MAT_230_Exam_Two - MAT 230 EXAM TWO This document is proprietary to Southern New Hampshire University It and the problems within may not be posted

MAT_230_Exam_Two - MAT 230 EXAM TWO This document is...

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MAT 230 EXAM TWO This document is proprietary to Southern New Hampshire University. It and the problems within may not be posted on any non-SNHU website. Keith Ellison 1
Directions: Type your solutions into this document and be sure to show all steps for arriving at your solution. Just giving a final number may not receive full credit. Problem 1 This question has 2 parts. Part 1: Suppose that F and X are events from a common sample space with P ( F ) 6 = 0 and P ( X ) 6 = 0. (a) Prove that P ( X ) = P ( X | F ) P ( F ) + P ( X | ¯ F ) P ( ¯ F ). Hint: Explain why P ( X | F ) P ( F ) = P ( X F ) is another way of writing the definition of conditional probability, and then use that with the logic from the proof of Theorem 4.1.1. From the conditional probability, P(X—F)=P(XF)/P(F) P(X—F)P(F)=P(XF) P(X—F)=P(XF’)/P(F) P(X—F)P(F)=P(XF’) P(X)=P(X) P(X)=P(X(FF)) =P((XF)(XF)) the probability is independent, P((XF)(XF))=P(XF)+P(XF) =P(X)P(F)+P(X)P(F) P(X)=P(XF)+P(XF) (b) Explain why P ( F | X ) = P ( X | F ) P ( F ) /P ( X ) is another way of stating Theorem 4.2.1 Bayes’ Theorem. From the conditional density function, P(F—X)=P(XF)/P(X) P(X—F)P(F)=P(XF) Used in the conditional equation, P(F—X)=P(XF)/P(X) P(F—X) =P(X—F)P(F)/P(X) Part 2: A website reports that 70% of its users are from outside a certain country. Out of their users from outside the country, 60% of them log on every day. Out of their users from inside the country, 80% of them log on every day. (a) What percent of all users log on every day? Hint: Use the equation from Part 1 (a). Use ’X’ as the total percentage of users that log on everyday F is the percentage of users from outside the country who log on everyday F F be the percentage of users from inside the country logging on everyday. The probability of the total percentage of users logging on everyday; P(X)=100x Probability of users log from outside the country; P(F)=70x Probability of users log from inside the country; P( F )=30x Probability of users log from outside the country everyday; P(X—F)P(F)=60/100(70x)=42x Probability of users log from inside the country everyday, P(X— F )P( F )=80/100(30x)=24x Total probability of users logging on everyday is given by using the proof of Part 1(a), P(X)=P(X—F)P(F)+P(X— F )P( F ) P(X)=42x+24x=66x P(X)=66x.

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