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Tasks: Using simulations. generate vector 0f normally distributed RVX with , mcun MX cquals the sum of digits of first student id und standard deviation &x cqua...

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Tasks: Using simulations. generate vector 0f normally distributed RVX with , mcun MX cquals the sum of digits of first student id und standard deviation &x cquals 10% of the sum of digits of second student id Example= First student id 124327, second student id 132598,then the meun is /+2+4+3+2+7 19 and the standard deviation is 0.1*(1+3+2+5+9+8) 2.8 Using simulated vectors, find the probability Pr(X 2.Smx) Compare with theory. Plot the edf und pdf of the generated vector . Hint: consider cdf

Tasks: Using simulations. generate vector 0f normally distributed RVX with , mcun MX cquals the sum of digits of first student id und standard deviation &x cquals 10% of the sum of digits of second student id Example= First student id 124327, second student id 132598,then the meun is /+2+4+3+2+7 19 and the standard deviation is 0.1*(1+3+2+5+9+8) 2.8 Using simulated vectors, find the probability Pr(X 2.Smx) Compare with theory. Plot the edf und pdf of the generated vector . Hint: consider cdf and ccdfhist functions



Answers

Let $X$ denote the proportion of allotted time that a randomly selected student spends working on
a certain aptitude test. Suppose the pdf of $X$ is
$f(x ; \theta)=\left\{\begin{array}{cc}{(\theta+1) x^{\theta}} & {0 \leq x \leq 1} \\ {0} & {\text { otherwise }}\end{array}\right.$
where $-1<\theta .$ A random sample of ten students yields yields data $x_{1}=.92, x_{2}=.79, x_{3}=.90$
$x_{4}=.65, x_{5}=.86, x_{6}=.47, x_{7}=.73, x_{8}=.97, x_{10}=.94, x_{10}=.77 .$
Obtain the maximum likelihood estimator of $\theta,$ and then compute the estimate for the
given data.

Okay, So we're given that the amount of time it takes for class one to end is normally distributed. The amount of time it takes for the class to to start is normally distributed. And the amount of time it takes to get from class one to class two is also normally distributed. And they have these normal distributions, I have been down here. X one is the time that it takes to get to The time it takes for the first class to end. X two is the time it takes for the second class to begin. An X three is the time It takes to get from Class 1 to class too. So what I'm gonna do is I'm gonna let t equal X one plus X three and X one plus X three is the time it takes for the first class to end, added to the time it takes to get from class one to class too. So this will be equal to This will be equal to do one of these lines is equal to a normal distribution where we add the main values and we add the variances and this is allowed because all three of these are normally distributed and they're all independent of each other. We can do that when they're independent of each other. So if we add 902-6 minutes we get 908 and if we add one squared to 1.5 squared 1.5 squared is 2 to 5. So this is 3 to 5. And now that we have this normal distribution of the time it takes to get to class two added to the time it takes for the first class to end. We can now set this up With this 3rd distribution. So what we're trying to find is we want this time this normal distribution to be less than at the time it takes for the class to to start. So this normal distribution, so we want t To be less than X two. And so what we're gonna do now is I'm gonna rewrite this by just minus ng ti from both sides. So you get zero Is less than X two -T. And now we can set this up as another normal distribution by minus ng the two means and then we still add the variances. So this would be normal Where we have 908. We're sorry, we have 9 10 minus 98 just to two minutes and we have 3.25 Added to one squared up here. So that is 4.25 as our variants. And now what we can do is just set up a Z score. So right now we want to find the probability that X two minus T is greater than zero, but I'm going to change this. So we've found the probability of Z is greater than 0-. Are you divided by our variance? I'm sorry, our standard deviation and this will let us use then the standard normal distribution table to find our probability. So if we let zero minus you, if we set up this, so we actually find zero minus, you will use to so we know zero minus you is negative two and the standard deviation, It's just the square root of our variance, which is 4-5. So we're trying to find the probability that Z is greater than -2 over the square to 4.25. And we can do this by setting this up a different way, we can say that this is equal to one minus the probability that Z is actually less than or equal to -2 over the square to 4.25. And this will make it. So we can actually look at our distribution table And find the value that we need to, which is the value at -2 divided by the square to 4.25. So if we simplify this further, We'll see that -2 divided by the square Of 4-5, goes to about -97. And now if we go and look at our Normal distribution table, we'll see that the value at negative .97 Is just about .16 60 And so now all we have to do is take one and subtract this value of .1660. And we'll get a resulting value of .834, which is then our probability.

