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The storm runoff X (in cubic meters pcr sccond, m3/s)from subdivision can bc modclcd by random variablc with thc following probability density function:f(x)=c(x- );...

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The storm runoff X (in cubic meters pcr sccond, m3/s)from subdivision can bc modclcd by random variablc with thc following probability density function:f(x)=c(x- );0<x<6Dctcrminc thc constant and skctch thc pdf: Sketch the cdf cumulativc distribution function") Thc runoff is carricd by pipelinc with capacity of 4 m3/s Ovcrflow occurs whcn thc runoff excccds pipelinc capacity. What is thc probability of overflow? If ovcrflow occurs aftcr storm what is thc probability that thc runoff in

The storm runoff X (in cubic meters pcr sccond, m3/s)from subdivision can bc modclcd by random variablc with thc following probability density function: f(x)=c(x- );0<x<6 Dctcrminc thc constant and skctch thc pdf: Sketch the cdf cumulativc distribution function") Thc runoff is carricd by pipelinc with capacity of 4 m3/s Ovcrflow occurs whcn thc runoff excccds pipelinc capacity. What is thc probability of overflow? If ovcrflow occurs aftcr storm what is thc probability that thc runoff in thc storm is less than 5 m3/s An cnginccr considers replacing thc pipelinc by largcr one with capacity for Sm3/s Supposc there is probability of 60% that the rcplacement would be completcd prior to thc next storm: Calculatc thc probability of an ovcrflow during the next storm Calculatc thc Expccted Valuc and thc Variancc of X



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In western Kansas, the summer density of hailstorms is estimated at about 2.1 storms per 5 square miles. In most cases, a hailstorm damages only a relatively small area in a square mile (Reference: Agricultural Statistics, U.S. Department of Agriculture). A crop insurance company has insured a tract of 8 square miles of Kansas wheat land against hail damage. Let $r$ be a random variable that represents the number of hailstorms this summer in the 8 -square-mile tract. (a) Explain why a Poisson probability distribution is appropriate for $r$. What is $\lambda$ for the 8 -square-mile tract of land? Round $\lambda$ to the nearest tenth so that you can use Table 4 of Appendix II for Poisson probabilities. (b) If there already have been two hailstorms this summer, what is the probability that there will be a total of four or more hailstorms in this tract of land? Compute $P(r \geq 4 | r \geq 2)$ (c) If there already have been three hailstorms this summer, what is the probability that there will be a total of fewer than six hailstorms? Compute $$ P(r<6 | r \geq 3) $$ Hint: See Problem 28

