Let X be a normally distributed random variable with mean 5 and variance 16. Find the following probabilities: 1. P₁ = P(X> 12. 28). 2. p₂ = P(-2.28 < X < 5). 3. P3 P(-2.28 < X < 1.36). = 4. P4 = P(8.64 ≤ X ≤ 12. 28). (P₁, P2, P3, P4) = 0.0344,0.4656,0.1470,0.1470
Let X be a normally distributed random variable with mean 5 and variance 16. Find the following probabilities: 1. P₁ = P(X> 12. 28). 2. p₂ = P(-2.28 < X < 5). 3. P3 P(-2.28 < X < 1.36). = 4. P4 = P(8.64 ≤ X ≤ 12. 28). (P₁, P2, P3, P4) = 0.0344,0.4656,0.1470,0.1470
A First Course in Probability (10th Edition)
10th Edition
ISBN:9780134753119
Author:Sheldon Ross
Publisher:Sheldon Ross
Chapter1: Combinatorial Analysis
Section: Chapter Questions
Problem 1.1P: a. How many different 7-place license plates are possible if the first 2 places are for letters and...
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How were the answers of 0.0344, 0.4656, 0.1470, and 0.1470 attained?

Transcribed Image Text:Let X be a normally distributed random variable with mean 5 and variance 16. Find the following probabilities:
1. P₁ =
P(X> 12. 28).
2. P₂ = P(-2.28 < X < 5).
3. P3 = P(-2.28 < X < 1.36).
4. P4 = P(8.64 < X < 12. 28).
(P₁, P2, P3, P4) = 0.0344,0.4656,0.1470,0.1470
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