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October 20th, 2013, 02:11 AM   #1
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Probability Question Help. Please!

You have been hired as a consultant by a strawberry farmer. After being awarded the con-tract, you receive the following letter.

To: Recently Hired Statistician
From: Morgan Hill Berry Growers
Please advise us on which company to use as our strawberry distributor. Four highly rec-ommended distributors have provided us with statistical data on the weekly prices for one load of strawberries per week for a ten-week period last year. Prices fluctuate according to availability, and we would like to use the company with the lowest overall price and the least amount of fluctuation. We would like your written report showing your results and a detailed recommendation as to which company we should choose. Thank you.

Data in attachment: [attachment=0:3qoxi2vd]maths.png[/attachment:3qoxi2vd]

Any help is appreciated. Thank you
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October 21st, 2013, 03:32 AM   #2
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Re: Probability Question Help. Please!

I would suggest that you find the mean and standard deviation for each of the companies

Week FP FnF always Berry
mean 327 318 323 318
sd 20.8 20.3 19.6 22.6

I did this and these were my results. (The table does not diplay properly). I make lots of mistakes so you should check.

FnF and Berry have the cheapest average price $318
Fnf have a smaller standard deviation than Berry so there price fluctuates less.
So FnF might be your choice.

always Ripe has the least fluctuation in price but they are not as cheap.

Maybe you are supposed to do more tests. Anyway, this is a starting point.
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October 21st, 2013, 04:37 AM   #3
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Re: Probability Question Help. Please!

Difficult one this with few data to go by. If fluctuation is your main criterion, why not go for the company with the least range in which case the forth column company is eliminated because it has a range of $70. Columns 2 and 3 give the least range.
At this stage i would go for either columns 2 or 3 since the median value is $320 and both have the same range of $60. The median value for column 4 is $315
I realise that using the median value in this case is a bit of anomaly but it will do for starters
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