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The random ector Xhas MN (4,2) distribution #hereuI-( {) 131Find the distribution of Y = 2X, + Xz - X3If the random vector T ofT:is such that Y; = 3X, + X;Yz = X, -...

Question

The random ector Xhas MN (4,2) distribution #hereuI-( {) 131Find the distribution of Y = 2X, + Xz - X3If the random vector T ofT:is such that Y; = 3X, + X;Yz = X, - 2X,, find the distribution IS]

The random ector X has MN (4,2) distribution #here uI-( {) 131 Find the distribution of Y = 2X, + Xz - X3 If the random vector T ofT: is such that Y; = 3X, + X;Yz = X, - 2X,, find the distribution IS]



Answers

Let $x$ be a continuous random variable with a standard normal distribution. Using Table $A,$ find each of the following. $$P(0 \leq x \leq 2.13)$$

You know that's probably been doing a table for the joint distribution of X and Y. And we would like to find a few expected values office now we want first the expected value of X. The expected value of X is equal to the sum of the excess times the probability of the arts. And so we have to inform So the probability of two if we add up everything in the two columns this 0.55 and so this is two times 2.55 And then four and the probability of 4.45. And so adding these together Gives us are expected value of 2.9. Mhm. Similarly for why the expected value of why is equal to the sum of all the wise times their individual probabilities. So we have 1, 3 and five Probability of 1.25. The probability of three is .5 And the probability of five This .25. And so when we had all this together, this gives us through. Now on a we want the expected value of to explain is 31. I used your properties of expected value. This is april 22 times the expected value of x -3 times the expected value of wine. Mhm. This is two times 2.9 -3 times three. She is a sinner expected value of -3.2. Now, if you would like to find the expected value of, that's why now. Only because these are independent and we're told they're independent, this is equal to the expected value of x times the expected value of why. And so this is 2.9 times three and 2.9 times three Is 8.7.

This probably were told to refer back to exercise 3.1. And in that problem we found the following probability density function. Now here, the first thing we would like to do is find the cumulative distribution. Now if X is zero then the probability that X is less than or equal to zero is just that first value there? Of 2/7. No, that's that's one. We just keep adding in the values and so we add in a 4/7s And to use this the 6th, 7th And then lastly affects us to the probability that which is less than a pillow to we had in 17 and that's one. And so here's a cumulative distribution. Now we want to find first on a The probability that X is one. Well, this is equal to f of 1 -50, which is 6/7 minus two, sevens, Which is 4, 7. So the probability acts as one is four sevens. Mhm. And on C we want to find the probability zero is less than X is less than Rachel 22 This is April the f of two -F of zero. There's 1 -27s, Which is five steps.


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