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PR-6: Solutions — Normal Distribution

Module 3 · Probability Distributions

How to use this page: Try each problem in the lesson before checking solutions here. If your answer doesn't match, read the solution carefully — especially the part that explains why common wrong answers are wrong. Understanding the error matters more than getting the right answer the first time.

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Section 5: Guided Practice Solutions

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Problem 1 — Standard Normal Probabilities (Generator)

  • Left-tail : read directly → 0.8997.
  • Right-tail : , so (more than half, since ✓).
  • Between : , , so .

The mistake that guarantees zero credit: adding left-tail areas for a “between” problem gives a result above 1. Always subtract the smaller left-tail from the larger.

Problem 2 — Standardize and Find Probability (Generator)

  • Less-than (exam scores, , , ): , .
  • Greater-than (heights, , , ): , .

Key rule: never look up a raw value in the z-table (it’s calibrated for ). Always convert first: .

Problem 3 — Inverse Normal (Generator)

  • Percentile (, 90th percentile): → closest body entry 0.8997 at ; .
  • Probability (, , ): closest entry 0.0505 at ; g.

Inverse vs. forward: forward problems look up a z-score in the margins and read a probability from the body; inverse problems look up a probability in the body and read a z-score from the margins. The direction is reversed.

Problem 4 — Classify the Distribution (All 5 Variants)

  • Variant 0 (age at first driver’s licence): not normal — right-skewed (clustered near 16, long right tail).
  • Variant 1 (blood pressure, healthy adults): normal — appropriate (continuous, symmetric, bell-shaped).
  • Variant 2 (complaints per day): not normal — discrete count data, right-skewed (Poisson-like).
  • Variant 3 (apple weights, one orchard): normal — appropriate.
  • Variant 4 (household incomes): not normal — heavily right-skewed (long upper tail).

Three conditions: (1) continuous (not counts/proportions); (2) roughly symmetric (no strong skew); (3) no heavy tails or dominating outliers.

Section 6: Independent Practice Solutions

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Problem 1 — Standard Normal Probability (Generator)

Between : , , so . (Left-tail: read directly. Right-tail: complement the left-tail.)

Problem 2 — Standardize, One-Tail (Generator)

Greater-than (temps, , , ): , so .

Problem 3 — Between Two Values (Generator)

(part diameters, , , ): , ; .

Four sub-steps: standardize lower → standardize upper → look up both left-tail areas → subtract smaller from larger.

Problem 4 — Inverse Normal (Generator)

Percentile (, , 84th percentile): → closest 0.8389 at ; mm.

Problem 5 — Empirical Rule Application (All 5 Variants)

  • Variant 0 (trout, , ): (a) → ≈95%; (b) → ≈16% above ().
  • Variant 1 (July temps, , ): (a) → ≈95%; (b) → ≈16% above.
  • Variant 2 (birth weights, , ): (a) → ≈95%; (b) → ≈0.15% below.
  • Variant 3 (reaction times, , ): (a) → ≈68%; (b) → ≈2.5% ().
  • Variant 4 (placement scores, , ): (a) top 16% begins at ; (b) → ≈99.7%.

Symmetry shortcut: when the rule gives the percentage inside a symmetric interval, split the remainder equally between the two tails (e.g. 5% outside → 2.5% per tail).

Problem 6 — Find the Error (Generator)

A student finds for () and writes , .

Error: the numerator should be , not . A below-mean value () must give a negative z-score. Correct: , — only 6.7% fall below 65 (expected, since 65 is below the mean).

Sign check: below the mean → negative z; above → positive z. A quick check catches the reversed-subtraction error.

Problem 7 — Multi-Step Synthesis: Bottling Plant

Bottles fill with ( mL); reject if or .

(a) , . Two-tail: (≈5.74%).

(b) Re-center to : , ; . Yes — rejection drops from 5.74% to 4.56%; centering makes the ±12 mL bounds symmetric (), giving equal, smaller tails.

(c) 1st percentile fill: → closest 0.0099 at ; mL.

Section 7: Mastery Check Solutions

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Problem 1 — Feynman Test: Why

The standard normal is perfectly symmetric about zero — the left half is a mirror image of the right. The point is 1.5 units left of center and is 1.5 units right, so the left tail below has exactly the same area as the right tail above . Formally, for any .

Check: and ✓.

Problem 2 — Apply: Commute Times

(); coverage for 27–51 minutes.

Part A: , ; (the interval is wider on the right — above vs. below).

Part B — 95th percentile: ; min.

Problem 3 — Error Analysis: Forgetting to Standardize

Error: the student looked up the raw value 112 directly in the z-table, which is defined only for . The raw value must be standardized first.

Correct for , , : , , so (~21%, since is under 1 SD above the mean).

Section 8: Boss Fight Solutions

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Path A — The Engineer: Bolt Manufacturing

( mm); accept within [9.7, 10.3] mm; 5,000 bolts/day.

Task 1 — : , ; (≈86.64%).

Task 2 — Rejections: rate (≈13.36%); expected daily bolts.

Task 3 — New upper threshold (top 5%): → ; mm.

Task 4 — Upgrade to : , ; . Acceptance rises 86.64% → 99.74%; rejections fall from 668 to ~13/day (saving 655 bolts). If each rejected bolt costs more than $50,000 ÷ 655 ≈ $76 in waste, the upgrade pays off within a day — usually justified.

Path B — The Analyst: Grade Cutoffs

(); top 15% earn an A.

Task 1 — A cutoff: → ; .

Task 2 — Percentile of 85: , → ~90th percentile (top 10%).

Task 3 — Max for cutoff ≥ 80: . If scores are more spread out () the top-15% threshold rises above 80 — counterintuitively easier to earn an A on a relative curve.

Task 4 — “1σ above mean” vs. “top 15%”: — ~15.87% earn an A, slightly more than the strict top-15% rule (, the 85th percentile). Setting the cutoff at is the 84.13th percentile.

Section 9: Challenge Problem Solutions

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Challenge 1 — Symmetry Proof and Numerical Verification

Proof that for : the density satisfies , so by symmetry . Applying the complement (continuous, so ): . Combining: .

Check (): , so the prediction is , and directly ✓. Thus each cut off 2.5% — the basis of the 95% CI (INF-2): .

Challenge 2 — Two-Equation Inverse Normal System

Given and with , find and .

Step 1 — z-scores: → ; → (equal magnitudes confirm 60 and 85 are equidistant from ).

Step 2 — System:

Step 3 — Solve: adding (1) and (2): . Substituting into (1): . So .

Check: is halfway between 60 and 85: ✓; and ✓.
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