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Question 3: The following schematic diagram depicts a production process used in manufacturing of aluminium cans. The initial input into the process consists of coiled aluminium sheets, approximately 0.25 mm thick. In a process known as ‘‘cupping’’, these sheets are uncoiled and shaped into can bodies, which are cylinders that are closed on the bottom and open on top. These can bodies are then washed and sent to the printer, which prints the label on the can. In practice, there will be several printers in a line and the schematic diagram shows a line with three printers. The printer deposits the cans onto pallets, which are wooden structures that hold 7140 cans each. The cans next go to be filled. Some fill lines can accept cans directly from the pallets, but others can accept them only from cell bins, which are large containers holding approximately 100,000 cans each. To use these fill lines, the cans must be transported from the pallets to cell bins, in a process called ‘‘depalletizing’’. Although in practice there are several fill lines, the schematic diagram presents a case where there are two fill lines, one of which will accept cans directly from the pallets, and the other of which will not. In the filling process, the cans are filled and the can top is seamed on. The cans are then packaged and shipped to distributors. It is desired to estimate the probability that the process will function for one-day without failing. Assume that the cupping process has probability 0.975 of functioning successfully for one-day. Since this process component is denoted by ‘‘A’’ in the schematic, we will express this probability as P(A) = 0.975. Assume that the other process components have the following probabilities of functioning successfully during a one-day period: P(B) = 0.93, P(C) = P(D) = P(E) = 0.97, P(F) = P(G) = 0.98, and P(H) = 0.92. If it is assumed that the process components function independently, compute the probability that the process functions successfully for one-day. Cup Wash Print Depalletize Fill

Question 3:
The following schematic diagram depicts a production process used in manufacturing of aluminium cans. The initial input into the process consists of coiled aluminium sheets, approximately 0.25 mm thick. In a process known as ‘‘cupping’’, these sheets are uncoiled and shaped into can bodies, which are cylinders that are closed on the bottom and open on top. These can bodies are then washed and sent to the printer, which prints the label on the can. In practice, there will be several printers in a line and the schematic diagram shows a line with three printers. The printer deposits the cans onto pallets, which are wooden structures that hold 7140 cans each. The cans next go to be filled. Some fill lines can accept cans directly from the pallets, but others can accept them only from cell bins, which are large containers holding approximately 100,000 cans each. To use these fill lines, the cans must be transported from the pallets to cell bins, in a process called ‘‘depalletizing’’. Although in practice there are several fill lines, the schematic diagram presents a case where there are two fill lines, one of which will accept cans directly from the pallets, and the other of which will not. In the filling process, the cans are filled and the can top is seamed on. The cans are then packaged and shipped to distributors.
It is desired to estimate the probability that the process will function for one-day without failing. Assume that the cupping process has probability 0.975 of functioning successfully for one-day. Since this process component is denoted by ‘‘A’’ in the schematic, we will express this probability as P(A) = 0.975. Assume that the other process components have the following probabilities of functioning successfully during a one-day period: P(B) = 0.93, P(C) = P(D) = P(E) = 0.97, P(F) = P(G) = 0.98, and P(H) = 0.92. If it is assumed that the process components function independently, compute the probability that the process functions successfully for one-day.

Cup
Wash
Print
Depalletize
Fill

Question 3:
The following schematic diagram depicts a production process used in manufacturing of aluminium cans. The initial input into the process consists of coiled aluminium sheets, approximately 0.25 mm thick. In a process known as ‘‘cupping’’, these sheets are uncoiled and shaped into can bodies, which are cylinders that are closed on the bottom and open on top. These can bodies are then washed and sent to the printer, which prints the label on the can. In practice, there will be several printers in a line and the schematic diagram shows a line with three printers. The printer deposits the cans onto pallets, which are wooden structures that hold 7140 cans each. The cans next go to be filled. Some fill lines can accept cans directly from the pallets, but others can accept them only from cell bins, which are large containers holding approximately 100,000 cans each. To use these fill lines, the cans must be transported from the pallets to cell bins, in a process called ‘‘depalletizing’’. Although in practice there are several fill lines, the schematic diagram presents a case where there are two fill lines, one of which will accept cans directly from the pallets, and the other of which will not. In the filling process, the cans are filled and the can top is seamed on. The cans are then packaged and shipped to distributors.
It is desired to estimate the probability that the process will function for one-day without failing. Assume that the cupping process has probability 0.975 of functioning successfully for one-day. Since this process component is denoted by ‘‘A’’ in the schematic, we will express this probability as P(A) = 0.975. Assume that the other process components have the following probabilities of functioning successfully during a one-day period: P(B) = 0.93, P(C) = P(D) = P(E) = 0.97, P(F) = P(G) = 0.98, and P(H) = 0.92. If it is assumed that the process components function independently, compute the probability that the process functions successfully for one-day.