So this particular problem, you've been given the conditions for the CDF off w So for the party, eh? You know that W is our non zero value. So therefore I concede that w was credit and only Puerto zero on duh in a non zero for this part. So therefore, for w less than zero the CDF off W we can write it as B or for capital of Lula's Annie Woodrow the small right w which cannot about it and as a CD off, off, off, off, doubling on duh that. But we need to simplify for different values now. So first I'm going to find reliability for W is equal with zero right into the first part of your wonder find So to find out probably W is equal to zero. The condition going in The question is that it is vey lesson. Is he really? It's between that distribute indignant from zero to me D and for the extension distribution, the value off effects is land on to eat is to minus lambda X, the X So if you apply the intimidation, this is Lambda is to mine is Land X a born minus lambda limiters from zero to Edie. So Lamda is cancelled on if I substitute the limits and get one minus is too minus lambda we be. So this is the Obama team and our music will do zero now. Next is I'm supposed to find the probability when, uh, the capital w well, I between the value zero and a small w Okay, so to start of it, that's what the conditions that are given in the question this we can write is equal probability. Off zero is less than capital W for this condition is K and will be minus video and less than the small value. Doubly So Now we can read out this part like this where we begin mother, black teens I really and then divide for by key. So the expression I don't know that we have zero less than we have gave me. Minus kennedy is less than w. Okay, so then further if I divide toward by key Ah, What we will have here is zero less than the minus. VD is less than w by K. Okay. And for them to add told with VD, Sylvia probability off really is less than the his less than this is W plus we have. Ah, I mean, this would be w by K plus Weedy. Okay, so now are based on this condition we have to integrate from VD toe. This is the blue place DVD upon. Okay. For the whole a lambda it is to mine. Islam dy ex d x. So if I continue with the indication I want to keep from the hazardous dilation is ears to mine. Islam decks upon. Minus lambda limits are from TD to Mrs W. Place Cady upon Cape. So Amanda gets cancel your and the value of an expression that we get now is, uh it is to my nest DVD. I'm sorry. This is Edie's to minus not Katie, land of VD and, uh, minus. We will have e j screw minus. I don't know, bless gave me the bike. So this was for the conditions between zero and part of you. Now we're talking about ah, the values for no w s president or equal to zero. Then, in this case, the CDF is they find us again. If off w will be re off w les and equal w So this part I can take this as POWs, zeal plus the value off between the Oran. Smart. Lovely. So we found what the values if I substitute surely one minus CDs to mine Islam W t plus, this is years to minus Lambda. Really minus. It is to minus the group less giving the of wonky, so you can see if we can save this park. So what remains is one minus is to minus new place gave India punky. So based on this, I can therefore conclude the CDF as a form W Can we return here as eagled for st abuse lessons, you know? And this is one minus beatings to minus the blue bliss que vd by G for double Ingrid and zero. Now we want to the part B. They're also trying to find the pdf for this particular part. So the pdf off, uh, W for w Great as you will be. Ah, form W We can take this as dead over two off the CDF. So finding the dead emitter off this expression with double is one minus easiest minus lambda. We have the blue bless que vd of one key. What's yours if I differentiated B minus off, please to minus Lambda into this w place DVD bikey on into the Internet of Atonement. Don't move me! Minus lambda by key to the final value for the sun Get in. Here it is Lambda by key in tow. Edie's to minus lambda W S K V d Bike key. So this is the pdf for Dylan particular part. Now we also know that and I lose a mixed distribution. So I can say that since W is mixed distribution eyes getting the question with the combination of discrete and continuous boat, it's now to find out the expectation for W uh when he was a concept off the street as a list continues. So then we're going to write expectation of W So for discreet, I'm gonna take this as X into the office. So Lazio and be off W is equal to zero. And for this partisanly continue. So you haven't gigolos Theo to infinity be w in tow, F w Theo into the blue. We've just found out. So that's actually Lambda my gay and oh, we have evens to minus lambda and this is w by gay plus really with need of you so I'm gonna get a lambda by gay Healy's tow. Give my understanding the racket. We have mine Islam, WD questions digging that part outside healed and in bigger list for the organ firmly between w ant eaters to minus lambda w by key. I don't know. So, uh, I blame the indignation by parts. Now what the hell are you going to get the blue in tow? Indignation for this is he is to mine, Islam. That of you by k upon my name is Linda by giving them a tron zero to infinity minus is integral zero to infinity, the inevitable that we will be one and again, indeed, Asian off eBay stores derided by Mayan Islam that w born gay, sodium to move this w sure really dealing with that. So Lamda McGee is a constant and quit Needham. So for that, huh? If you take my Islam DeBakey comment well too well, have minus okay by land Outside will get cancelled on dhe. We will have Edie's to mine Islam Davidi as just on in the bracket here, the parliament. This will change to zero because it just reminds and pretty zero and W Walter white hunters use the force. Thomas is you know directly. This would change to Plus, we have here on Lander. I'm keeping London. Keep your eyes off now because that I could cancel anyway. And for this they are hindering in delusion. So you can have years to minus lambda doubling by key upon mine is lambda by P. Yodo Infinite. So I can gather this year on. And now, after putting the limits, the Lambda by K it is true minus lambda VD and the Parliament is to minus. Infinity is zero and mine is he? This was you. Is one Ondas divided by minus one minus one gets cancels on. Duh. There is a connection of yours. Just doesn't cancel into frankly can want apply. So his land on my key. And do I have key square by land a square and the nominee elements, he'll have one minus zero social. Now, simply for this part. The final answer for the execution of doubling that I'm getting is give our land up years to minus lander VD. So this is the final answer for expectation off W

