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08-22-2011 , 03:00 PM
So i guess this is the forum where i should post this . If any1 wants to help me please explain how to solve this ( i got a exam in like a week and i would like to know how to solve this )

A multiple choice examination has 25 questions, each with 5 possible answers,
only one of which is correct. Suppose that one of the students who takes the examination answers each of the question with an independent random guess.
a) Write the random variable representing the number of correct answers.
b) Which is the expected number of correct answers?

i will post more after
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08-22-2011 , 03:49 PM
Quote:
Originally Posted by Dannyz0r
So i guess this is the forum where i should post this . If any1 wants to help me please explain how to solve this ( i got a exam in like a week and i would like to know how to solve this )

A multiple choice examination has 25 questions, each with 5 possible answers,
only one of which is correct. Suppose that one of the students who takes the examination answers each of the question with an independent random guess.
a) Write the random variable representing the number of correct answers.
b) Which is the expected number of correct answers?

i will post more after
http://forumserver.twoplustwo.com/sh...35&postcount=3
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08-22-2011 , 03:52 PM
a) B(25, 1/5)
b) 5
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08-23-2011 , 07:08 AM
Ok thanks for the help on that one .
The next one i know how to solve but i don't know to write the formula for the random variable


The telephone lines serving an information office are all busy about 60% of the
times. Which is the expected number of tries for succeeding a reservation?

So i'm thinking like this :
0.6+0.6*0.6+0.6*0.6*0.6=0.6+0.36+0.216 >>> 1

so this means in 3 tries i should get a reservation but how does the random variable look like ?
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08-23-2011 , 07:43 AM
Quote:
Originally Posted by Dannyz0r
Ok thanks for the help on that one .
The next one i know how to solve but i don't know to write the formula for the random variable


The telephone lines serving an information office are all busy about 60% of the
times. Which is the expected number of tries for succeeding a reservation?

So i'm thinking like this :
0.6+0.6*0.6+0.6*0.6*0.6=0.6+0.36+0.216 >>> 1

so this means in 3 tries i should get a reservation but how does the random variable look like ?
This is a geometric distribution with probability of success 0.4, and the expected value is 2.5.

You need to go back to your course teacher and ask them to go over this stuff with you again. Not trying to be rude but it's clear from the two questions you asked that you have no idea how to solve these, or recognise standard probability distributions.
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08-23-2011 , 08:05 AM
ok so i was the wrong way
thought p=0.6 not 0.4
this means ev = 1/p=10/4 right ?
thanks for the help , im understand / have idea but i didnt realize the value of p
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08-23-2011 , 08:19 AM
Yes, p is the probability of a 'success'. Getting a busy line is a failure and happens 60% of the time as per the question, so getting through occurs the other 40% of the time.
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