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8. In a certain city, the number of power outages per year is a random variable, having a distribution with mean 11.6 and standard deviation 3.3. If this distribution is normal, what is the probability that: a) There will be at least 15 outages in any year? P(x >= 15) = 1- P(x = 15) = 1 – ø(1.03) = 1-0.8485= 0.1515 b) There will be between 12 and 15 outages in any year? P(12< X <15) = P(X = 15) – P(X = 12) = $(15-11.6)-$(12-11.6) 3.3 = 0.8485-0.5478 = 0.3007 3.3

8. In a certain city, the number of power outages per year is a random variable, having a distribution with mean 11.6 and standard deviation 3.3. If this distribution is normal, what is the probability that: a) There will be at least 15 outages in any year? P(x >= 15) = 1- P(x = 15) = 1 – ø(1.03) = 1-0.8485= 0.1515 b) There will be between 12 and 15 outages in any year? P(12< X <15) = P(X = 15) – P(X = 12) = $(15-11.6)-$(12-11.6) 3.3 = 0.8485-0.5478 = 0.3007 3.3

A First Course in Probability (10th Edition)
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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8. In a certain city, the number of power outages per year is a random variable, having a distribution with mean 11.6 and standard deviation 3.3. If this distribution is normal, what is the probability that: a) There will be at least 15 outages in any year? P(x >= 15)=1- P(x = 15) = 1− ¢(1.03) = 1-0.8485 = 0.1515 b) There will be between 12 and 15 outages in any year? P(12< X <15) = P(X = 15) – P(X=12) = $(¹5–11.6)-$(¹2-11.6, 3.3 = 0.8485-0.5478 = 0.3007 3.3
Transcribed Image Text:8. In a certain city, the number of power outages per year is a random variable, having a distribution with mean 11.6 and standard deviation 3.3. If this distribution is normal, what is the probability that: a) There will be at least 15 outages in any year? P(x >= 15)=1- P(x = 15) = 1− ¢(1.03) = 1-0.8485 = 0.1515 b) There will be between 12 and 15 outages in any year? P(12< X <15) = P(X = 15) – P(X=12) = $(¹5–11.6)-$(¹2-11.6, 3.3 = 0.8485-0.5478 = 0.3007 3.3
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