Okay, so for this problem, we are assuming a null hypothesis of the mean difference of zero and or alternative hypothesis is that the main difference is less than zero. Since we're looking at runoff. Okay, The average difference is going to be -015. The standard deviation is going to be a low 01373 Are calculated, T. Statistic is -20.204. Our degrees of freedom is going to be one less than the sample size, which is 341. However, in the table, we only go up by 50th tool around town to about 300 instead of going up to 400 for fun, we probably could average out the next two numbers and sort of interpolate the two. But why make extra work for ourselves? We see that the P value is less than 1%, which is what our significance level is. And therefore we reject the null hypothesis. This means that we reject that. And in turn in favor of the alternative, the mean difference is less than zero, Not for the student's T distribution. We find that we are looking at a significant level of 0.005. The T. star statistic is 1650 from the table. And then our mhm plus or minus part is 1.650 times 0.1373 divided by the square 342. This gets us approximately 0.0123. And then we'll add that to the D. Bar J. Giving us a 99% confidence interval Of negative 1.6 negative .16-3 and negative .1377 or 99% confidence that the runoff is between these two numbers. If we put this into our instagram, it'll look approximately like this.

As usual are null hypothesis is always that the two averages are equal. So anyone is equal to be a two, supposed to be a colon or semicolon. All of those two and turn all hype. Alternative hypothesis is that one of them is greater than the other. So we'll say that when you do absolutely nothing. Yeah. Okay. U one which is going to be a logged area and it is going to be greater than or equal to than youtube indicating that we will be using a right tail test, chop it off their shaded region, rejected region. So uh formulating all and alternatives that we have applied the Will cooks one signed rank test to perform the required hypothesis test at the 1% significance level. So alpha is .1 and you have a sample size of however many are provided in the data disk. Okay. Using these information right here, you will be able to go to table for and find out what your critical value is right here. And then performing the test. You will find that the Lance in this area. You fail to reject the null hypothesis. Otherwise you will reject the null hypothesis. If you landed this shaded region and then you will apply a paired T. Test which is given by this formula up here. Oh all right. Yeah. Right here it's gonna be this formula right here for the T. Test and the reason freedom are giving given by this. Well, she'll find in Table four compare that test statistic and then which test is going to be better? The parent detest is going to be best when your data when the differences are normally distributed. Someone like this. Um If you got some outlier points out here and here um then you're definitely gonna want to opt for the T. Test because that looks more approximately symmetrical if you just try to connect those lines. So T. Test for normal distribution amongst differences W test otherwise if it at least looks somewhat symmetrical

In this problem, the amount of rainfall in inches on a tropical island in the month of august is having the probability density function effects equal to one by 16 X. Multiplied by six minus X. On the close to interview. L 0 to 6 we have to find the probability that the amount of rainfall in august building less than two inches, probability that the amount of rainfall in august will be less than two inches is equal to probability of X less than two. To find this probability we will integrate the given function when by 60 next multiplied by six minus X. D X from the lower limit in this case, which is equal to zero two, the given number that is too as we have to fight the probability that the amount of rainfall will be less than two inches. So we will integrate it from 0 to 2. This is equal to as one by 16 is a construct. It will remain as it is So one by 16 integration from 0 to 2 66 minus x square D X. This is equal to one by 16. Yeah 6 28 cities integration of X. Dx is X squared by two minus integration of x square is X cubed by three. So minus X cube by three. Between the limits. Zero and two. This is equal to one by 16. Six by two will become three. So three multiplied by two square is four. So three multiplied by four minus two. QB Said. So eight by three. Then again my nurse, three in 20 square is 3 to 0 which is zero minus zero. Q b 00 by 30 So again zero and zero minus zero is zero. So this is equal to one by 16. Multiplied by three into four is 12 minus eight by three. Yeah. Now we will simplify this to get our final answer. This is equal to one by 16. My flight bag 12 threes are 36 and 36 minus eight is 28 so this is 28 divided by three. We will simplify this. We have 44 to 16 47 28 so this is equal to in the numerator, we have seven and in the terminator we have four in 23 which is equal to 12. So the probability that the amount of rainfall in august will be less than two inches is equal to seven by 12, which is our required answer.


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