AP Stats Practice FRQ Responses & Feedback (Unit 4)

The Practice FRQ Prompt

• i. What is the probability that the person lives in County A and prefers Party Z

• ii. What is the probability that the person lives in County A or prefers Party Z?
• iii. What is the probability that the person lives in County A, given that they prefer Party Z?

FRQ Writing Samples and Teacher Feedback

Student Response 1

Well done! All three parts show strong command of statistical notation, and clear communication. Your answers are all correct and would all earn credit.

Student Response 2

In all three parts, your answers are correct and presented with accompanying work and appropriate notation. You would get full credit for each part! Part ( c ) provides a good contrast to your work on Unit 4 FRQ #2 - by saying “the probability for those who prefer Party Y is the same at 0.510,” you provide an in-context reason for the scenario being binomial. Well done!

Student Response 3

• C. The probability that exactly 2 of them prefer Party Y is .306.

1. Each trial has 2 possible outcomes (Party Y or not party Y)
2. There are fixed number of trials (n=5)
3. Outcomes are independent
4. Probability of success is the same for each trial (p=.510)

All calculations and explanations are correct, with appropriate work shown. However, in part (b), you make a small notation error. P(County B | Party X) = P(County B) = 0.6. The probabilities themselves should not be inside the P( ) notation. Think of it like input/output function notation - inside the P( ) is what probability you’re finding (in words or symbols), while outside on the other side of the equals sign is the numerical answer.

Student Response 4

Small thing: for part a-ii, you provide the correct answer and use correct notation. However, you don’t show the numbers that go into the calculation, so on some rubrics that would be considered a “naked” answer in that you didn’t show the numbers along the way. Taking the extra step to show calculations with probability scenarios is always helpful. Everything else should receive full credit.

Student Response 5

Nice job on your calculations (and super shout-out for going by hand - or at least writing it all out by hand - on the binomial calculations). Be really careful with your notation on part (b). Probability notation is kind of like function notation f(x) from algebra. The “stuff” inside the parentheses is like an input - and should represent what we’re finding the probability of. This will include variables and equality/inequality statements in some cases, but not just “plain numbers”, since those are the “output” in this situation. So P(.108) isn’t a thing - it should be P(B and X) = .108. That is, you already wrote the notations correctly - just put the actual numbers on your second line for part (b), and you’ll be good to go to say yes for independence.

Student Response 6

On part (a) and part ( c ), you’ve done all calculations correctly. For part (b), you’ve defined what we need to find to show independence… but your calculation P(B | X) is incorrect. You should get 0.108 / 0.18 = 0.6, so it would be independent. Double check what you did. Given what probability you got, you made an appropriate conclusion, so you’d still be eligible for partial credit.

Student Response 7

Good on all three parts!

Student Response 8

In part (b), you should do a little more to show where the P(B)*P(X) = 0.108 comes from. P(B and X) comes directly off the table so there’s no additional work to show, but P(B)*P(X) required a calculation, so you should show the numbers you multiplied. All other parts have appropriate work and correct answers.

Student Response 9

All answers are correct. However your work is not shown in part a-ii or a-iii, and should be, since it involved calculations of some kind. Anytime you need to add/subtract/multiply/divide to get an answer, you should show that work. Parts (b) and ( c ) show appropriate work.

Student Response 10

Solid work - in part (a), you give all probabilities correctly with supporting work. In part (b), you give the appropriate definition of independence, but unfortunately do not execute correctly - you gave P(b and x) and not P(b | x), which would be .108/.180 = 0.6… so the actual answer would be “yes”. You would get partial credit for defining independence, but making a mistake with execution. Part ( c) is all good.

Student Response 11

Perfect! Full credit all around!

Student Response 12

1. 2 Outcomes: Could prefer party y or could not (Good)
2. Independent Trials: Outcomes are independent of each other (Good)
3. Constant prob: The prob is constant ( 0.510) (Good)
4. Fixed Number of Trials (Good)

Nice work! All parts have correct answers with work shown. If you do get asked a probability question that would require a binomial distribution (like part ( c )), you do not need to do a “conditions check”, as long as you clearly label the parameters for a binomial distribution. So something like “this scenario is binomial, with p = 0.510, n = 5 trials, and we want x = 2 successes.” [then show calculation] would get full credit. You only need to check the binomial conditions when explicitly asked if a binomial distribution is appropriate.

Student Response 13

• a) i. P(A and Z) = 0.152

• ii. P(A or Z) = P(A) + P(Z) - P(A and Z)
• 0.71-0.152=0.558

• iii. P(A|Z) = P(A and Z)/P(A) = 0.152/0.4 = 0.38

• b) Is P(B) equal to P(B|Z)?
• P(B) = 0.6
• P(X|B) = P(X and B)/P(X) = 0.108/0.18 = 0.6
• Since the probability of preferring Party X given that they live in County B is equal to the probability of living County B, the events are independent.
• c) This is a binomial distribution. There is success of failure because they either prefer Party Y or do not, there is a fixed number of trials (n = 5), each voter outcome is independent of one another, and there is a constant probability (p = 0.51) that the voter prefers Party Y.
• binompdf (trials = 5, p = 0.51, successes = 2) = 0.306 probability that exactly 2 of the voters prefer Party Y.

In part (a), you’ve correctly calculated the and / or probabilities, though you appear to have your conditional probability (the ‘given that’) probability backwards. The general formula is P(A | B) = P(A and B) / P(B); that is, the “given that” goes in the denominator. You put the wrong thing in the denominator, and so part (a) as a whole would be marked partially correct.

In part (b), check your notation - you mention P(B | Z) but the question asked about County X (you then go on to do the calculation involving X). You actually end up getting the right answer (“yes”), but it happens to be because you reversed the probability you should have been finding, P(B | X), into P(X | B), and then made the same reversal as in part a-iii, so you actually did end up calculating P(B | X) and reaching an appropriate conclusion. I’m not sure how something like this would be scored, but it would either get partial or no credit.

In part ( c ), you appropriately label the scenario as binomial, and execute the calculation correctly.

Student Response 14

All work in parts (a) and ( c ) are correct. In part (b), you’ve got the right idea, but calculate the wrong probabilities - we should do P(b) and P(b|x) -or- P(x) and P(x|b). That is, one of the probabilities should be “regular” and the other should be “conditional” on the other variable. If THOSE are equal, then we’re independent.