The bubba chump shrimp company processes and packages shrimp for sale to wholesale seafood distributors. the shrimp are transported to the main plant by trucks that carry 1,000 pounds (lb) of shrimp. once the continuous flow processing of the shrimp begins, no inventory is allowed in buffers due to spoilage and all of the shrimp must be processed within 12 hours to prevent spoilage. the processing begins at the sorter, where the trucks dump the shrimp onto a conveyor belt that feeds into the sorter, which can sort up to 500 lb per hour. the shrimp then proceed to the desheller, which can process shrimp at the rate of 400 lb per hour. however, after 3 hours and 45 minutes of processing, the desheller must be stopped for 15 minutes to clean out empty shrimp shells that have accumulated. the veins of the shrimp are then removed in the deveining area at a maximum rate of 360 lb per hour. the shrimp proceed to the washing area, where they are processed at 750 lb per hour. finally, the shrimp are packaged and frozen. what is the daily process capacity of the processing plant (excluding the packaging and freezing)?

a.

P(A > 2,659)= 0.5239

P(B > 2,659)= 0.4325

Process A will be investigated more often.

b.

P(A > 2,659) = 0.8186

P(B > 2,659)= 0.6406

Process A will be investigated more often.

d.

K= 5.7741

P(B> 5.77409)= 0.51595

Step-by-step explanation:

Hello!

The cost variance is defined as the difference between a budget cost and an actual cost.

There are two processes of interest:

Process A: stable: easily controlled production process with little fluctuation invariance.

Process B: random events: the equipment is more sensitive and prone to breakdown, the raw material prices fluctuate more,...

Since he's paying more attention to B, even though both processes have the same probability of developing inefficiency, he decides to investigate each process if their variances exceed $2659 regardless of what the process is.

In control:

Process A

Mean cost variance (in control) $ 3

The standard deviation of cost variance (in control) $5,473

Process B

Mean cost variance (in control) $ 1

The standard deviation of cost variance (in control) $9,743

Out of control

Process A

Mean cost variance (out of control) $7,651

The standard deviation of cost variance (out of control) $5,473

Process B

Mean cost variance (out of control) $ 6,169

The standard deviation of cost variance (out of control) $9,743

a) The cost variances of both processes have a normal distribution, they will be investigated if the cost variance is greater than $2,659, symbolically:

Both are "in control" so you have to use the given data for this characteristic.

P(A > 2,659)

And

P(B > 2,659)

Since both variables have a normal distribution you can calculate these probabilities using the standard normal distribution as:

For process A

P(A > 2,659) = P(Z> )

P(Z>-0.06)= 1 - P(Z ≤ -0.06) = 1 - 0.47608= 0.52392

For process B

P(B > 2,659)= P(Z>)

P(Z> 0.17)= 1 - P(Z ≤ 0.17)= 1 - 0.56749= 0.43251

Process A will be investigated more often.

b. For this item you have to calculate the probability that the process are going to be investigated, asuming the cost variances have normal distribution, but the processes are out of control. This means you have to use the values of the second data set.

For process A

P(A > 2,659) = P(Z> )

P(Z>-0.91)= 1 - P(Z ≤ -0.91) = 1 - 0.18141= 0.81859

For process B

P(B > 2,659)= P(Z>)

P(Z> -0.36)= 1 - P(Z ≤ -0.36)= 1 - 0.35942= 0.64058

Process A will be investigated more often.

c. Looking at the probabilities of both process "out of control", process A has a grater probability than process B of having a variance cost greater than $2.659. That's why process A will be investigated more often.

d.

In this case, they want to know what is the number of cost variance that leads process B "in control" to have a probability of 0.3121 of being investigated. In this case, you have to do a reverse standardization, this means, you know the probability and need to look for its corresponding value.

P(Z>k)= 0.3121

P(Z>k) = 1 - P(Z ≤ b)

1 - P(Z ≤ k)= 0.3121

P(Z ≤ k)= 1 - 0.3121

P(Z ≤ k)= 0.6879

Now you have to look in the table for the number "b"that corresponds to the cummulative probability of 0.6879,

k= 0.49

Now k=

0.49*9.743= K - 1

4.77409 + 1 = K

K= 5.77409

Process B will be investigated if its cost variance is greater than $5.77409

Using this new policy you need to calculate the probability of B "out of control" to be investigated:

P(B> 5.77409) = P(Z>)

P(Z> -0.04)= 1 - P(Z ≤ -0.04)= 1 - 0.48405= 0.51595

I hope it helps!

Main route:

Smelting --> Rolling --> Converting --> Sheared Sheet

Secondary route     -->

Smelting --> Rolling -->  rolled sheet

1) Smelting trasnferred materials into Rolling

2) It will be part of that department work in process inventory

"WIP SConverting debit then factorty overhead credit"

Later it will be transferred out as a complete process therefore,

Finished good Inventory - Shared sheet

3) the Smelting department transfer the entire of his output into Rolling department

4) the finished good will become cost of good solg once they are sold.

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Explanation:

We have to read he description of how the processing system works and check to whichdeparmtent are the goods being transferred or sold.