Respuesta :
Answer:
The standard deviation of the mean weight of the salmon in the boxes sold to restaurants is of 2 lbs.
The standard deviation of the mean weight of the salmon in cartons sold to grocery stores is of 1 lb.
The standard deviation of the mean weight of the salmon in in pallets sold to discount outlet stores is of 0.3636lb.
Step-by-step explanation:
The Central Limit Theorem estabilishes that, for a normally distributed random variable X, with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the sampling distribution of the sample means with size n can be approximated to a normal distribution with mean [tex]\mu[/tex] and standard deviation [tex]s = \frac{\sigma}{\sqrt{n}}[/tex].
In this question, we have that:
[tex]\sigma = 4[/tex]
To restaurants:
Boxes of 4 salmon, so [tex]n = 4[/tex]
Then
[tex]s = \frac{\sigma}{\sqrt{n}} = \frac{4}{\sqrt{4}} = 2[/tex]
The standard deviation of the mean weight of the salmon in the boxes sold to restaurants is of 2 lbs.
To grocery stores:
Cartons of 16 salmon, so [tex]n = 16[/tex]
Then
[tex]s = \frac{\sigma}{\sqrt{n}} = \frac{4}{\sqrt{16}} = 1[/tex]
The standard deviation of the mean weight of the salmon in cartons sold to grocery stores is of 1 lb.
To discount outlet stores:
Pallets of 121 salmon.
Then
[tex]s = \frac{\sigma}{\sqrt{n}} = \frac{4}{\sqrt{121}} = 0.3636[/tex]
The standard deviation of the mean weight of the salmon in in pallets sold to discount outlet stores is of 0.3636lb.
