Tutorial Exercise Given that is a standard normal random variable, compute the following probabilities. (a) P-1.95 $260.44) (b) P(0.55 s2s 1.23) (c) P(-1.85 s2s-1.05) Step 1 (a) P(-1.95 20.46) For 2 to be a standard normal variable means that the random variable z follows a normal distribution with mean 0 and standard deviation 1. The normal distribution is a bell-shaped curve that is symmetric about the mean. The graph of a standard normal curve is below.

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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