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

If
are indentically distributed and independent, like in the situation of the problem, then the variance of
will be the sum of the variances, in other words, it will be n times the variance of
.
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
and as a result, the standard deviation of
is the standard deviation of
divided by
.
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

- Grocery stores: Each carton has 49 salmon, thus the standard deviation is

- Discount outlet stores: Each pallet has 64 salmon, as a result, the standard deviation is

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.
1. 4
2. 6
3. 2
I had got it wrong on study island and it gave me the right answer
1. Check the picture attached. We can calculate the area of the orange region G by dividing it into right triangles with both legs equal to 1.
Area of one triangle is 1/2*1*1=1/2 units square.
There are 9 in total, so area G = 9/2 units square.
2. The square has side length equal to 10, so the square area is 100 units square.
3. P(landing in G)=Area of G/total area= (9/2)/100=9/200= 0.045