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Compute the following probabilities:If Yis distributed N ( - 4,9), Pr (Ys -3) =(Round your response to four decimal places:)...

Question

Compute the following probabilities:If Yis distributed N ( - 4,9), Pr (Ys -3) =(Round your response to four decimal places:)

Compute the following probabilities: If Yis distributed N ( - 4,9), Pr (Ys -3) = (Round your response to four decimal places:)



Answers

Solve the given problems. In Exercise $49,$ find the probability that at least one engine functions properly (the sum of the terms containing $0.95^{4}, 0.95^{3}, 0.95^{2}$ and $0.95^{1}$ ). Round to the third decimal place.

All right So we're going to try to determine the cumulative distribution function. The C. D. F. Of the binomial random variable. So it's gonna be been sample of three. Yeah. And a probability of 25%. Yeah. Mhm. So that means we'll have just a couple of probabilities to calculate here. This is going to mean that our probability mass function which we're going to get from the binomial is going to be the F. F. X. R. P M. F. Of three choose X. And then we get our probability 0.25 braised to the power of X. And the probability of 1 -254.75 Raise the power of three -X. So that's what our probability is gonna be. And this is going to be four X equals one. Wait actually we're gonna be doing zero as well. So 01 two 10 3. Okay so that's gonna look like this. It's gonna be zero for everything less than zero because that's how we start off. Okay. And then for the ones between zero and 1 that is going to be Which is exactly zero. That's gonna be the f. of zero Which is going to equal 3-0 which is just gonna be one time 0.25 to the power of zero. Which is just gonna be one. And then we'll have 075 cubed. And that's going to equal 42 18 75. Mhm. And then we'll add that on to the next probability uh that's gonna be between zero and one Including zero excluding 1. Mhm. So we're gonna have the f. of one And that's going to equal to three choose one which is gonna be three Uh times .25 Times .75 Squared. And that is going to be equal to surprisingly the same thing. And that means that the next one oh is going to be equal to R. F. Of zero plus R. F. One for the accumulation of zero point The 2nd 184 three. Mhm. And there's gonna be another one. There were my calculator goal here it is .84375. Oh. Uh huh. So we're going to have that probability for one and two Are not including two. Then our half of two. That's gonna be three choose to yeah. Okay Squared 075. And that's gonna be it's gonna be 14 0625. We're just going to add that onto this one. It's gonna be 0.84375 plus 0 14 6 to 5. Which is going to be equal to zero 984375. And the last one is gonna be one. So summarizing it up into our a cumulative distribution function, We're gonna have zero for all x less than zero Rene have 042 1875 For our exes between zero and up to one. Mhm. We're going to have a 0.84375 accumulation probability Forex between one and 2 And between two and 3. We're going to have a 0.984375. Okay. Yeah. And then for anything greater than X is equal to three. It's gonna be one. Yeah.

Okay, we want to come up with uh the cumulative distribution function for the Bino mule rain and variable X. Where and the number of trials is three and P to profit. P the probability of success is 1/4, which equals .25. Uh First I'd like to say. Uh I'd like us to find out what is the probability if we have three trials with the probability of zero successes? Was the probability of exactly one success. What's the probability of exactly two successes? And what is the probability of exactly three successes? We need these values before we try to create the cumulative distribution function. Uh The cumulative distribution function we will call it F F X. We'll talk more about that in a minute. All right, so with a little help from the uh probability applet, we're drawing the probability mass function. This is not the cumulative distribution function. This is the probability mass function of the binomial distribution parameters and is three ps 30.25. Now um I need to know what is the probability of 012 and three successes. So the probability of X equaling zero is 00.4 to 19, Probability of exactly one success. Also .4- 19. Yeah, Probability of having exactly two successes. .146. And the probability of exactly three successes. .0156. If you add up all these individual probabilities you will get a total of one as you should now a cumulative distribution function F of X. Okay, F of zero F. of zero means to probably the probability of having zero or fewer successes at most. zero successes. Um so the same thing as probability to x equals 0.4219 F of one cumulative distribution function F of one. The probability that we have one or less success is at most one. That's the probability that we have zero plus the probability that we have one success, So 10.4 to 19 plus 190.4 to 19 .8438 F 02 F is our cumulative distribution function that we are being used to fight F to. The probability of At most two successes, two or fewer? Well FF two is going to be the probability of zero successes plus the probability of one success plus the probability of having two successes, so 0.4 to 19 plus 190.4 to 19. Which of course currently is 190.8438 plus 0.1406. So F two His .9844. The probability of having at most two successes is .9844, Last but not least f. of three. The probability of having at most three successes, that means zero or 1 or two or 3 successes. Uh that means we have to add up the probabilities of having zero or one or two or three successes. So all these numbers get added basically is 2.9844 plus the probability of having the three successes, so .9844 plus .0156 Equals a total of one. So F of x is a cumulative distribution function F of three. The probability that you have three or fewer successes is exactly one. F up to the probability that you have two or fewer successes 20.9844 F. of one probability that you have at most one success one or fewer .8438 probability that you have zero or fewer successes would be .4-1 night. So here is our a cumulative distribution function.

