1. Question Details DevoreStat9 3.AE.003. [3232943]. Two gas stations are locate
ID: 3048959 • Letter: 1
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1. Question Details DevoreStat9 3.AE.003. [3232943]. Two gas stations are located at a certain intersection. Each one has six gas pumps. Consider the experiment in which the number of pumps in use at a particular time of day is determined for each of the stations. An experimental outcome specifies how many pumps are in use at the first station and how many are in use at the second one. One possible outcome is (2, 2), another is (4,1), and yet another is (1, 4). Define rv's X, Y, and U by Example 3.3 X = the total number of pumps in use at the two stations Y = the difference between the number of pumps in use at station 1 and the number in use at station 2 U = the maximum of the numbers of pumps in use at the two stations ) so we If this experiment is performed and s = (2, 3) results, then x(( 2, 3)) = 2 ? 3- say that the observed value of X was . y = 2 71 3 Similarly, the observed value of Y would be and the observed value of u would be u = max( 2, 3) = 2. Question Details DevoreStat9 3.AE.008. [3272020]. Example 3.8 Six boxes of components are ready to be shipped by a certain supplier. The number of defective components in each box is as follows 1 2 3 4 5 6 Box Number of defectives 02 012 0 One of these boxes is to be randomly selected for shipment to a particular customer. Let X be the number of defectives in the selected box. The three possible X values are 0, 1, and 2. Of the six equally likely simple events, three result in X-0, one in X 1, and the other two in X-2. Then p(0)-P(X-0)-P(box 1 or 3 or 6 is sent) . p(1) PX 1) P(box 4 is sent) p(2) PX 2)P(box 2 or 5 is sent) (Enter your answer as a fraction.) (Enter your answer as a fraction.) (Enter your answer as a fraction.) That is, a probability of 0.500 is distributed to the X value on the X value values of X along with their probabilities collectively specify the pmf. If this experiment were repeated over and over again, in the long run X-0 would occur one-half of the time, X- 1 one-sixth of the time and X = 2 one-third of the time a probability of 0.167 is placed and the remaining probability, 0.333, is associated with the X value 2. TheExplanation / Answer
3.3
X(2,3)=2+3=5
observed value of X was x=5
observed value of Y was y=2-3=-1
observed value of U was u=max(2,3)=3
3.8
each of the boxes are equally likely to be selected and so each box has probability of being selected equal to 1/6, hence :
P(X=0)=P(box1 or box3 or box6 selected)=P(box 1)+P(box 3)+P(box 6)=3/6=0.5
P(X=1)=P(box 4)=1/6=0.167
P(X=2)=P(box 2 or box 5) = P(box 2)+P(box 5)=2/6=0.333
probability 0.500 associated with x=0, 0.167 associated with x=1, and the rest 0.333 is associated with x=2
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