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y 1 2 4 8 16 p(y) 0.05 0.10 0.35 0.40 0.10

y 1 2 4 8 16 p(y) 0.05 0.10 0.35 0.40 0.10

MATLAB: An Introduction with Applications
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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An individual who has automobile insurance from a certain company is randomly selected. Let Y be the number of moving violations for which the individual was cited during the last 3 years. The pmf of Y is the following. Find the following.

(a) E[Y ]

(b) V[Y]

 
The image displays a table representing a probability distribution as follows: - The first row indicates the values of the random variable \( y \), which are: 1, 2, 4, 8, and 16. - The second row lists the corresponding probabilities \( p(y) \) for each value of \( y \): - \( p(y=1) = 0.05 \) - \( p(y=2) = 0.10 \) - \( p(y=4) = 0.35 \) - \( p(y=8) = 0.40 \) - \( p(y=16) = 0.10 \) The table illustrates how the probability is distributed across different values of a random variable. Each probability value represents the likelihood of the corresponding value of \( y \) occurring.
Transcribed Image Text:The image displays a table representing a probability distribution as follows: - The first row indicates the values of the random variable \( y \), which are: 1, 2, 4, 8, and 16. - The second row lists the corresponding probabilities \( p(y) \) for each value of \( y \): - \( p(y=1) = 0.05 \) - \( p(y=2) = 0.10 \) - \( p(y=4) = 0.35 \) - \( p(y=8) = 0.40 \) - \( p(y=16) = 0.10 \) The table illustrates how the probability is distributed across different values of a random variable. Each probability value represents the likelihood of the corresponding value of \( y \) occurring.
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