5 answers

QUESTION 31Y is random varible that is distributed N(19,0.3600). Find Prob(20.704 <Y < 20.716)QUESTION 32Y is random varibl that is

Question:

QUESTION 31 Y is random varible that is distributed N(19,0.3600). Find Prob(20.704 <Y < 20.716) QUESTION 32 Y is random varibl that is distributed N(19,32.4900}. Find Prob(24.073 <Y < 30.571). QUESTION 33 Y is random varible that is distributed N(19,34.8100}. Find Prob(20.475 < Y < 27.378). QUESTION 34 Y isa random varibe that is distributed N(19,0.8100). Find Prob(l9.15 < YertORY 22.27). QUESTION 35 Y is random varible that is distributed N(19,34.8100}. Find Prob(29.62 < Y < 36.523). QUESTION 36 Y is random varibl that is distributed N(19,1.6900). Find Prob(l9.247 < Y < 20.651)


Answers

Random numbers Let Y be a number between 0 and 1 produced by a random number generator. Assuming that the random variable Y has a uniform distribution, find the following probabilities:(a) $P(Y \leq 0.4)$ (b) $P(Y<0.4)$ (c) $\mathrm{P}\{0 . \mathrm{I}<\mathrm{Y} \leq 0.15 \text { or } 0.77 \leq \mathrm{Y}<0.88$

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For this question because USAA random number and between the around one and fought all uni from distribution, you can have ah, graph like this. That is a zero. And here is a one and the heart also be one. Okay, so we can see for a when why you called last in point. You're that Yukos.

Who have we here as your old 0.4 in for prints a reality? How point for Because dressy coats upon for and also for me, we can see well why life standpoint for the answer is also reporting for and for C first, I can see that it's a different one. We also use described Thio explain the answer So first this room's 0.12 point 15 So maybe here is 0.1 and here is point, But Cy ends with this area that is point 12 point 15 and class on 77 Maybe here 2.88 And also this area. So we Adam, together we can see the property here dressed Pecos who 0.1 side of my nose point one and then class find h eight miners point salve in salad. You can see breath dress ICO 2.16.

So this normal distribution has I mean 30. We don't even have to worry about the standard deviation. So if you pick something from this distribution, what is the probability that its value is greater than 30? So obviously 50% off? The values for this distribution are greater than 30. That's what is the definition off mean, right? Mean is the central value So 50% of values are less than 30 50% of values are greater than 30. So correct Answer is 50% are in decimal city zero point.

So you don't fight. Thank you..

All right. So we're given a normally distributed variable with the meat of 50 and standard deviation of five. Now, part eh wants us to sketch this distribution. Or should I should say, sketch the graph? I'm not the straightest line in the world, but whatever Eccles wrote to hit a little bit, I actually think that made it worse. Let's just redraw it.

There we go. All right, So 50 is gonna be in the middle. It's right here. It also wants us to just take it off every five. So 45 55 60 40 35 in 65.

Right. So this right here is gonna be in the middle. Then if we compare this a figure 6.4 in the textbook, you know, it looks something like this. It's not perfectly symmetrical. At least it's drawing isn't.

Ah, there we go. That's someone more accurate, and I'm just gonna work off these standard deviations. So 45 55 or one standard deviation away from the mean 40 and 60 are two standard deviations and 35 65 or three standard deviations knowing this. So for part B, we want to find the probability that this variable lies between Ah, sorry. It wants to find the probability that lies between 45.

I leave it to find some random variables. X no, it doesn't swell. We're now defining the random variable is X, and that's one standard deviation away. So we know that that's 68.3%. This is true for any normal distribution and then part See, it wants us to find the probability that the value is between 40 and 60.

That's two standard deviations away. So he gets 95.4 and there you have it..

We go from a pdf to a CDF. The first one is simply going to be zero for all values less than one. And then after that between one and two we had our first value And then we're gonna add to get 2.14 and the .38 plus the next one. And then we're gonna repeat this process for all the other values. And this is going to go all the way down to X is greater than or equal to 15.

So for all values access greater than your equal to 15, that's gonna be um all the values that we have..

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