L-PR-6: Solutions — Normal Distribution

A tag such as (C2 + C3) after a title names the Core Concepts parts the problem draws on, so you know where to look back.

Example 1 — Fully Worked: Three Probability Types for Z

(a) Left tail

(b) Right tail

The displayed table lists positive z-scores, so begin with:

Use symmetry for the negative score:

Because is below the mean, a right-tail probability above 0.50 is sensible.

(c) Between two values

Example 2 — Standardize and Find Probability

For , . We want .

About 11.5% of scores in this model are above 118.

Example 3 — Inverse Normal: 90th Percentile

For , . The 90th percentile satisfies:

The closest course-table entry is 0.8997 at .

The table-based 90th-percentile score is approximately 80.24.

Example 4 — Empirical Rule and Table Precision

The bounds are two standard deviations from the mean:

The Empirical Rule therefore gives an estimate of about 95%.

Using the displayed four-decimal table:

The displayed table gives about 95.44%. A higher-precision calculation rounds to approximately 95.45%; the difference is caused by the precision of the table entries.

A tag such as (C2 + C3) after a title names the Core Concepts parts the problem draws on, so you know where to look back.

Problem 1 — Standard Normal Probabilities (C2)

Representative generated draw: find .

The course table gives the left-tail area:

The requested area is the complementary right tail:

For any generated draw, first identify what the left-tail table stores, then decide whether to read, complement, or subtract.

Problem 2 — Standardize and Find Probability (C3)

Representative generated draw: ; find .

The raw value is above the mean, so the positive z-score and a right-tail result below 0.50 are consistent.

Problem 3 — Inverse Normal (C4)

Representative generated draw: and .

The table body maps 0.8997 to .

For any draw: target cumulative area → → raw-scale value.

Problem 4 — Classify the Distribution (C5)

The generated scenario supplies direct evidence about the observed distribution. Use this decision sequence:

1. Check whether the values are continuous or sufficiently fine-grained.
2. Examine symmetry, number of modes, boundary pile-up, and tail behaviour.
3. Decide whether a normal curve would preserve the shape relevant to the question.
4. Justify the verdict with the stated evidence, not with the variable’s name or sample size.

Representative judgment: a continuous histogram that is symmetric, unimodal, tapered on both sides, and free of extreme outliers is reasonably modeled as approximately normal. A long one-sided tail, two separated peaks, a binding boundary, or much heavier tails is evidence against that model.

A tag such as (C2 + C3) after a title names the Core Concepts parts the problem draws on, so you know where to look back.

Problem 1 — Standard Normal Probability (C2)

Representative generated draw: find .

The generator may instead ask for a left or right tail. Read a left tail directly, complement for a right tail, and subtract for a between-values area.

Problem 2 — Standardize a Raw Value (C3)

Representative generated draw: ; find .

Problem 3 — Recover a Percentile Value (C4)

Representative generated draw: and .

If the generated target probability is below 0.50, is negative and the requested value must lie below the mean.

Problem 4 — Judge the Model (C5)

Apply the same evidence-first decision rule as Guided Practice Problem 4. A large sample does not make the original values normal. A continuous variable is not automatically normal. The verdict must follow from the described shape, its boundaries and tails, and the purpose for which the model will be used.

Problem 5 — Estimate a Central Band (C1)

Representative generated draw: , , and the interval is 60 to 100.

The interval is , so the Empirical Rule gives:

The live generator may choose one, two, or three standard deviations; match the band to approximately 68%, 95%, or 99.7%.

Problem 6 — Compare Two Quality-Control Settings (C2 + C3)

Representative generated draw: limits are 485 to 515 mL, mL, and the original mean is 505 mL. Compare that process with one re-centred at 500 mL.

For the original process:

After re-centring at 500 mL:

Re-centring lowers rejection because the fixed limits become equally distant from the mean while the spread stays unchanged.

Error-Analysis Review — Locate the First Broken Step

The generator rotates three error families:

• Right-tail error: the student reports instead of .
• Between-values error: the student adds two left-tail areas instead of subtracting.
• No-standardization error: the student looks up a raw value in a table for .

Representative generated draw: a student reports .

The table lookup is correct, but reporting it as the right tail is the first broken step.

