Bring ideas together without being told which method to use. Show the reasoning that makes your result valid.
Suggested session length: 70 minutes, as an unvalidated planning estimate. This collection contains 2 tasks; choose a manageable set and return for further sessions. Completing the whole collection may take substantially longer. This is a unit assessment, not a full course or AP examination.
State the target population and sampling assumptions before substituting a sample count.
Keep inclusive and strict error events distinct. A worst-case upper bound is not the actual probability.
Attempt the changed designs before revealing the public solutions; written arguments require review.
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0 of 2 tasks have recorded responses. No overall mastery score is assigned.
A binary measurement process samples 400 independent units, each with success probability p, then stores five identical copies of every result. An analyst treats the 2,000 rows as independent. Prove that the average of the stored rows is the average of the 400 sampled units. Derive the correct worst-case variance and a Chebyshev upper bound for error at least 1/20. Find the least number of independent units certified by this bound to make that failure probability at most 0.1. Compare the number of stored rows if five copies are retained with the number needed when each stored row is a fresh independent unit. Explain why this probability guarantee does not repair selection bias relative to a different target population.
Structure your argument and preserve revisions
These optional planning fields save locally. Recording your response saves an immutable copy of the argument and these fields. Older records may contain only first/latest responses; missing intermediate revisions are not reconstructed.
Each field allows 2,000 characters. Keep a specific claim, state where it applies, cite the facts you use, and explain what still needs checking. Blank fields are allowed and are not evidence of mastery.
Compare designs A and B for seven fair binary observations. A uses seven independent bits. B uses the seven nonzero binary linear combinations of three independent fair seed bits. In B, the number S of ones is 0 with probability 1/8 and 4 with probability 7/8; prove this using the seed mechanism. Both averages have mean 1/2 and variance 1/28. Find the exact inclusive-tail probabilities P(|S/7−1/2|≥ε) for both designs at ε=3/14 and ε=1/2. Derive the common Chebyshev bounds. Does equal variance imply equal tail probabilities? Does either of these two designs have smaller inclusive tails at both tolerances? Explain why the variance theorem remains useful despite these differences.
Structure your argument and preserve revisions
These optional planning fields save locally. Recording your response saves an immutable copy of the argument and these fields. Older records may contain only first/latest responses; missing intermediate revisions are not reconstructed.
Each field allows 2,000 characters. Keep a specific claim, state where it applies, cite the facts you use, and explain what still needs checking. Blank fields are allowed and are not evidence of mastery.
Review and revisit
Use each rubric to identify a valid argument, a missing condition or a step to revise. For independent evaluation, ask a teacher or reviewer to examine the complete reasoning. Self-review and recorded responses are not verified assessment results.
Coverage and targeted lesson returns
A binary measurement process samples 400 independent units, each with success probability p, then stores five identical copies of every result. An analyst treats the 2,000 rows as independent. Prove that the average of the stored rows is the average of the 400 sampled units. Derive the correct worst-case variance and a Chebyshev upper bound for error at least 1/20. Find the least number of independent units certified by this bound to make that failure probability at most 0.1. Compare the number of stored rows if five copies are retained with the number needed when each stored row is a fresh independent unit. Explain why this probability guarantee does not repair selection bias relative to a different target population.
Compare designs A and B for seven fair binary observations. A uses seven independent bits. B uses the seven nonzero binary linear combinations of three independent fair seed bits. In B, the number S of ones is 0 with probability 1/8 and 4 with probability 7/8; prove this using the seed mechanism. Both averages have mean 1/2 and variance 1/28. Find the exact inclusive-tail probabilities P(|S/7−1/2|≥ε) for both designs at ε=3/14 and ε=1/2. Derive the common Chebyshev bounds. Does equal variance imply equal tail probabilities? Does either of these two designs have smaller inclusive tails at both tolerances? Explain why the variance theorem remains useful despite these differences.
