PR-6 Solutions: Normal Distribution

Problem 1 — Standard Normal Probability (Generator IP1)

The generator randomly selects one of three problem types. Below is a representative worked example of the between two values type — the most involved case.

Example: Find \(P(-0.52 < Z < 1.96)\).

Step 1 — Left-tail area at upper bound: row 1.9, column 0.06 → \(P(Z < 1.96) = 0.9750\).

Step 2 — Left-tail area at lower bound: row −0.5, column 0.02 → \(P(Z < -0.52) = 0.3015\).

Step 3 — Subtract: \[ P(-0.52 < Z < 1.96) = 0.9750 - 0.3015 = \mathbf{0.6735} \]

For a left-tail problem (e.g., \(P(Z < 1.28)\)): read directly from the table → 0.8997.

For a right-tail problem (e.g., \(P(Z > -0.84)\)): look up left-tail (0.2005), then complement → \(1 - 0.2005 = 0.7995\).

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Problem 2 — Standardize and Find One-Tail Probability (Generator IP2)

Below is a representative worked example for the greater than direction.

Example: Daily temperatures (°F) follow \(\mu = 70\), \(\sigma = 8\). Find \(P(X > 82)\).

Step 1 — Standardize: \[ z = \frac{82 - 70}{8} = \frac{12}{8} = 1.50 \]

Step 2 — Left-tail lookup: \(P(Z < 1.50) = 0.9332\).

Step 3 — Complement: \(P(X > 82) = 1 - 0.9332 = \mathbf{0.0668}\). About 6.7% of days exceed 82°F.

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Problem 3 — Between Two Values (Generator IP3)

The generator produces a between-two-bounds problem requiring standardization at both ends. Below is a representative worked example.

Example: Machine part diameters (mm) follow \(\mu = 100\), \(\sigma = 15\). Find \(P(77.5 < X < 118.75)\).

Step 1 — Standardize lower bound: \[ z_1 = \frac{77.5 - 100}{15} = \frac{-22.5}{15} = -1.50 \]

Step 2 — Standardize upper bound: \[ z_2 = \frac{118.75 - 100}{15} = \frac{18.75}{15} = 1.25 \]

Step 3 — Look up both left-tail areas: \(P(Z < -1.50) = 0.0668\) and \(P(Z < 1.25) = 0.8944\).

Step 4 — Subtract: \[ P(77.5 < X < 118.75) = P(-1.50 < Z < 1.25) = 0.8944 - 0.0668 = \mathbf{0.8276} \]

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Problem 4 — Inverse Normal (Generator IP4)

Below is a representative worked example for the percentile phrasing.

Example: Monthly rainfall (mm) follows \(X \sim N(500, 400)\) (\(\sigma = 20\)). Find the 84th percentile.

Step 1 — The 84th percentile means \(P(X < x) = 0.8400\). Search the z-table body for 0.8400. The closest entry is 0.8389 at \(z = 1.00\). So \(z^* = 1.00\).

Step 2 — Unstandardize: \[ x = \mu + z^* \cdot \sigma = 500 + (1.00)(20) = \mathbf{520} \text{ mm} \]

The 84th percentile of monthly rainfall is 520 mm — about 84% of months receive less than 520 mm.

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Problem 5 — Empirical Rule Application (All 5 Variants)

Variant 0 — Trout lengths, \(X \sim N(42, 25)\) (\(\sigma = 5\) cm):

• (a) \(32 = 42 - 2(5) = \mu - 2\sigma\) and \(52 = 42 + 2(5) = \mu + 2\sigma\). By the Empirical Rule: approximately 95% of trout are between 32 cm and 52 cm (precisely: 95.45%).
• (b) \(47 = 42 + 1(5) = \mu + \sigma\). The Empirical Rule says 68% are within 1\(\sigma\); by symmetry, 16% are above \(\mu + \sigma\). Answer: approximately 16% (precisely: \(P(Z > 1.00) = 1 - 0.8413 = 0.1587\)).

Variant 1 — Montreal July temperatures, \(X \sim N(27, 16)\) (\(\sigma = 4\)°C):

• (a) \(19 = 27 - 2(4) = \mu - 2\sigma\) and \(35 = 27 + 2(4) = \mu + 2\sigma\). Answer: approximately 95% of July days.
• (b) \(31 = 27 + 1(4) = \mu + \sigma\). By symmetry: approximately 16% of days exceed 31°C (precisely: \(P(Z > 1.00) = 0.1587\)).

