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Quostion 5 Cousidler experituent which Iezulta of threr prnsibk: outcome> with outcotue oceurting with probability Pa =1,23a DR Suppose that indepeudent replicat...

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Quostion 5 Cousidler experituent which Iezulta of threr prnsibk: outcome> with outcotue oceurting with probability Pa =1,23a DR Suppose that indepeudent replicatious of this expetient Ae petformed nld let X; deuote the nutubcr of titles outcolue #PPen> Deterqine' the: prolmbility HHLS fuuctiou o tle rdol !tr [Or Xb) Derive the joint montent gencratiou fuuctiou a (XI. X2 X3). Are raudom vuriable 4,,4z X iudepeudeut? WZwv wht not ?Shew tht the cotrelation cocflicieut betweru 4 aid X i g

Quostion 5 Cousidler experituent which Iezulta of threr prnsibk: outcome> with outcotue oceurting with probability Pa =1,23a DR Suppose that indepeudent replicatious of this expetient Ae petformed nld let X; deuote the nutubcr of titles outcolue #PPen> Deterqine' the: prolmbility HHLS fuuctiou o tle rdol !tr [Or X b) Derive the joint montent gencratiou fuuctiou a (XI. X2 X3). Are raudom vuriable 4,,4z X iudepeudeut? WZwv wht not ? Shew tht the cotrelation cocflicieut betweru 4 aid X i gitu by P Va-PJ( Show Llutt tlu: couditlouudistributiou o 4, given that X <"I binomial Derluce the conditional HUnu Audvzriiuuce af . gitu tlwt 4 =



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Show that Theorem 2.6 , the additive law of probability, holds for conditional probabilities. That is, if $A, B,$ and $C$ are events such that $P(C)>0,$ prove that $P(A \cup B | C)=P(A | C)+P(B | C)-P(A \cap B | C) .$

We have a binomial distribution with N equals five trials and probability of success P equals 0.75 We want to find p of x equals four. That is the probability of finding exactly four successes in these trials. Using first a binomial probability function B p F. And secondly, a binomial probability table. To start. Let's solve it with the B p f. Remember that GPS is defined as P of X equal to the binomial coefficient and choose X times P. E. To the X times one minus P to the n minus X. Remember that a binomial coefficient has the following definition. And with this we can plug in our X equals four and equals five and P equals 0.75 To obtain the following expression, this reduces down with some algebra 20.396 Our probability of X equals four. Next use binomial probability table. Well, either google the term or look it up in a textbook. Doing so allows us to obtain P X equals four of 0.396 And at this point, we note that these probabilities are equal, i. E. You attain the same probability as using a Bps, as you do with a binomial probability table.

In problem 128. We want to prove the following for a if the probability of a given that b equals probability of a given that b compliment, then e and be or independent events from the law of Total Probability. We know that the probability of the equals the probability of the given that B was employed by the probability of B plus probability of a given that given the accompaniment multiplied by the probability of b compliment. As long as we know that the probability of a given B equals probability of giving the commitment, then this equals this and we can take it as a common factor. Then the ability of a given b multiply it, boy, we have from the first remaining the probability of B and from the second term, we have the probability of people compliment from the law. Total probability. We know that an event with its complement, their probabilities are one. Then it equals the probability of a given B, and we can see that the probability of A is the same as the probability of a given B, which means the Event B has no effect on the event A or the events E and be the events E N B or and dependent for part B. If the probability of a given C is greater than the probability of B given C and the probability of a given C compartment is greater than the probability of B given C compartment, we can conclude that the probability of A is greater than the probability of being. Let's think we need to get the probability of a from the left hand side of the inequality, and we want to get the availability of beef from the right hand side from the fastest inequality. We can get the probability of e I'm a blind boy, both sides with the probability of C. Then we will multiply the first inequality by the probability of C, and as long as any probability is greater than zero and smaller than a month, the inequality sign will not change. Then we will have the possibility of a given. C multiplied by the probability of C is greater than the probability of B given C. Multiply by the relative see and call it two all three Equality number three We will do the same for the second inequality here, the second inequality. We multiply it by the possibility of C compliment. Then we will have the probability of a given C compartment multiplied by. The probability of C compartment is greater than the probability of B. Given the compliment, multiply it by probability of C complement, and we can call it for by adding three and four. You can reach the probability of A and the probability would be by adding three and four. Then we have the probability of a given that c multiplied by the probability of C given multiplied by the variety C plus. The probability of a given see compliment multiplied by the probability of C compartment is greater than the probability of the given. C multiplied by the probability of C plus the probability of P. Kevin See compliment deployed by the viability of C compartment from the law of total probability. We can see that this two terms, or the probability of a and these two terms or the probability of B as we have written here the same, but they please be with C. Then the probability of A is greater than the probability of B, which is the required to be proven. And this is the relevance

So this question we are asked for the values of the standard normal variables East are where the probability that see is less than the star has a certain area given here. So we're just going to look up these areas being our table. So in 14 This area corresponds to a value of -2.434 disease starts, 5.985 corresponds to 2.17 from the table, Part seeing an area of .8997. We look at the table, we're going to get 1.28 Which has that area and part D. We're going to backtrack the area 4.011 and the value here is -2.29. So we have our four values of z star.


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SUMMARY OUTPUTRegression Statistics Multiple R 0.93757569 R Square 0.879048175 Adjusted R 0.874728467 Square Standard 46.18909524 ErrorObservations 30ANOVASignificancedfSSMS2.27552E- 434147.3895 434147.3895 203.4971275 14RegressionResidual Total28 2959736.11054 2133.432519 493883.5Standard Coefficients Errort StatP-valueIntercept16.29149581 33.61615683 0.484632907 0.631706505 48.187734 3.377982246 14.26524194 2.27552E-14
SUMMARY OUTPUT Regression Statistics Multiple R 0.93757569 R Square 0.879048175 Adjusted R 0.874728467 Square Standard 46.18909524 Error Observations 30 ANOVA Significance df SS MS 2.27552E- 434147.3895 434147.3895 203.4971275 14 Regression Residual Total 28 29 59736.11054 2133.432519 493883.5 Stand...

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