Possible Solutions, 2015 AP Statistics free response questions, Draft 2

Posted on May 17, 2015 by roughlynormal

Hi Colleagues!   Thanks for the edit suggestions.  Most edits were typos and cleaning up details.  Here is draft #2:  Possible Solutions 2015 AP FRQ

I welcome any critiques, alternate solutions, questions or criticism.

About roughlynormal

I have been a math/statistics teacher for 25 years. I currently teach at an independent school in southern California. I also coach teaching fellows for Math for America - Los Angeles chapter. I love my career, my colleagues, and my friends & family.
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3 Responses to Possible Solutions, 2015 AP Statistics free response questions, Draft 2

  1. Rich Gill says:
    May 18, 2015 at 12:12 pm

    I explained Q6 (f) quite differently than you, and thought I would share it. Frankly, I didn’t really ‘get’ your reason (not saying its not correct, I just don’t get it!); I’m just not convinced that variability is a big issue for this part? Here’s my explanation, let me know your thoughts:

    “If the sample is large enough, then Method 1 would be more likely to result in a sample mean of 6″. A large enough sample should result in a distribution that’s very similar to the population’s distribution. In the population, 50% of the points fall between 5.9″ (Line A) and 6.1″ (Line B) but only 25% of the points fall below 5.9″ and above 6.1″. Method 1 should result in a similar distribution of tortilla diameters, so the sample mean should fall comfortably between 5.9″ and 6.1”. Using Method 2, on the other hand, would result in an approximately 50% chance of a sample mean below 5.9″ (if Line A is selected) or above 6.1″ (if Line B is selected).

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  2. mickeyyogi says:
    May 19, 2015 at 9:41 am

    I agree with your answers. With regard to number 6, I thought perhaps one could estimate the daily standard deviation for method 1 through the histogram with the values given at bottom of the graph. Given the daily standard deviation off all (.11) and thus resulting variance, then one could solve for the standard deviation of method 2 by working with the sum of variances (given my approximation from the histogram) and solving the resulting equation. Repeating the process, and assuming the standard deviation for both the 5.9 day and 6.1 day would be the same, then one could come up with the standard deviation for each. It gave mathematical verification to my answers which agreed with your analysis. Simple idea from an algebra teacher who somehow got stuck teaching AP Stat.

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  3. Jack says:
    May 19, 2015 at 5:01 pm

    Small detail on number 2 part c on draft 2- it reads that it provides convincing evidence p doesn’t equal 0.02 when it should be it provides convincing evidence p doesn’t equal 0.2.

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