Variant 2 — Birth weights, \(X \sim N(3400, 90000)\) (\(\sigma = 300\) g):

• (a) \(2800 = 3400 - 2(300) = \mu - 2\sigma\) and \(4000 = 3400 + 2(300) = \mu + 2\sigma\). Answer: approximately 95%.
• (b) \(2500 = 3400 - 3(300) = \mu - 3\sigma\). The Empirical Rule: 99.7% within 3\(\sigma\), so 0.3% outside; by symmetry, 0.15% are below 2500 g (precisely: \(P(Z < -3.00) = 0.0013\)).

Variant 3 — Athlete reaction times, \(X \sim N(180, 400)\) (\(\sigma = 20\) ms):

• (a) \(160 = 180 - 1(20) = \mu - \sigma\) and \(200 = 180 + 1(20) = \mu + \sigma\). Answer: approximately 68%.
• (b) \(140 = 180 - 2(20) = \mu - 2\sigma\). By symmetry: approximately 2.5% qualify as elite (precisely: \(P(Z < -2.00) = 0.0228\)).

Variant 4 — Math placement scores, \(X \sim N(500, 10000)\) (\(\sigma = 100\)):

• (a) The top 16% begins at \(\mu + \sigma = 500 + 100 = \mathbf{600}\). By the Empirical Rule, 68% are within 1\(\sigma\), so 32% are outside; by symmetry, 16% are above \(\mu + \sigma\).
• (b) \(200 = 500 - 3(100) = \mu - 3\sigma\) and \(800 = 500 + 3(100) = \mu + 3\sigma\). Answer: approximately 99.7%.

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Problem 6 — Find the Error (Generator)

The generator produces a generateStandardize problem with an error in the standardization or lookup step. The most common error type is a sign mistake in the numerator — computing \((mu - x)\) instead of \((x - mu)\).

Representative error scenario: A student finds \(P(X < 65)\) for \(X \sim N(80, 100)\) (\(\sigma = 10\)) and writes: \(z = (80 - 65)/10 = 1.50\). Therefore \(P(X < 65) = P(Z < 1.50) = 0.9332\).

The error: The numerator should be \(x - \mu\), not \(\mu - x\). The student flipped the subtraction order, turning a below-mean value (\(65 < 80\)) into a positive z-score, which should be negative. A value below the mean must yield a negative z-score.

Correct solution:

Step 1 — Standardize (correctly): \[ z = \frac{65 - 80}{10} = \frac{-15}{10} = -1.50 \]

Step 2 — Left-tail lookup: \(P(Z < -1.50) = 0.0668\).

\(P(X < 65) = \mathbf{0.0668}\) — only 6.7% of values fall below 65, not 93.3%. Since 65 is well below the mean of 80, a small left-tail probability is expected ✓.

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Problem 7 — Multi-Step Synthesis: Quality Control at a Bottling Plant

Bottles are filled with \(X \sim N(500, 36)\) (\(\mu = 500\) mL, \(\sigma = 6\) mL). Rejection if \(X < 490\) mL or \(X > 514\) mL.

Part (a) — Probability that a bottle is rejected:

Standardize both bounds: \[ z_1 = \frac{490 - 500}{6} = \frac{-10}{6} \approx -1.67 \qquad z_2 = \frac{514 - 500}{6} = \frac{14}{6} \approx 2.33 \]

Rejection is a two-tail event: \[ P(\text{rejected}) = P(Z < -1.67) + P(Z > 2.33) \] \[ = 0.0475 + (1 - 0.9901) = 0.0475 + 0.0099 = \mathbf{0.0574} \]

About 5.74% of bottles are rejected.

Part (b) — Re-centering to \(\mu = 502\) mL (same \(\sigma = 6\) mL):

New bounds: \[ z_1 = \frac{490 - 502}{6} = \frac{-12}{6} = -2.00 \qquad z_2 = \frac{514 - 502}{6} = \frac{12}{6} = 2.00 \]

\[ P(\text{rejected}) = P(Z < -2.00) + P(Z > 2.00) = 0.0228 + 0.0228 = \mathbf{0.0456} \]

Yes — the rejection rate drops from 5.74% to 4.56%. By centering at 502 mL, the bounds ±12 mL straddle the mean symmetrically (\(z = \pm 2.00\)), giving equal and smaller tails than the asymmetric setup at \(\mu = 500\) (where \(z_1 = -1.67\) and \(z_2 = +2.33\)).