Spaced Review — Z-Scores and the Empirical Rule

Representative generated draw: , , and .

The 68% Empirical-Rule interval begins at:

The z-score for 110 is:

The right-tail probability is:

Spaced Review — Binomial Parameters

Representative generated draw: and .

A binomial model requires a fixed number of independent trials, two outcomes per trial, and a constant success probability.

A tag such as (C2 + C3) after a title names the Core Concepts parts the problem draws on, so you know where to look back.

Question 1 — Feynman Test

The standard normal curve is a mirror image across zero. The point is the reflection of , so the area farther into the left tail beyond matches the reflected area farther into the right tail beyond :

This explanation uses symmetry alone; no table values are required.

Question 2 — Novel Application

Representative generated draw: a bottling process follows and rejects fills below 480 mL or above 530 mL.

Lower tail:

Upper tail:

The two rejection events are disjoint, so add them:

Question 3 — Diagnose a Flawed Analysis

This generator rotates the same three error families used in the independent error-analysis review: reporting a left-tail entry as a right tail, adding two left-tail areas, and looking up a raw value without standardizing.

Representative generated draw: and a student tries to find by looking up 115 in the z-table.

The first broken step is using a raw value in a table defined for . Standardize first:

For a right-tail draw, the first invalid step is failing to subtract the table entry from 1. For a between-values draw, the first invalid step is adding the two left-tail areas instead of subtracting the lower one from the upper one.

A tag such as (C2 + C3) after a title names the Core Concepts parts the problem draws on, so you know where to look back.

Path A — The Engineer

The process follows , so mm. Original acceptance limits are 9.7 to 10.3 mm and production is 5,000 bolts per day.

Task 1 — Acceptance probability

About 86.64% of bolts are accepted.

Task 2 — Rejections

Task 3 — New upper threshold

Rejecting only the top 5% on the upper end means 95% lie below the new threshold. Interpolating between nearby table entries gives .

The new upper threshold is approximately 10.33 mm.

Task 4 — Equipment upgrade

With mm:

The table-based acceptance rate is about 99.74%, so the rejection rate is about 0.26%.

Under the stated assumptions, cumulative savings pass the upgrade cost during the seventh production day. Maintenance, downtime, production volume, and the saving per avoided rejection would change this estimate.

Model reflection: Reducing packs more of the normal curve close to the unchanged mean. The fixed limits therefore sit more standard deviations from the mean, so less modeled area remains outside them and the rejection rate falls.

Path B — The Analyst

Scores follow , so . The top 15% earn an A.

Task 1 — A cutoff

The top 15% leaves 85% below the cutoff:

The closest course-table entry is 0.8508 at .

The table-based cutoff is approximately 82.4, or 83 if whole-number scores must meet or exceed it.

Task 2 — Percentile of 85

A score of 85 is at approximately the 90.32nd percentile: about 90.32% of modeled scores are lower.

Task 3 — Theoretical lower bound on spread

Using the course-table approximation :

The theoretical lower bound is , reported as approximately 7.69 when rounded to two decimal places. Using the exact 85th-percentile z-value gives a boundary of approximately 7.72. There is no maximum: increasing moves this upper-percentile cutoff farther above the unchanged mean.

Task 4 — One-standard-deviation policy

About 15.87% of students qualify, which is 0.87 percentage points more than the top-15% policy.

Model reflection: The cutoff is , so increasing moves this upper-percentile score farther above the unchanged mean. The fraction qualifying stays at 15% because the policy fixes a percentile; only the raw score attached to that percentile changes.

A tag such as (C2 + C3) after a title names the Core Concepts parts the problem draws on, so you know where to look back.

Challenge 1 — Mirror Tails

For the standard normal density, symmetry gives:

Therefore, the area left of equals the area right of :

Because the normal distribution is continuous:

Combining the two statements:

For :

The course table also gives , confirming the identity.

Challenge 2 — Recovering a Distribution

Representative generated draw: equal tail probabilities satisfy

Equal tails place the mean at the midpoint:

The tail probability 0.1003 corresponds to on the lower side and on the upper side.

Using the upper cutoff:

Therefore, this representative draw is:

The live challenge may generate a different centre, spread, or tail z-score. The invariant method is midpoint → tail z-score → divide the half-width by .