 steustatiushistory.orgSolutions Probcapacity and also Statistics for Engineering and also Science, 9th Edition

Authors: Jay L. Devore

ISBN-13: 978-1305251809

See our solution for Concern 3E from Chapter 5 from Devore's Probability and Statistics for Engineering and also Science.

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Problem:1E2E3E4E5E6E7E8E9E10E11E12E13E14E15E16E17E18E19E20E21E22E23E24E25E26E27E28E29E30E31E32E33E34E35E36E37E38E39E40E41E42E43E44E45E46E47E48E49E50E51E52E53E54E55E56E57E58E59E60E61E62E63E64E65E66E67E68E69E70E71E72E73E74E75E76E77E78E79E80E81E82E83E84E85E86E87E88E89E90E91E92E93E94E95E96E
A certain industry has actually both an express checkout line and a superexpush checkout line. Let X1 signify the number of customers in line at the expush checkout at a particular time of day, and let X2 represent the variety of customers in line at the superexpress checkout at the same time. Suppose the joint pmf of X1 and also X2 is as offered in the accompanying table. ... a. What is P(X1 = 1, X2 = 1), that is, the probcapability that tbelow is exactly one customer in each line? b. What is P(X1 = X2), that is, the probability that the numbers of customers in the two lines are identical? c. Let A represent the event that there are at least 2 more customers in one line than in the other line. Express A in regards to X1 and also X2, and also calculate the probcapability of this event. d. What is the probability that the full number of customers in the two lines is precisely four? At leastern four?
Tip 1 of 2

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Step 2 of 2

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Step 1We are provided through a joint probcapability distribution functionStep 2: (a) The probcapability that tright here is exactly one customer in each line: The probcapacity corresponds to row and column matching to \$X_1=1\$ and also \$X_2=1\$Step 3: (b) The probability that variety of customers are exact same in both lines is provided by the amount of diagonal elements of the joint pdf:<eginarraylPleft( X_1 = X_2 ight) = pleft( 0,0 ight) + pleft( 1,1 ight) + pleft( 2,2 ight) + pleft( 3,3 ight)\ = 0.08 + 0.15 + 0.10 + 0.07\ = 0.40endarray>As such, Tip 4: (c)Let A signify the event that there are at leastern two more customers than other line: The possible combinations of variables are:<eginarraylX_2 = 0:,,,X_1 = 2,,3,,4\X_2 = 1:,,,X_1 = ,3,,4\X_2 = 2:,,,X_1 = 0,,4\X_2 = 3:,,,X_1 = 0,1endarray>Required Probability:<eginarraylPleft( A ight) = pleft( 0,3 ight) + pleft( 0,4 ight) + pleft( 1,3 ight) + pleft( 2,0 ight) + pleft( 3,0 ight) + pleft( 3,1 ight) + pleft( 4,0 ight) + pleft( 4,1 ight) + pleft( 4,2 ight)\ = 0.04 + 0.00 + 0.04 + 0.05 + 0.00 + 0.03 + 0.00 + 0.01 + 0.05\ = 0.22endarray>As such, Tip 5: (d) Let A denote the event that there are 4 customers in both lines combined: The feasible combicountries of variables are:<eginarraylX_2 = 1:,,,X_1 = ,3\X_2 = 2:,,,X_1 = ,2\X_2 = 3:,,,X_1 = 1\X_2 = 4:,,,X_1 = 0endarray>The compelled probcapacity is:<eginarraylPleft( A ight) = pleft( 1,3 ight) + pleft( 2,2 ight) + pleft( 3,1 ight) + pleft( 4,0 ight)\ = 0.04 + 0.10 + 0.03 + 0.00\ = 0.17endarray>Because of this, Step 6: Let A signify the occasion that tbelow are at least 4 customers in both lines combined: The possible combinations of variables are:<eginarraylX_2 = 1:,,,X_1 = ,3\X_2 = 2:,,,X_1 = ,2,3\X_2 = 3:,,,X_1 = 1,2,3\X_2 = 4:,,,X_1 = 0,1,2,3endarray>The forced probcapacity is:<eginarraylPleft( A ight) = pleft( 1,3 ight) + pleft( 2,2 ight) + pleft( 2,3 ight) + pleft( 3,1 ight) + pleft( 3,2 ight) + pleft( 3,3 ight) + pleft( 4,0 ight) + pleft( 4,1 ight) + pleft( 4,2 ight) + pleft( 4,3 ight)\ = 0.04 + 0.10 + 0.06 + 0.03 + 0.04 + 0.07 + 0.00 + 0.01 + 0.05 + 0.06\ = 0.46endarray>Thus,