Part (c) — Find the 1st percentile fill level (with \(\mu = 500\), \(\sigma = 6\)):

Step 1 — \(P(X < x) = 0.01\). Search the z-table body for 0.0100. The closest entry is 0.0099 at \(z = -2.33\). So \(z^* = -2.33\).

Step 2 — Unstandardize: \[ x = 500 + (-2.33)(6) = 500 - 13.98 = \mathbf{486.02} \text{ mL} \]

The 1st percentile fill level is approximately 486.02 mL — only 1% of bottles are filled below this amount.

Question 1 — Feynman Test: Why \(P(Z < -1.5) = P(Z > 1.5)\)

The standard normal distribution is perfectly symmetric about zero — the curve to the left of zero is an exact mirror image of the curve to the right. The point \(z = -1.5\) is exactly 1.5 units to the left of center, and \(z = +1.5\) is exactly 1.5 units to the right of center. Because the two halves of the curve are congruent mirror images, the area in the left tail below \(-1.5\) is identical to the area in the right tail above \(+1.5\). Symmetry means the two shaded regions are the same shape and the same area.

More formally: by symmetry, \(P(Z < -a) = P(Z > a)\) for any \(a > 0\). This is one of the most useful properties of the standard normal — it means you can always convert a left-tail at a negative value into a right-tail at the corresponding positive value.

Numerical check: From the z-table, \(P(Z < -1.5) = 0.0668\) and \(P(Z > 1.5) = 1 - P(Z < 1.5) = 1 - 0.9332 = 0.0668\). ✓

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Question 2 — Apply: Commute Times

Commute times follow \(X \sim N(35, 64)\) (\(\mu = 35\) min, \(\sigma = 8\) min). The policy covers commuters between 27 and 51 minutes.

Part A — \(P(27 < X < 51)\):

Standardize both bounds: \[ z_1 = \frac{27 - 35}{8} = \frac{-8}{8} = -1.00 \qquad z_2 = \frac{51 - 35}{8} = \frac{16}{8} = 2.00 \]

Look up both left-tail areas: \(P(Z < -1.00) = 0.1587\) and \(P(Z < 2.00) = 0.9772\).

Subtract: \[ P(27 < X < 51) = P(-1.00 < Z < 2.00) = 0.9772 - 0.1587 = \mathbf{0.8185} \]

About 81.85% of commuters fall in this range. The interval is asymmetric: the upper bound is 2\(\sigma\) above the mean but the lower bound is only 1\(\sigma\) below, so the interval is wider on the right side.

Part B — 95th percentile commute time:

Find \(z^*\) such that \(P(Z < z^*) = 0.95\). The z-table body: 0.9495 at \(z = 1.64\) and 0.9505 at \(z = 1.65\). Using \(z^* = 1.645\) (midpoint): \[ x = 35 + (1.645)(8) = 35 + 13.16 = \mathbf{48.16} \text{ min} \]

The 95th percentile commute time is approximately 48.16 minutes.

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Question 3 — Error Analysis: Forgetting to Standardize

The error: The student used the raw \(X\) value (112) as though it were a z-score, looking it up directly in the z-table. The z-table is defined only for \(Z \sim N(0, 1)\). A z-score of 112 would be absurdly extreme; the table returns a value near 1.0000 because 112 is far to the right of the standard normal scale — but this is entirely meaningless. The raw value must be converted to a z-score first using \(z = (x - \mu)/\sigma\).

Correct solution for \(P(X > 112)\) where \(X \sim N(100, 225)\), \(\sigma = 15\):

Step 1 — Standardize: \[ z = \frac{112 - 100}{15} = \frac{12}{15} = 0.80 \]

Step 2 — Left-tail lookup: \(P(Z < 0.80) = 0.7881\).

Step 3 — Complement: \(P(X > 112) = 1 - 0.7881 = \mathbf{0.2119}\).

About 21.2% of values exceed 112 — far from zero. Since \(z = 0.80\) is less than one standard deviation above the mean, a substantial fraction of the distribution naturally lies above it.

