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0.40.30.20.124252627280.200.15d) 0.100.05...

Question

0.40.30.20.124252627280.200.15d) 0.100.05

0.4 0.3 0.2 0.1 24 25 26 27 28 0.20 0.15 d) 0.10 0.05



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$$ \begin{array}{|c|cccc|}\hline X & {20} & {30} & {40} & {50} \\ \hline P(X) & {0.05} & {0.35} & {0.4} & {0.2} \\ \hline\end{array} $$

Today we will be continuing our discussion of probability distributions with an example of a distribution and determining if it is a probability distribution for not now. Before we determine if this is or is not a probability distribution, we first need to review the definition of probability distribution as well as the two rules that accompany it. No, start with probability. Distribution is a table or an equation that links each outcome of his statistical experiment with its probability of occurrence. And, of course, the two rules that a company that our number one the all probabilities in the probable distribution must be between zero and one. We cannot have negative probabilities, and we cannot have probabilities larger than one. We can't suddenly have a probability of two because that's just not possible. And her second rule is that the sum of all probabilities must equal one. You're determining the probabilities of the entire sample space essentially, So all the probabilities combined must equal one because you can't just have extra probabilities that don't contribute to the overall instances occurring. Now the distribution will be looking at today is her ex sir X over our probabilities of x three day. Oh, our exes are negative. Five negative three. There we go. 30 two and four. Sorry, we don't need those lines there. My mistake. Here we go. And our probabilities are 0.1 to your 0.3 0.2 0.30 point one. Now, first thing we see after writing it out, writing out her distribution, we see that number Rule number one has followed. All of our probabilities are between zero and one. We don't have any negatives. And we don't have anything greater than one. No. For rule number two, which is the sum of all our probabilities must equal zero total. This quit one plus 0.3 so well right down here. 0.4 went to 1.3 point five 0.1 0.5 plus 0.4 is not 0.9. Sorry. Plus 0.1. Our total does indeed equal ones. Based on that rule number two is also followed. So this distribution is probability distribution. All our rules were followed and it is table that links each outcome of whatever this statistical experiment ISS with its probabilities of occurrence

We want to find the university's two by two matrix. Using a calculator. I'm going to use the decimals. Matrix calculators. We first type are click on the new matrix. Other calculators will be somewhat similar. We have to define the matrix And so we put in the elements negative .5 0.5 three and two. And then you can hit enter so that it is defined. And then you can type Matrix A. And then hit the inverse key. And we see that as a negative .8.21 .2.2. Mhm.

So to start off the subtraction, we will distribute AR minus sign or negative symbol to each of term inside the parentheses. So we'll have 0.9 a cubed plus point two a minus five and then we'll flip the sign of each of these terms. So we get negative when 70 to the fourth plus went 15 a plus went one. Now all it is do is combine like terms. So we do not have another a cube term and we do not have another A to the fourth term. We do have to a terms and two constants. So we will then begin writing negative 0.7 a fourth. That remains unchanged plus 0.9 a cube that remains unchanged. Then we have point to a plus 0.15 a. And that becomes waas 0.35 a. And then we have minus five plus 0.1 and that will become negative or went nine. So this is theano, sir.

All right today will be continuing our discussion of probability distributions. Before we look at our example, I just want to remind us of definition of probability astray shin just a table or an equation that links each outcome of a statistical experiment with its probability of occurrence. And we remember our two rules for probability distribution, first of which is that all proud buildings in the probable distribution must be between zero and one, and that the son of all probabilities in the distribution must equal one. Now, the example we will be looking at today to determine if this distribution is a probably distribution or not is as follows it is for a our X and her p of x way we go way ever excel Use of 15 16 20 and 25 under probabilities of 0.20 point by five, no. 0.7 and negative 0.8. Now, just from drawing that out, we can immediately see why. One of the reasons why this cannot possibly be a probability distribution. It's just distribution. And the reason for that is, as we can see in our first rule, all probabilities in the probability distribution must be between zero and one, and we have a negative 0.8 here, which falls outside of those parameters. Now let's ignore that for a second. Let's pretend we just forgotten about Rule one and let's add up the total four are probabilities. Anyway, we have pointed to and point fire, which gives us 0.0.7 at another 0.72 That equals 1.4. And then if we add our negative point and we still end up with a total 1.6, which still doesn't equal one. So both of our rules are not being followed with this distribution, so we know this example absolutely cannot be a probability distribution.


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