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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Transcribed Image Text:Tutorial Exercise
Given 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 -1
0
mean, "
z
-2.0
-1.9
1
This 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 = -1.95
2
3
0.00
0.01 0.02 0.03 0.04 0.05 0.06 0.07 0.08
0.09
0.0228 0.0222 0.0217 0.0212 0.0207 0.0202 0.0197 0.0192 0.0188 0.0183
0.0287 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 0.0294
0.00
0.6179
Step 2
To 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 to
the 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) = 0.6528
0.02
0.6255
0.04
0.05
0.06
0.07
0.08
0.03
0.6293
0.6331 0.6368 0.6406 0.6443 0.6480
0.3
0.01
0.09
0.6217
0.6517
0.6554 0.6591 0.6628 0.6664 0.6700 0.6736 0.6772 0.6808 0.6844 0.6879
0.5 0.6915 0.6950 0.6985 0.7019 0.7054 0.7088 0.7123 0.7157 0.7190 0.7224
0.4
-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
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