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A food company sells salmon to various customers. The mean weight of the salmon is 31 lb with a standard deviation of 2 lbs. The company ships them to restaurants in boxes of 4 salmon, to grocery stores in cartons of 16 salmon, and to discount outlet stores in pallets of 100 salmon. To forecast costs, the shipping department needs to estimate the standard deviation of the mean weight of the salmon in each type of shipment. Complete parts (a) and (b) below. Find the standard deviations of the mean weight of the salmon in each type of shipment. Find the standard deviation of the mean weight of the salmon in the boxes sold to restaurants. SD (ybar) = (Round to two decimal places as needed.) Find the standard deviation of the mean weight of the salmon in the cartons sold to grocery stores. SD (ybar) = (Round to two decimal places as needed.) Find the standard deviation of the mean weight of the salmon in the pallets sold to outlet stores. SD (ybar) = (Round to two decimal places as needed.) The distribution of the salmon weights turns out to be skewed to the high end. Would the distribution of shipping weights be better characterized by a Normal model for the boxes of pallets? Explain. Choose the correct answer below. The boxes, because, regardless of the underlying distribution, the sampling distribution of the mean approaches the Normal model as the sample size increases. The pallets, because, as long as the underlying distribution is Normal, the sampling distribution of the mean approaches the Normal model as the sample size increases. The pallets, because, regardless of the underlying distribution, the sampling distribution of the mean approaches the Normal modal as the sample size increases. The boxes, because, as long as the underlying distribution is Normal, the sampling distribution of the mean approach the Normal model as the sample size increases.

Respuesta :

Answer:

a) 1 lb , 0.5 lb, 0.2 lb

b) The distribution of shipping weights (as discussed above) will be better characterized by  a Normal model for the pallets than others, since,

regardless of the underlying distribution, the sampling distribution of the mean approaches the Normal model as the sample size increases.

Step-by-step explanation:

Standard deviation of the variable Y( weight of each salmon)

=σ = 2 lbs.

Standard deviation of the variable , mean weight of each salmon in boxes( sent to restaurants), from the formula,

[tex]\overline {Y_{n}}= \frac{\sigma}{\sqrt{{\textrm {(sample size = n)}}}}[/tex]-----------(1)

is obtained as [tex]\frac{2}{2}[/tex]  lbs

                        = 1 lb -------------------(2)

Standard deviation of the variable , mean weight of each salmon in cartons( sent to grocery stores), from the formula (1)

is given by,

  [tex]\frac{2}{4}[/tex] lbs

= 0.5 lb ---------------------(3)

Standard deviation of the variable , mean weight of each salmon in pallets ( sent to discount outlet stores), from the formula (1) , is given by,

  [tex]\frac{2}{10}[/tex]  lbs

= 0.2 lb -----------------(4)

The distribution of shipping weights (as discussed above) will be better characterized by  a Normal model for the pallets than others since,

regardless of the underlying distribution, the sampling distribution of the mean approaches the Normal model as the sample size increases.

 

                   

As sample size increases, the distribution tends to normal distribution, regardless of underlying distribution. Part(a): The mean and standard deviation for boxes, cartons, and pallets are:

  • Standard deviation of the mean weight of salmon in boxes = 1 lb
  • Standard deviation of the mean weight of salmon in cartons = 0.5 lbs
  • Standard deviation of the mean weight of salmon in pallets = 0.2 lbs

Part(b): The pallets, because, regardless of the underlying distribution, the sampling distribution of the mean approaches the normal modal as the sample size increases.

How does the sampling distribution behaves if size of sample increases?

Doesn't matter what the underlying distribution is, if the sample size is denoted by 'n', then, as [tex]n \rightarrow \infty[/tex], the sampling distribution tends to normal distribution.

How to find the standard deviation of sample from the population standard deviation if population is normally distributed?

Supposing the population is pertaining normal distribution.

Suppose we've:

  • [tex]\sigma[/tex] = population standard deviation
  • n = sample size
  • s = sample standard deviation

Then, we can estimate value of s by:

[tex]s = \dfrac{\sigma}{\sqrt{n}}[/tex]

For this case, we've the population standard deviation as [tex]\sigma = 2[/tex] lbs.

Finding standard deviation for all three cases:

  • Case 1: Boxes of 4 salmon

We've n = 4, thus:

[tex]s = \dfrac{\sigma}{\sqrt{n}} = \dfrac{2}{\sqrt{4}} = 1 \: \rm lb[/tex]

  • Case 2: Cartons of 16 salmon

We've n = 16, thus:

[tex]s = \dfrac{\sigma}{\sqrt{n}} = \dfrac{2}{\sqrt{16}} = 0.5 \: \rm lbs[/tex]

  • Case 3: Pallets of 100 salmon

We've n = 100, thus:

[tex]s = \dfrac{\sigma}{\sqrt{n}} = \dfrac{2}{\sqrt{100}} = 0.2 \: \rm lbs[/tex]

Since sample size of pallets is biggest, thus, it is closest among other samples for behaving as normal distribution.

Thus, we get the evaluation of sub parts as:

Part(a): The mean and standard deviation for boxes, cartons, and pallets are:

  • Standard deviation of the mean weight of salmon in boxes = 1 lb
  • Standard deviation of the mean weight of salmon in cartons = 0.5 lbs
  • Standard deviation of the mean weight of salmon in pallets = 0.2 lbs

Part(b): The pallets, because, regardless of the underlying distribution, the sampling distribution of the mean approaches the normal modal as the sample size increases.

Learn more about normal distribution here:
https://brainly.com/question/12421652

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