Question
The given tablc displays the probability distribution for discrete random variable X_05LZ00100.200.400.150.10What is thc mcan for X?Rbr{ 10 Points03.75Hagtss021001.65400
The given tablc displays the probability distribution for discrete random variable X_ 05 LZ0 010 0.20 0.40 0.15 0.10 What is thc mcan for X? Rbr{ 10 Points 03.75 Hagtss 0210 01.65 400


Answers
Determine whether the distribution is a discrete probability distribution. If not, state why. $$\begin{array}{lc} x & P(x) \\ \hline 10 & 0.1 \\ \hline 20 & 0.23 \\ \hline 30 & 0.22 \\ \hline 40 & 0.6 \\ \hline 50 & -0.15 \end{array}$$
Mhm. According to the definition, the probability that the random variable X. Is greater than we called you, too, because The probability that equals two plus the probability the articles. Mhm. Which vehicles .05 plus .05 Read trickles .1. Mhm. Mhm. And the expectation he calls the south of exciting times its probability. And so there's Eco's zero times 0.8, last one times, mhm 0.1 plus two times .05 Plus three times .05. And this tickles .35 Yeah, participation.
Mhm. The probability That the random variable is greater than or equal to two Equals the probability that it equals to plus the probability that it equals three plus the probability that it equals four And this equals .05, 1.05 Plus zero Robot, Their vehicles .1 thought, and the expectation echoes the sun of the value of X I times its probability metric, als zero times 00.7 Plus one times .15 plus two points times 0.5 plus three times 30.5 and plus four times 40.5 Hence their sickles, points 15 last 0.1 Plus .15 plus point to Which vehicles .6.
We will come to new Madrid in the current problem we are given X. And the corresponding probabilities. And we have to conclude that if this could be a random variable, uh discrete probability distribution, our description, random variable or not. So here we have the value their skin. 0.1 20. 0.23 could be And 0.22 14 Which is 0.6 and 50 Which is -0.15. So for that we have to verify to form uh conditions. 1st 1 is zero. Listen, he goes to P X less than it goes to one for all X. That is all the countries should be refuge and zero but less than equals to one. And some of all the PX for the entire range of it should be one. Now, by the moment I wrote this, you could see that for this value. It is a negative value. So this does not. This rejects this first condition itself, so do to that this cannot be considered has uh huh discrete random variable, more discrete distribution distribution. So I hope you could understand it. So even though there's some suppose it comes to the one even then we will not consider it because it is having a negative that so I hope I could explain this to you between you if you have any questions.
Oh, welcome to enumerate. In the current problem. We are given a list of values of X. Along with its probability. It comes like zero Is 0.3. one is 3.1. Mm hmm. This is from milk. He It is 0.20. This city is going to 15 and five. It is still a kind beautiful in order to this becoming uh robert list remission the first and foremost in the series. All the properties should lie between 0 to 1 for all X. So if this would be P two. If P two is also lying between 0 to 1, we can accept it to be probably the distribution. Unless Until and unless we also see the 2nd from uh conditions being met, which is the sum of all this value should be one. So this total should be one. So now you can see if we add this will be this plus this plus this plus this. Plus this, correct? So P two plus addition of the all the numbers. So +555 it is five and carry over one will be there. Get it. So then we have 345678. So we have 0.85. Therefore P two would be 0.1 Faith and 0.15 definitely it lies between 0 to 1. So if P two is equal to 0.15 not one, not letting this, not letting more. Exactly this value, then this will be called. This will be called uh discreet probability distribution. So I hope you could understand this. Let me know if you have any questions.