Probability using box and whisker

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SUMMARY

The discussion centers on calculating the probability that two randomly selected salmon lengths from a box and whisker plot are both greater than the lower quartile value. With 39 unique salmon lengths, the first selection has a probability of 29/39, and the second selection, after one salmon is removed, has a probability of 28/38. The final calculated probability is approximately 54.8%. An alternative estimation method using 0.75^2 yields a probability of 56.3%, which is also considered reasonable for standardized test scenarios.

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  • Understanding of box and whisker plots
  • Basic probability concepts
  • Familiarity with quartiles and their significance
  • Knowledge of standardized testing formats
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  • Study the properties of box and whisker plots in detail
  • Learn advanced probability calculations involving combinations
  • Explore the implications of quartiles in statistical analysis
  • Review standardized test strategies for open-ended questions
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Students preparing for standardized tests, educators teaching statistics, and anyone interested in understanding probability through practical examples.

joel amos
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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 over-complicating the problem. What would you give as the answer?
 
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I would do the same.

If 1-2% do not matter, 0.75^2 = 56.3% gives a reasonable estimate.
 
Yeah, that's what a lot of people did. This was a standardized test open-ended question. I wonder which response they're looking for.
 

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