. [14 points)] An individual who has automobile insurance from a certain company
ID: 2921911 • Letter: #
Question
. [14 points)] An individual who has automobile insurance from a certain company is randomly selected. Let X-the number of moving violatioms for which the individual was cited during the last 3 years. The probability mass function of X is given below px) 050 0.20 015 0.10 0os (a) Calculate the probability of each of the following events. [2 points each] ) Exactly one moving violation (ii) At most one moving violation (ii) More than two moving violations (iv) Between 1 and 3 (inclusive of the endpoints) moving violations (b) Find the cumulative distribution function Fx) Be sure to write your answer in the appropriate way [6 points 4. [12 points] Refer to Problem 1 (a) Find the mean number of moving violations14 points (b) Find the variance among the values of X. [5 points] (c) If an individual has X moving violations their insurance premium will change by sSox-$20. Find the mean amount of change to their premium, ic. find E(SSox-$20). 3 points]Explanation / Answer
4)
a) E(X) = 0 * 0.5 + 1 * 0.2 + 3 * 0.15 + 4 * 0.1 + 5 * 0.05 = 1.3
b) E(X2) = 02 * 0.5 + 12 * 0.2 + 32 * 0.15 + 42 * 0.1 + 52 * 0.05 = 4.4
Var(X) = E(X2) - (E(X))2 = 4.4 - 1.32 = 2.71
c) E(50X - 20) = 50*E(X) - 20 = 50 * 1.3 - 20 = 45
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