Path A — The Engineer: Bolt Manufacturing

Bolt diameters follow \(X \sim N(10.0, 0.04)\) — \(\mu = 10.0\) mm, \(\sigma = 0.2\) mm. Rejection interval: outside \([9.7, 10.3]\) mm. Production: 5,000 bolts/day.

Task 1 — Probability that a bolt is accepted:

Standardize both acceptance bounds: \[ z_1 = \frac{9.7 - 10.0}{0.2} = \frac{-0.3}{0.2} = -1.50 \qquad z_2 = \frac{10.3 - 10.0}{0.2} = \frac{0.3}{0.2} = 1.50 \]

By symmetry: \[ P(9.7 < X < 10.3) = P(-1.50 < Z < 1.50) = P(Z < 1.50) - P(Z < -1.50) = 0.9332 - 0.0668 = \mathbf{0.8664} \]

About 86.64% of bolts are accepted.

Task 2 — Rejection rate and expected daily rejections:

Rejection rate: \(1 - 0.8664 = \mathbf{0.1336}\) — about 13.36% of bolts are rejected.

Expected daily rejections: \(5000 \times 0.1336 = \mathbf{668}\) bolts per day.

Task 3 — New upper threshold rejecting only the top 5%:

We want \(P(X > x^*) = 0.05\), i.e., \(P(X < x^*) = 0.95\).

Find \(z^*\): the z-table entry closest to 0.9500 is 0.9505 at \(z = 1.65\). So \(z^* \approx 1.645\).

Unstandardize: \[ x^* = \mu + z^* \cdot \sigma = 10.0 + (1.645)(0.2) = 10.0 + 0.329 = \mathbf{10.329} \text{ mm} \]

The new upper threshold is approximately 10.33 mm. Bolts above this diameter represent the top 5% by size.

Task 4 — Upgrade to \(\sigma = 0.1\) mm: change in acceptance rate and cost-benefit:

With \(\sigma = 0.1\) mm and the same bounds \([9.7, 10.3]\): \[ z_1 = \frac{9.7 - 10.0}{0.1} = -3.00 \qquad z_2 = \frac{10.3 - 10.0}{0.1} = 3.00 \]

\[ P(\text{accepted}) = P(-3.00 < Z < 3.00) = 0.9987 - 0.0013 = \mathbf{0.9974} \]

Acceptance rate improves from 86.64% to 99.74%. Rejection rate drops from 13.36% to 0.26% — a reduction from 668 to just 13 rejected bolts per day.

Cost-benefit: the upgrade saves 655 bolts/day. If each rejected bolt costs more than \(\$50{,}000 \div 655 \approx \$76\) in waste, the investment pays off within a single day of operation. For most manufacturing contexts, this precision upgrade is economically justified.

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Path B — The Analyst: Grade Cutoffs

Scores follow \(X \sim N(72, 100)\) (\(\mu = 72\), \(\sigma = 10\)). Top 15% earn an A.

Task 1 — Minimum score for an A (top 15%):

We want \(P(X > x^*) = 0.15\), i.e., \(P(X < x^*) = 0.85\).

Find \(z^*\): the z-table entry closest to 0.8500 is 0.8508 at \(z = 1.04\). So \(z^* = 1.04\).

Unstandardize: \[ x^* = 72 + (1.04)(10) = 72 + 10.4 = \mathbf{82.4} \]

A score of approximately 82.4 is the minimum to earn an A (or 83 if rounding to a whole number).

Task 2 — Percentile for a score of 85:

Standardize: \(z = (85 - 72)/10 = 13/10 = 1.30\).

Left-tail lookup: \(P(Z < 1.30) = 0.9032\).

A score of 85 is at approximately the 90th percentile — the student scored higher than about 90.3% of the class. Plain language: "A score of 85 puts this student in the top 10%, comfortably within A territory."

Task 3 — Maximum \(\sigma\) so that the A cutoff is at least 80:

We need the 85th percentile (\(z^* = 1.04\) for top 15%) to be at least 80: \[ \mu + z^* \cdot \sigma \geq 80 \] \[ 72 + (1.04)\sigma \geq 80 \] \[ (1.04)\sigma \geq 8 \] \[ \sigma \leq \frac{8}{1.04} \approx \mathbf{7.69} \]

The standard deviation must be at most approximately 7.69 for the A cutoff to be at 80 or higher. If scores are more spread out (\(\sigma > 7.69\)), the top 15% threshold rises above 80, lowering the barrier — which counterintuitively makes it easier (not harder) to earn an A when graded on a relative curve.