Were given this probability density function, and then we were asked to find the accumulative distribution function and its inverse. Then write a program to simulate 10,000 values from this distribution and then to compare the sample mean and standard deviation from our simulated values to the theoretical values. So we'll begin with this CDF. That's part a. Yeah, find that definition. CDF is given as this integral, and this comes out two x squared over four. Now, to find the inverse, we can set the CDF equal to you and then solve for X. So that's two times the square root of you. And that is our inverse CDF function. Now for part B, we restaurants, um, code that simulates 10,000 valleys from this distribution. So our code that will perform the simulation is given right here. It's only one line of code disfunction produces 10,000 vary. It's from a uniformed random variable, and then we have minus two times the square root of these random variables. So X will have 10,000 random values taken from this distribution, and then for part C, we're asked to compare the theoretical mean and standard deviation of this distribution to what we obtain from the simulated values. So let's first do the theoretical values. The expected value is the integral from 0 to 2. The pdf is only defined on 0 to 2. It is zero elsewhere. It is only non zero from zero attitude. This integral is X cubed over six, and we evaluate from 0 to 2 and this comes out to 4/3. Now for the standard deviation, let's first find the expected value of X squared. So that is extra. Explain it. Four divided by eight, evaluated over 0 to 2 and this comes out to two. Now the variance of X is the expectation on X squared minus the expected value squared comes out to about 0.222 which means that the standard deviation of X is the square root of that which is 0.471 now, to find the mean and standard deviation of the simulated values that I've produced in our I run the following commands. And so for my simulation, these came out to one point 337 and 0.467 So we can see that using 10,000 simulated values from the distribution, the mean and standard deviation are fairly close to the theoretical values

Yeah, that's probably going to find the mean variance and standard deviation. Yeah, starting with the mean, the mean is a song of all the X values times the probability of each of them. So here we want to go from X0. So so the zero Times .059, that's the probability of zero Plus one times .1-2. That's the probability of one Plus all the way up to seven, I was .53, that's the probability of something. So then it's just a matter of adding all these together. Uh huh. This just 3.349 when we do and so that is army. Now in order to find the variants. First thing we need to find is the expected by events squared. This is the sum as extras from 0 to 7 of that squared pivots And so similar to what we did above this is zero square in times 0.059. We just square all these terms first. Just one square times .1- two. All the way up to plus seven, spread Times .53. This is not a very answer is just something that we have to find in order to find the very and I'm gonna do this business that he have X squared The 17.5- two. The variance is the expected value of X squared minus the expected value of that square. We just found that he have X squared is 17.5 to 2. And above we found the expected value of X is 3.349. And so we'll take it. That's right. And so this gives us 6.306 when we evaluate this, so that's our parents. Yeah, our standard deviation, there's always the square root of parents Until I take the square root of 6.3 or six, That's for 6.306. This 2.511. So that's the standard deviation. Yeah. No, I mean, yeah. That's for the interpretation. The average number of extracurriculars. Just 3.349. Her student, Yeah. Which varies on average By 2.511 standard deviation.

The following is a solution for number 25. And this looks a grade school enrollment numbers and these are listed in thousands. So 3635, that's not 3635 students. That's 3,635 1st graders in the year 2000 were asked to find the probability distribution First. So what we need here is the total enrollment. And if we just add these all together, all these frequencies together, that gives us the total enrollment of 29 164. Okay. And to find the probability is you just take the favorable over the total. So it's the frequency 36 35 divided by the total 29 1 64 and we get .1246. And we do that for all these. So these are all basically going to be the same number just to tick off here and there sometimes there may be like an inflation for you know, grade level or whatever. But But these are all how you find it. You just take the favorable over the total. Some 1.1254 and then point 1243 And then .1- 1 1. Okay. And a good way to double check this. If you were to add all these together, it should equal basically one. It should equal exactly one. Now I rounded slightly different. They gave me a .9999. So I mean you can just round one of those up I guess and you should get one, you know to the nearest 10,000 but the next part is to do the hist a gram and unfortunately this is it's really hard because they're basically all at that point 0.12 you know like they deviate a little bit here and there but they're so close to the same thing, you can always change your scale I guess. But um they're all basically the same. It's it's a uniform distribution or at least it appears to be I want to find the mean. What we're gonna do is we're gonna take the summation of X times P. Of X. Okay, so at one times this 10.1246 and then plus two times 12.1246 Plus three times .1259 Plus four times .1-7 one Plus five times .1269 Plus six times .1254 Plus seven times .1243 And then plus eight times .1- 1 1. And whenever you add all those together mm You should get 4.4 ah eight 57. That's the average. And you know the interpretation there is the average grade level for all the students in the US from grades one through 8 is 4.4857. Yeah. Okay and then the standard deviation is a lot like it it's the square root of the variance. And the way you find variances you take the X squared times the p of X. Okay, so I'm not gonna do all these just because it'll take way too long, but it's um the X squared, so one squared times 10.12462 squared plus two squared times 1246 plus three squared times .1259, you know, etc. All the way up to plus eight squared times .1-1 1. And then what we do is we subtract the mean squared. Okay, so this actually finds the variance, so point minus 4.4857 squared. And that's gonna give us the variance and variance is 5.1962. But we square root that to the square root of that is the actual standard deviation 2.28. So the standard deviation of this probability distribution is 2.28.


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