Well, here’s my first draft of possible solutions.
You can access the questions here at AP Central.
Disclaimer: I construct these as a service for both students and teachers to start discussions. There is nothing “official” about these solutions. I certainly can’t even guarantee that they are correct. They probably have typos and errors. If you catch some, let me know! But if they generate discussion and help others, then I’ve succeeded.
The link to my solutions is here: Possible Solutions 2015 AP FRQ
Thoughts about the questions:
#1. Part a was straightforward. Part b will require students to construct a pretty sophisticated criterion for preferring either company. It will be interesting to see how “convincing” students’ arguments need to be.
#2. A great, simple question that will require precise communication of how confidence intervals work. I like how students must explain why a lack of evidence for claim does not imply evidence that its negation is true.
#3. This should, hopefully, be a slam dunk for kids. This is a good indicator of whether your students are understanding the formulas you use, or simply mimicking things that were done in previous problems.
#4. A straight up inference test for the difference in two population proportions. I anticipate students not being specific enough in stating that volunteers were randomly assigned to treatments.
#5. Again a great litmus test to see if students understand the tools they use. This seems almost too simple for #5.
#6. I think that this was a great, challenging problem. It’s a great problem to use in teaching sampling distributions in the future. It requires students to consider the distribution of a population, the distribution from a sample from that population, and the distribution of the sampling distribution of the sample means. I especially like how the oft-ignored requirement of simple random sampling comes to the surface here. I worry that too many students will overlook the questions posed and write something that is simplistic and irrelevant.
Thank you for posting this!!!! I just worked the problems myself and only find one disagreement. I haven’t been teaching statistics as long so maybe you can help clarify. #3a-c, we are spot on with one another. but 3d …. Since you are finding the expected value knowing that at least one ATM is open, doesn’t that mean you calculate the expected counts for all but 0 ATMS, So (1*.21)+(2*.4)+(3*.24). I thought they would be the same. Is this a rookie mistake?
Thanks for reading!
Notice that in your calculation, you are computing a weighted average of only 85% of the “possible outcomes” in the original. I believe that you would want to divide by 0.85 to scale this to a probability between 0 and 1.
Oh okay. That does make sense. Just needed an explanation. Thanks! Looking forward to following your posts.
5b needs to be fixed
Thanks: That will be resolved in the next draft!
I solved 3(d) using a different method. I created a second probability distribution, one in which at least one of the ATMs working.
Number of ATMs working given that at least one is working: 1 2 3
Probability .247 .471 .282
I then calculated the expected value to be 2.035.
Thanks for sharing, Heidi! I agree that works as well. Did you divide each probability for X = 1,2, or 3 from the original distribution by 0.85, and then re-compute the expected value? If so, then our processes are pretty similar. You divided by 0.85 before taking the expected value, and I divided afterwards. In this particular case, the sequence of those steps doesn’t matter.
itsy bitsy – should 2c be 0.2 not .02?
Also, thank you for providing this. It was nice to compare your approaches. I agree that these questions seemed fair and took a good cross section of the topics discussed this year.
Thanks for the work you have put into this and making this space for discussion! Great for a first year teacher like me. I don’t mean to nit pick, but in 1a, I think the outliers should be described as “5-year employees” instead of “entry level employees” to convey an understanding between the 2009 and 2014 contexts of this problem.
Thanks David! Good point.