Task 4 — "1σ above mean" grading policy vs. "top 15%":

Under the \(X > \mu + \sigma\) policy: \[ P(X > \mu + \sigma) = P(Z > 1.00) = 1 - P(Z < 1.00) = 1 - 0.8413 = \mathbf{0.1587} \]

About 15.87% of students earn an A — slightly more than the strict "top 15%" policy.

Comparison: the "top 15%" policy requires \(z^* \approx 1.04\) (85th percentile cutoff), which is slightly higher than \(z = 1.00\). Setting the cutoff at exactly \(\mu + \sigma\) corresponds to the 84.13th percentile, giving 15.87% of students an A versus exactly 15% under the stricter rule.

Challenge 1 — Symmetry Proof and Numerical Verification

Proof that \(P(Z < -a) = 1 - P(Z < a)\) for any \(a > 0\):

The standard normal density satisfies \(f(-z) = f(z)\) for all \(z\) — the curve is symmetric about zero. By this symmetry, the area to the left of \(-a\) is equal to the area to the right of \(+a\): \[ P(Z < -a) = P(Z > a) \]

Now apply the complement rule. Since total area = 1 and \(P(Z = a) = 0\) for a continuous distribution: \[ P(Z > a) = 1 - P(Z \leq a) = 1 - P(Z < a) \]

Combining both steps: \[ P(Z < -a) = 1 - P(Z < a) \quad \square \]

Numerical verification for \(a = 1.96\):

From the z-table: \(P(Z < 1.96) = 0.9750\).

Prediction: \(1 - P(Z < 1.96) = 1 - 0.9750 = 0.0250\).

Direct z-table lookup: \(P(Z < -1.96) = 0.0250\). ✓

The values match. This confirms the familiar fact that \(\pm 1.96\) each cut off exactly 2.5% in their tails — the foundation of the 95% confidence interval, covered in INF-2. The "between" formula gives \(P(-1.96 < Z < 1.96) = 0.9750 - 0.0250 = 0.9500\) — the standard 95% band.

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Challenge 2 — Two-Equation Inverse Normal System

Given \(P(X > 85) = 0.10\) and \(P(X < 60) = 0.10\) with \(X \sim N(\mu, \sigma^2)\), find \(\mu\) and \(\sigma\).

Step 1 — Translate each probability into a z-score:

\(P(X > 85) = 0.10\) means \(P(X < 85) = 0.90\). From the z-table, the closest entry to 0.9000 is 0.8997 at \(z = 1.28\). So the standardized value of 85 is \(z^* = 1.28\).

\(P(X < 60) = 0.10\). From the z-table, the closest entry is 0.1003 at \(z = -1.28\). So the standardized value of 60 is \(z^* = -1.28\).

By symmetry, the two z-scores are equal in magnitude — a confirmation that the distribution is symmetric with 60 and 85 equidistant from \(\mu\). ✓

Step 2 — Write the system of equations: \[ \frac{85 - \mu}{\sigma} = 1.28 \quad \Rightarrow \quad \mu + 1.28\sigma = 85 \tag{1} \] \[ \frac{60 - \mu}{\sigma} = -1.28 \quad \Rightarrow \quad \mu - 1.28\sigma = 60 \tag{2} \]

Step 3 — Solve:

Add (1) and (2): \[ 2\mu = 145 \quad \Rightarrow \quad \mu = \mathbf{72.5} \]

Substitute into (1): \[ 72.5 + 1.28\sigma = 85 \quad \Rightarrow \quad 1.28\sigma = 12.5 \quad \Rightarrow \quad \sigma = \frac{12.5}{1.28} \approx \mathbf{9.77} \]

Answer: \(\mu = 72.5\), \(\sigma \approx 9.77\). The distribution is \(X \sim N(72.5,\ 95.45)\).

Check — two ways:

• Symmetry: \(\mu\) should be exactly halfway between 60 and 85: \((60 + 85)/2 = 72.5\). ✓
• Verify equation (1): \((85 - 72.5)/9.77 \approx 12.5/9.77 \approx 1.28\). ✓