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2. (20 points) A certain market has both an express checkout line and a superexpress checkout line. Let X1 denote the number customers in line at the express checkout at a particular time of day and let X2 denote the number of customers in line at the superexpress checout at the same time. Suppose the joint pmf of X1 and X2 is x2 p(x, y) 0 1 2 3 0 0.08 0.07 0.04 0 x1 1 0.06 0.15 0.05 0.04 2 0.05 0.04 0.1 0.06 3 0 0.03 0.04 0.07 4 0 0.01 0.05 0.06 (a) What is the probability exactly one customer is in each checkout line? (3 points) (b) What is the probability the same number of customers are in each checkout line? (5 points) (c) Let A denote the event that there are at least two more customers in one line than the other. Express A in terms of X1 and X2 and calculate the probability of this event. (7 points) (d) What is the probability the total number of customers in the two lines is at least four? (

          2. (20 points) A certain market has both an express checkout line and a superexpress checkout line. Let X1 denote the number customers in line at the express checkout at a particular time of day and let X2 denote the number of customers in line at the superexpress checout at the same time. Suppose the joint pmf of X1 and X2 is x2 p(x, y) 0 1 2 3 0 0.08 0.07 0.04 0 x1 1 0.06 0.15 0.05 0.04 2 0.05 0.04 0.1 0.06 3 0 0.03 0.04 0.07 4 0 0.01 0.05 0.06 (a) What is the probability exactly one customer is in each checkout line? (3 points) (b) What is the probability the same number of customers are in each checkout line? (5 points) (c) Let A denote the event that there are at least two more customers in one line than the other. Express A in terms of X1 and X2 and calculate the probability of this event. (7 points) (d) What is the probability the total number of customers in the two lines is at least four? (

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Elementary Statistics a Step by Step Approach

Allan G. Bluman 9th Edition

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02/19/2023

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2. (20 points) A certain market has both an express checkout line and a superexpress checkout line. Let X1 denote the number customers in line at the express checkout at a particular time of day and let X2 denote the number of customers in line at the superexpress checout at the same time. Suppose the joint pmf of X1 and X2 is x2 p(x, y) 0 1 2 3 0 0.08 0.07 0.04 0 x1 1 0.06 0.15 0.05 0.04 2 0.05 0.04 0.1 0.06 3 0 0.03 0.04 0.07 4 0 0.01 0.05 0.06 (a) What is the probability exactly one customer is in each checkout line? (3 points) (b) What is the probability the same number of customers are in each checkout line? (5 points) (c) Let A denote the event that there are at least two more customers in one line than the other. Express A in terms of X1 and X2 and calculate the probability of this event. (7 points) (d) What is the probability the total number of customers in the two lines is at least four? (

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Transcript

00:01 Hello everyone.

00:03 So in this question the two variables x1 and x2.

00:08 X1 denotes the number of customers, number of customers in line at the express checkout whereas x2 denote the number of customers in line at super express checkout number of customers in line at super express checkout.

01:00 Check out.

01:06 Okay, so in the question, part a, b and c are solved.

01:11 Coming to part, second part of c that is probability of a.

01:17 So c, in c we have to find probability of a.

01:23 Now event a is defined as x1 is greater than equals to 2 plus x2.

01:40 Union x2 greater than equals to 2 plus x1 right now we have to find the probability of this event a which will be now let us note down for the outcomes possible for each of these two cases now case one is x1 is greater than equals to 2 plus x2 now in this case let these be the values of x1 and these be the values of x2.

02:17 So if x2 equals to 0, so we get x1 is greater than equals to 2.

02:24 So possible values are 2, 3, 4.

02:38 Okay.

02:41 Now, if x2 equals to, okay, the values of x1 range from 0 to 4, whereas the values of x2 range from 0 to 3.

02:57 Right? so now, if x2 equals to 0, the values of 0, of x1 possible are 2, 3, 4.

03:03 Now, if x2 equals to 1, then our equation becomes x1 is greater than equals to 3, that is 3 and 4.

03:10 If x2 equals to 2, then our equation becomes x1 is greater than equals to 4, so 4 is 4 is possible.

03:18 Now, coming to next case, that is x2 is greater than equals to 2 plus x1.

03:29 Again, doing in the same way, if we take x1 equals to, this is our x1 values.

03:37 This is our x2 values.

03:38 If we take x1 equals to 0, the x2 is greater than equals to 2, that is 2 and 3...

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