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bazaltina [42]
2 years ago
8

A food company sells salmon to various customers. The mean weight of the salmon is 37 lb with a standard deviation of 2 lbs. The

company ships them to restaurants in boxes of 9 ​salmon, to grocery stores in cartons of 49 ​salmon, and to discount outlet stores in pallets of 64 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.
Mathematics
1 answer:
brilliants [131]2 years ago
7 0

Answer:

The standard deviation for the mean weigth of Salmon is 2/3 lbs for restaurants, 2/7 lbs for grocery stores and 1/4 lbs for discount order stores.

Step-by-step explanation:

The mean sample of the sum of n random variables is

\overline{X} = \frac{X_1+X_2+...+X_n}{n}

If X_1, ..., X_n are indentically distributed and independent, like in the situation of the problem, then the variance of X_1 + .... + X_n will be the sum of the variances, in other words, it will be n times the variance of X_1 .

However if we multiply this mean by 1/n (in other words, divide by n), then we have to divide the variance by 1/n², thus \overkine{X} = \frac{V(X_1)}{n} and as a result, the standard deviation of \overline{X} is the standard deviation of X_1 divided by \sqrt{n} .

Since the standard deviation of the weigth of a Salmon is 2 lbs, then the standard deviations for the mean weigth will be:

  • Restaurants: We have boxes with 9 salmon each, so it will be \frac{2}{\sqrt{9}} = \frac{2}{3}
  • Grocery stores: Each carton has 49 salmon, thus the standard deviation is \frac{2}{\sqrt{49}} = \frac{2}{7}
  • Discount outlet stores: Each pallet has 64 salmon, as a result, the standard deviation is \frac{2}{\sqrt{64}} = \frac{1}{4}

We conclude that de standard deivation of the mean weigth of salmon of the types of shipment given is: 2/3 lbs for restaurants, 2/7 lbs for grocery stores and 1/4 lbs for discount outlet stores.

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