Cup
Wash
Print
Depalletize
Fill

Added by Sarah D.

Elementary Statistics a Step by Step Approach

Allan G. Bluman 9th Edition

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Solved by Expert Frank Deng

09/07/2022

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Question 3: The following schematic diagram depicts a production process used in manufacturing of aluminium cans. The initial input into the process consists of coiled aluminium sheets, approximately 0.25 mm thick. In a process known as ‘‘cupping’’, these sheets are uncoiled and shaped into can bodies, which are cylinders that are closed on the bottom and open on top. These can bodies are then washed and sent to the printer, which prints the label on the can. In practice, there will be several printers in a line and the schematic diagram shows a line with three printers. The printer deposits the cans onto pallets, which are wooden structures that hold 7140 cans each. The cans next go to be filled. Some fill lines can accept cans directly from the pallets, but others can accept them only from cell bins, which are large containers holding approximately 100,000 cans each. To use these fill lines, the cans must be transported from the pallets to cell bins, in a process called ‘‘depalletizing’’. Although in practice there are several fill lines, the schematic diagram presents a case where there are two fill lines, one of which will accept cans directly from the pallets, and the other of which will not. In the filling process, the cans are filled and the can top is seamed on. The cans are then packaged and shipped to distributors. It is desired to estimate the probability that the process will function for one-day without failing. Assume that the cupping process has probability 0.975 of functioning successfully for one-day. Since this process component is denoted by ‘‘A’’ in the schematic, we will express this probability as P(A) = 0.975. Assume that the other process components have the following probabilities of functioning successfully during a one-day period: P(B) = 0.93, P(C) = P(D) = P(E) = 0.97, P(F) = P(G) = 0.98, and P(H) = 0.92. If it is assumed that the process components function independently, compute the probability that the process functions successfully for one-day.

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00:02 Now here in this question, basically we look at the diagram, which shows kind of a stream line process, right? you have several operations in this process, copying, washing, print, and this, right? and obviously, cede, if any one of them works, they works, right? and while for this, you know, is similar, right? and if f works or both g and h works and then this can work, right? you ask you to calculate the probability, actually, the process functions successfully, right? now, for this purpose, let's first work out the probability for this printer process to work, right? and that, of course, obviously, as not as one of the cde works, then the whole operation work, the printing work, right? so the probability, of course, is given by the success rate, actually, is given by one minus the probability that none of this works, right? so that's, of course, given by, you know, the probability none of them works actually is given by, you know, 0 .03, which is 1 minus 0 .97, right, to the part of 3, right? and you'll find this actually to be given by 1 minus 0 .03 to the cubic, to the cubic, yes.

01:15 And that will be actually 0 .999, right? so actually it's very close to one, basically.

01:25 And how about the process of, how about this, you know, this operation works, what's probably for this operation works? well, that, of course, is going to be, we need to work out, you know, the first the probability of this works, right? well, that's, of course, given by, if you look at this, the probability for this to work is simply the product of this two together, right? so p .g times p .h, that's actually 0 .98 and times 0 .92, right? now, and then, of course, the, we need to find, then the probability that, you know, this process works, of course, is given by 1 minus the probability that this fails, right? and that is given by the probability of f fails and multiplied probability that these two fails, right? so that's given by the probability of this f fares is given by 1 minus 0 .98, that 0 .02 and times the probability that this fails, that's, of course, given by 1 minus 0 .98.

02:31 Times 0 .92, right? so you calculate this, you found this to be given by 1 minus 0 .02 times 1 minus 0 .98 times 0 .92.

02:46 And actually gives us 0 .9748...

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