Inverse Normal Distribution Examples

Problem: SAT math scores are approximately normally distributed with a mean of 528 and a standard deviation of 120. What score corresponds to the 75th percentile?

Solution

Step 1: Identify the given information.

• Distribution: Normal with μ = 528, σ = 120
• Probability: p = 0.75 (left cumulative, since the 75th percentile means 75% of scores fall at or below this value)
• Mode: Left tail

Step 2: Find the Z score.

Using the inverse normal function: z = Φ⁻¹(0.75) ≈ 0.67449

Step 3: Convert the Z score to the original scale.

x = μ + z × σ = 528 + 0.67449 × 120 = 528 + 80.94 = 608.94

Answer: A score of approximately 609 corresponds to the 75th percentile. This means about 75% of test-takers score at or below 609.

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Problem: A pharmaceutical company produces tablets with a mean weight of 250 mg and a standard deviation of 5 mg. Quality control requires that no more than 2% of tablets should weigh above the upper specification limit. What is the upper specification limit?

Solution

Step 1: Identify the given information.

• Distribution: Normal with μ = 250, σ = 5
• Right tail probability: p = 0.02 (2% of tablets exceed this value)
• Mode: Right tail

Step 2: Convert to left cumulative probability.

Left cumulative = 1 − 0.02 = 0.98

Step 3: Find the Z score.

z = Φ⁻¹(0.98) ≈ 2.05375

Step 4: Convert to the original scale.

x = 250 + 2.05375 × 5 = 250 + 10.27 = 260.27 mg

Answer: The upper specification limit should be set at approximately 260.27 mg. Only 2% of tablets will weigh more than this amount.

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Problem: Blood cholesterol levels in adults are approximately normally distributed with a mean of 200 mg/dL and a standard deviation of 40 mg/dL. Find the range that contains the middle 95% of cholesterol values. This range is often used as a clinical reference range.

Solution

Step 1: Identify the given information.

• Distribution: Normal with μ = 200, σ = 40
• Middle area probability: p = 0.95
• Mode: Middle area

Step 2: Find the tail areas.

Each tail = (1 − 0.95) / 2 = 0.025

Step 3: Find the Z scores.

• Lower Z: Φ⁻¹(0.025) ≈ −1.95996
• Upper Z: Φ⁻¹(0.975) ≈ 1.95996

Step 4: Convert to the original scale.

• Lower limit: x = 200 + (−1.95996) × 40 = 200 − 78.40 = 121.60 mg/dL
• Upper limit: x = 200 + 1.95996 × 40 = 200 + 78.40 = 278.40 mg/dL

Answer: The middle 95% of cholesterol values fall between approximately 121.60 and 278.40 mg/dL. Values outside this range might warrant further medical investigation.

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Problem: A researcher is performing a two-tailed hypothesis test at the 0.01 significance level. Find the critical Z values that define the rejection regions.

Solution

Step 1: Identify the given information.

• Distribution: Standard normal (Z distribution)
• Total tail area: α = 0.01
• Mode: Two tails

Step 2: Find each tail area.

Each tail = α / 2 = 0.01 / 2 = 0.005

Step 3: Find the cumulative probabilities.

• Lower cutoff: cumulative probability = 0.005
• Upper cutoff: cumulative probability = 1 − 0.005 = 0.995

Step 4: Find the Z scores.

• Lower Z: Φ⁻¹(0.005) ≈ −2.57583
• Upper Z: Φ⁻¹(0.995) ≈ 2.57583

Answer: The critical values are ±2.576. The researcher rejects the null hypothesis if the test statistic is less than −2.576 or greater than 2.576.

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Problem: A statistics student needs to find the Z score such that 80% of the standard normal distribution lies to the left.

Solution

Step 1: Identify the given information.

• Distribution: Standard normal (μ = 0, σ = 1)
• Left cumulative probability: p = 0.80
• Mode: Left tail

Step 2: Apply the inverse normal function directly.

z = Φ⁻¹(0.80) ≈ 0.84162

Since this is already the standard normal distribution, no further conversion is needed.

Answer: The Z score is approximately 0.8416. This means 80% of the standard normal distribution lies to the left of z = 0.8416.

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