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Let X be the number of months between successive doctor visits Let X have the following CDF: I < 1 1<I < 3 .48 3 <I < 4 F(c) = 4 <I < 6 6 <I...

Question

Let X be the number of months between successive doctor visits Let X have the following CDF: I < 1 1<I < 3 .48 3 <I < 4 F(c) = 4 <I < 6 6 <I < 12 I 212a) Using just the CDF compute P(3 < X < 6) and P(X 24). What is the expected value of X?

Let X be the number of months between successive doctor visits Let X have the following CDF: I < 1 1<I < 3 .48 3 <I < 4 F(c) = 4 <I < 6 6 <I < 12 I 212 a) Using just the CDF compute P(3 < X < 6) and P(X 24). What is the expected value of X?



Answers

An insurance company offers its policyholders a number of different premium payment options. For a randomly selected policyholder, let $X=$ the number of months between successive payments. The cdf of $X$ is as follows:
$$F(x)=\left\{\begin{array}{cc}{0} & {x<1} \\ {.30} & {1 \leq x<3} \\ {.40} & {3 \leq x<4} \\ {.45} & {4 \leq x<6} \\ {.60} & {6 \leq x<12} \\ {1} & {12 \leq x}\end{array}\right.$$
(a) What is the pmf of $X ?$
(b) Using just the cdf, compute $P(3 \leq X \leq 6)$ and $P(4 \leq X)$ .

The key to finding the expected value of X is too look at the score that X would do. You know? So if you can score one or 2 A three or 4 and then you want to look at the probability of that happening. So that's what I just right now P here I know that it's not really fancy enough. But The probability of getting a 1.1, the probability of two is point to the probability of 3.5, The probability of 4.2. So as we go through those options we multiply down. I can do this and read like one times point want to be a .1 and then two times a point to this 20.0.4. So I'm multiplying and then the expected value is the sum of all those three times 1.3 times 0.5 is 1.5 And then four times a .2.8. So as you add those together, what I would really do since one plus 10.7 point 4.1 is 0.5 plus, this will be two plus 20.8. Well give me 2.8. So the expected outcome in this situation, we randomly played this game uh you would expect to score 2.8

So the whole theme in this problem is that your expected value? You know, I'll straight X. And P. Is this is the probability that you'll land on zero or you'll get zero points Maybe like you're playing a dark game, uh if you land on a certain spot is 50% of the time, you get zero points And then you're 20% of the time, you get one And then 20% of the time you get to. And then but The remaining 10% music would land on three. So the whole premise of the expected value of getting that is multiplying. So you would expect to get zero points 50% of the time. You expect to get 1.20% of the time And then uh two points 20% of the time and then three points 10% of the time. This is how you approach the problem. And this isn't too bad because zero times anything is just that thing and one times anything is just that thing. So 0.2 plus 0.4 plus 0.3, I would expect to get just ran through it at the board or whatever game it is Point to .4.6 plus .3, That's how you get the answer.

Well, since the problem just tells you to go use a graphing calculator, that's exactly what I'm going to do. And perps did that backwards. And in this program, you can actually go to the functions under miscellaneous. And you can actually just find the answer right here and plugging in the values 48 to 60 of the function. Now, we could work this out if anybody's interested. But again, I'm following the directions. So 139 No, please. Um, T minus 48. Okay, Weird. Jeez, um, just double check that you typed it incorrectly, and this is your answer as a decimal. But the way the problem set up is that they want you to Oh, I'm breaking this at times. You know, if you take that value multiplied by 100 that your percent 47.72%. That's what we're looking for anyway. No need for the graph. All you need is the calculator

We're told that a random variable X has a continuous uniform distribution whose probability density function is shown here. And then we're told that we take 12 independent samples from this distribution. So each of these 12 observations is independent and identically distributed. Were asked what is the probability distribution of the sample average, -6. Let's start with sample average, It's made up of 12 independent and identically distributed random variables, so it is approximately normal shaped approximately. Now the mean of X is equal to the mean of the distribution from which the samples come, which is for a uniform distribution, He over B divided by two where A is the bottom of the interval and be at the top of the interval. So this is simply .5. Now the variance of the sample average will be equal to the variance of the distribution from which the sample comes from divided by the sample size. The variance is b minus a squared over 12. And sample sized it's 12 Comes out 2.00694. And from that we can derive the sample average standard deviation, which is just the square root of this value, And this is .0833. So now, to find the mean of x minus six, this is equal to the mean of x minus six, this is -5.5. And the standard deviation of X -6 is equal to the standard deviation of x average What was .0833. So this all means that our sample average averages are approximately normally distributed with a mean Equal to negative 5.5 and the standard deviation, Sorry, this should be Sample average -6 With a mean of negative 5.5, And the sample average -6 standard deviation Of 0.0833.


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