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.

You are watching: A certain market has both an express

**Problem:**1E2E3E4E5E6E7E8E9E10E11E12E13E14E15E16E17E18E19E20E21E22E23E24E25E26E27E28E29E30E31E32E33E34E35E36E37E38E39E40E41E42E43E44E45E46E47E48E49E50E51E52E53E54E55E56E57E58E59E60E61E62E63E64E65E66E67E68E69E70E71E72E73E74E75E76E77E78E79E80E81E82E83E84E85E86E87E88E89E90E91E92E93E94E95E96E

A certain industry has actually both an express checkout line and a superexpush checkout line. Let

*X*1 signify the number of customers in line at the expush checkout at a particular time of day, and let

*X*2 represent the variety of customers in line at the superexpress checkout at the same time. Suppose the joint pmf of

*X*1 and also

*X*2 is as offered in the accompanying table. ... a. What is

*P*(

*X*1 = 1,

*X*2 = 1), that is, the probcapability that tbelow is exactly one customer in each line? b.

**What is**

*P*(

*X*1 =

*X*2), 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

*X*1 and also

*X*2, 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 1**We are provided through a joint probcapability distribution function

**Step 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,