This is another problem that will be ending with us needing to use a news, the normal CDF function from a calculator to be able to solve it out. If you have your own calculator with the normal CDF function, then you want to get that. If you do not, you'll need to find one that does you condone. Find them online if you to search online calculator with normal CDF function. But make sure you have found some way to be able to use that at the end of this. Now your first use the normal CDF function. We will need to have a low, ah, high, a mean and standard deviation. That's what that's actual work that we need to do for. This problem is figuring out what those four things are going to be Lo and hi. We should be able to find pretty quickly without having to do any actual math. Really Just read the question. For example, with this problem, we're being asked to find the probability of, at most 100 successes. Well, at most ah, 100 successes would mean that 100 100 is the highest amount of success is that we are allowed to have. So 100 is our high. The question then is what is our low? Remember, this is supposed to be rooted in a real situation. We are running in this case 125 trials because it's told us that N is equal to 125. Well, it really doesn't matter. The point is for the low. You can't run a negative number of trials. So the most amount of success there, the least amount of successes we could possibly have would be zero right. There is no such thing as though I had a negative amount of success is where I ran a negative amount of trials. So zero was our low 100. Is there high? We do need to find our being and our standard deviation. They do not give those to us, but because they tell us that N is equal to 125 and they tell us the P is equal to 6/7. We will be able to find these two things ourselves for the mean. As you can see here, we find the mean by taking N Times P, which in this case would be 125 times 6/7. If you take 100 25 times 6/7, it doesn't come out perfectly, so well after around a little bit. But it will approximately come out to 107 0.14 That's that's approximately what we get for our mean. Okay, center deviation is a little bit more involved. We do for standard deviation. We need to take the square root of end times p times the quantity of one minus p. So we've got our squared. We already know that end times p is 107 0.14 We just did that, right, so you don't really need to do that again. One minus p means we'd be taking one minus 6/7. Which means effectively, we're doing the square root 107.14 times 17 because one minus 6/7 would be won over seven. If you plugged in your calculator, we should get approximately 3.91 again. You're going to need to round. Alright, It's not exactly 3.91 That's just roughly what we're getting here. Hopes did not back up for area, but that is roughly our standard deviation. So now that we've been given those two things or not, we found those two things I should say. We already to plug this into our calculator, so go to your calculator, whether it's yours or whether you're using the thing online or wherever you're using it. Put in your low of zero your high of 100 your mean of 107.14 your standard deviation of 3.91 And assuming that you plug a lot stiffing correctly, you should get back from the calculator. 0.339 We were asked around these, uh, answers off to three decimal points so that would be and we would round this off, too. 0.34 Or, in other words, saying there's a 3.4% chance 3.4% probability that we would get at most 100 successes

To find the probability of an event. We take the number of favorable outcomes and divided by the total number of outcomes. So in this problem, we're finding the probability that Donovan visits an out of state college if you randomly selects from 10 out of state colleges and foreign state colleges. So the number of favorable outcomes would be 10 because he has 10 out of state colleges on his list and the total number of outcomes would be 14. Because there are 14 colleges on his list altogether. Let's reduce that fraction, and we get 5/7 and that's the probability that he will visit and out of state college. And then we can convert that to a decimal. We're going to have to round it to three decimal places, and we get 30.714 as the probability of visiting and out of state college


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