Random variable?

Question

Random variable?

Answer 1

I trust the different members, yet i am going to attempt for my personal phraseology: the point of a random varioable is partly, as you're saying, that its distribution of values is probabilistic. although, that stands proud as the case for a non-random variable, too. it ought to take successive values, celebration by technique of celebration, from a given set, following some predetermined progression. in case you had no knowledge of the progression, or in case you had no knowledge of the heritage of its habit, the distribution of obtainable values for a given case is in basic terms as probabilistic as that of an particularly random variable. A random variable's values may have an identical distribution, yet at the prompt are not predictable; there is no progression and no thanks to imagine its fee in a given celebration by technique of interpreting its heritage. note that my description of a non-random "probabilistic variable" describes the pseudorandom variables we use so generally in computing. they don't look in any respect random yet are a perfect change for most makes use of. If the set of regulations is sufficient, the sequence of values generated would fulfill any analytic attempt for randomness. There are situations in which pseudorandom purposes do not artwork, and do not artwork spectacularly: imagine 2 tactics utilizing an identical pseudorandom generator for a purpose that relies upon on their independence, and what ought to take position if both generated sequences "sync up" so as that they produce matching values indefinitely. it is a renowned situation in Ethernet collision and backoff tactics.

Answer 2

Mean Y = Mean X1 + Mean X2 (but both these are Mean X

So Mean Y = 2 Mean X

Var Y = Var X1 + Var X2 (as above so...)

= 2Var X

Answer 3

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Answer 4

So we have a random variable x which can give values from 1 to 5. The probability of getting a 2 is 2 times the probability of getting 1. The probability of getting a 3 is 3 times the probability of getting 1, and so on, up to five.

The sum



k + 2k + 3k + 4k + 5k = 1



because one of these five outcomes must result from every trial. If we add all the terms we get:



15k = 1

so,

k = 1/15



The distribution (x,p) is



1,1/15

2,2/15

3,1/5

4,4/15

5,1/3



We can get the mean by multiplying each outcome by its probability and then adding them together:



x(bar) = 1(1/15) + 2(2/15) + 3(1/5) + 4(4/15) + 5(1/3) = 3 2/3



The variance is the sum of:



(x-x(bar))^2/n



for each outcome.



1(1-3 2/3)^2/15 + 2(2-3 2/3)^2/15 + 3(3-3/2/3)^2/15 + 4(4-3 2/3)^2/15 + 5(5-3 2/3)^2/15 = 1 5/9



Take the square root of this and you have the standard deviation. I get 1.247





They want to know the probability of getting 6 when 2 of these variables are added. The following outcomes can lead to a sum of 6.



1+5

5+1

2+4

4+2

3+3



We need to add together the probability of all the above outcomes:



2(p1)(p5) + 2(p2)(p4) + (p3)(p3) =



2(1/15(1/3) + 2(2/15)(4/15) + (3/15)^2 = 7/45