Solving Percentile Question: Find Canoe Carrying Capacity

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Homework Help Overview

The problem involves determining the carrying capacity of a canoe based on the weight distribution of individuals on a Pacific island, which is normally distributed with a specified mean and standard deviation. The goal is to ensure the canoe can accommodate 85% of the population's weight.

Discussion Character

  • Exploratory, Assumption checking, Mathematical reasoning

Approaches and Questions Raised

  • Participants discuss the use of percentiles and Z-scores to find the carrying capacity. There is uncertainty about the correct setup and calculations involved in determining the appropriate weight limit for the canoe.

Discussion Status

Some participants have provided calculations based on the Z-score and the normal distribution parameters, while others express doubt about the accuracy of the results. The discussion is ongoing, with different interpretations of the calculations being explored.

Contextual Notes

Participants are navigating the constraints of using statistical methods for a real-world application, questioning the assumptions behind the normal distribution and the interpretation of the Z-score in this context.

needhelp83
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I am not exactly sure what I am supposed to do to solve this problem:

The weight of people in a certain pacific island is normally distributed with a mean of 175 lb and a standard deviation of 33 lb. They want to design a one person canoe that will be able to serve 85% of the island's people. What should be the carrying capacity of the canoe?

Ok I know I wouldn't use a CI to solve this problem, but I am not sure what the exact setup would be. I have a hunch that I would use the percentiles where the Z critical value would be 1.04 using Z 15.

Where do I go from there?
 
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Use the percent given to find the corresponding z score. You have population mean & standard deviation, as well as the z score, so you can solve for the unknown
 
Ok...

[tex]z= \frac{(x-\mu)}{\sigma}[/tex]

[tex]1.04= \frac{(x- 175)}{33}[/tex]

x = 209.32

This doesn't sound right to me at all with the s.d. and mean?
 
It looks right to me because if you were to go only one standard deviation up (z = 1), the x score would be 175 (M) + 33 (1 standard dev) = 208. At the 85th percentile, your z score (1.04) is just a little bit higher than that.
 

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