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3 Paove %3 ex/litit computatzon that me exbechatron of Random Variable wilh pavameteos m/ } is mp. Bomial...

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3 Paove %3 ex/litit computatzon that me exbechatron of Random Variable wilh pavameteos m/ } is mp. Bomial

3 Paove %3 ex/litit computatzon that me exbechatron of Random Variable wilh pavameteos m/ } is mp. Bomial



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3-51. Determine the mean and variance of the random variable in Exercise $3-17 .$

Here we are asked to find the mean and the variance of the random variable given in exercise. 3 18. Mhm. Now the probability mass function for that random variable is given here and as we can see the distribution continues on to infinity. However, if we look at the probability masses for the first five integers that the function can take on, we can see that These five probability masses Add up to more than 0.999. So it means that the majority of the distribution occurs within the first five instances of the values that X can take on. So we can safely ignore the probability masses for the valley is 567 and so on. Up to infinity and still have a fairly accurate answer. Now to solve for the mean, we use the formula for the expected value of X, which is the summation overall X of X times the probability mass function. And this comes out to zero 33 to approximately. Now to find the variance of X, we can use this formula And this variance comes out to approximately zero 436. So we have the means of X is .332 and the variance of X is .436.

For this exercise, we are asked to determine the main and variance of the random variable that was given an exercise 3 19. The probability mass function for that random variable is shown in this table here. Yeah. And so to solve for the mean, we can use the formula for the expected value of X, which is the summation overall X of X times the probability mass function. And this comes out to 1.57. And then to solve for the variance of X, we can use this formula, the summation overall X of x squared times the probability mass function minus the mean squared, And this gives us a variance of .311. And so we have a mean of x of 1.57 and a variance of X of .311.

For this exercise, we are asked to find the cumulative distribution function for the random variable whose probability distribution is defined by the probability Nasa's shown on the left. Now, these probabilities all add up to one. These are the only values for X, for which the probability mass is non zero. For any other value of X, the probability mass is zero. So we can start by saying that For X is less than 217. The cumulative probability is zero. That is because Below 217, there are no points for X for which it has non zero probability mass. This is the lowest number of X that has probability mass. Then we can say once X is at least as big as 217, But X is still smaller than 218, Which is the second smallest value for which excess probability mass that is non-0. We can say that the cumulative probability will be .005 and similarly for X is at least as big as 218, But still smaller than 231. We can say that the cumulative probability is .008 And that value the probability of .008 is the cumulative sum Of these two numbers. So for X at least as big as 231, but still smaller than 255. We then add Probability of .045, 2 hour, a cumulative probability And that gives us .053 and then we just continue in this manner through the remaining through the remainder of the probability masses on the left side. And this defines our cumulative distribution function for X.

In this problem, we are given a random variable X. And the probability distribution of it is also given, we need to find new D X. Now, since the random variable is a discrete variable, this will become equal to summation D X, f X and g x is given to be two X plus one whole square. So let us expand the submission the first value of x minus three. So we have two times minus three plus one whole square times f minus three. The next value of X is six, so we have two times six plus one whole square times at of six and the last value of x is nine. So we have two times nine plus one rule square times F of nine, so two times minus three is minus six minus six, plus one is minus five. So we have minus five old square which is 25. And with that we have f of minus three, which is given to be equal to one by six. The next one is two times six which is 12, 12 plus one is 13. So we have 13 square which is equal to 169. And we have with this multiplied F of six, which is given to be equal to half. And the next one is two times nine, which is 18 plus one, which is 1919 square, which is 361. With that we multiply F of nine Which is given to be won by three. Now the value of this can be calculated to be equal to 209. So the required value of beauty x will be equal to 209.


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