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Probability using box and whisker 
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#1
Jan1013, 04:24 PM

P: 70

The question:
A box and whisker plot is made from the lengths of 39 salmon. No two lengths are the same. Two of the salmon are picked at random. What is the probability that they will both be longer than the lower quartile value. My solution: x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x x The x's stand for the 39 lengths. The underlined ones are (respectively) the minimum, lower quartile, median, upper quartile, and maximum. There are 29 x's above the lower quartile value (including the second x that was averaged to find that value). Therefore, the probability for the first pick would be 29/39. For the second pick, however, the quantities have been lowered by one since one salmon has already been picked. So the second probability is 28/38. When multiplied together, I got a final probability of 54.8% Is this correct, or am I somehow overcomplicating the problem. What would you give as the answer? 


#2
Jan1013, 04:48 PM

Mentor
P: 12,045

I would do the same.
If 12% do not matter, 0.75^2 = 56.3% gives a reasonable estimate. 


#3
Jan1013, 04:55 PM

P: 70

Yeah, that's what a lot of people did. This was a standardized test openended question. I wonder which response they're looking for.



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