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jhevtan
Aug 13, 2014, 07:19 PM
You want to determine how many bunches of radishes you want to buy each day. Here's what you know.
1. You sell radishes (the sale price) for a dollar a bunch.
2. You purchase radishes (the cost price) for 50 cents a bunch to sell
3. A farmer will purchase the old radishes for a dime a bunch to grind up for his farm animals.
4. You sell about 20 bunches a day.
5. The standard deviation (that give and take of how many you sell every day... some days 17, some days 25, some days 19) is 3.
So, you'll want to know based on the above, how many bunches your next order should be. (50 points).
First, find the cost per unit or sale price minus cost price = C(u)
Second, find the difference of the cost price minus the price the famer will pay = C(o)
Third, find the purchase requirements by doing these steps:
a. C(u)/C(o) +C(u) = P (only go out to four decimal places) (HINT: This must be less than one

smoothy
Aug 13, 2014, 07:32 PM
Since we don't do your assignments, we only help once you have shown us your work FIRST. WHat is your answer.

jhevtan
Aug 13, 2014, 08:42 PM
Thank you for responding, the answer I came up with is incorrect since it has to be under 1
1.00-.50=.50 for c(u)
.50-.10=.40 c(o)

.50/.40+.50= 1.75

ArcSine
Aug 17, 2014, 08:47 AM
Kind of late for the party, but FWIW, your probability "ceiling" for a problem like this is given by the ratio
L / (L + P)

where L and P are, respectively, the net loss incurred on an unsold unit and the net profit earned on a sold unit (here, 0.4 and 0.5).

You're coming out of the starting gate a little sideways by (1) using P instead of L in the numerator; and (2) mis-stating your denominator (avoided by paying close attention to parentheses).

Ah-ite, back to you.

(Ya know, on second thought, I suppose you could use P in the num'r, just keep in mind that you're then getting the complement of the probability "ceiling", and so handle your z-score calcs accordingly. But the bit about the denominator stands.)