The tensile strength of paper can be modelled by a normal distribution with a mean of 240kPa and standard deviation
15kPa.

If the paper fails to meet a minimum strength it is discarded as waste. What is the maximum value that can be specified as the minimum strength so that it wastes only 10% of production?

Question

The tensile strength of paper can be modelled by a normal distribution with a mean of 240kPa and standard deviation
15kPa.

If the paper fails to meet a minimum strength it is discarded as waste. What is the maximum value that can be specified as the minimum strength so that it wastes only 10% of production?

BlackTom AI · Accepted Answer

We want the 10th percentile x of a normal distribution N(μ = 240 kPa, σ = 15 kPa), i.e. the cutoff satisfying P(X < x) = 0.10.

1. Relevant knowledge points and concepts
- For a normal random variable X ~ N(μ, σ^2), percentiles are found by standardizing: z = (x − μ)/σ, where z follows the standard normal distribution N(0,1).
- The 10th percentile corresponds to the z-value z0.10 = Φ^{-1}(0.10) ≈ −1.28155 (Φ is the standard normal CDF).

2. Complete reasoning process (step by step)
- Set P(X < x) = 0.10, so Φ((x − μ)/σ) = 0.10.
- Solve for x: (x − μ)/σ = z0.10, so x = μ + z0.10 · σ.

3. Calculations with all intermediate steps
- μ = 240 kPa, σ = 15 kPa.
- z0.10 = Φ^{-1}(0.10) ≈ −1.28155.
- Multiply: z0.10 · σ = (−1.28155) × 15 = −19.22325 kPa.
- Add to mean: x = 240 + (−19.22325) = 220.77675 kPa.
- Round to two decimals: x ≈ 220.78 kPa.

4. Why the correct answer is correct
This x is the value below which 10% of the normally distributed tensile strengths fall. Using the standard normal inverse for 0.10 gives the exact z-score; scaling and shifting by σ and μ yields the corresponding tensile-strength cutoff.

5. Why other choices would be incorrect
Any other numeric choice would not produce P(X < x) = 0.10 when standardized: for example a larger cutoff (closer to the mean) would waste more than 10%; a smaller cutoff would waste less. Only the value computed from z = −1.28155 scaled by σ gives the required 10% tail.

6. Conclusion and confirmation
The maximum minimum strength that wastes only 10% of production is 220.78 kPa, which matches the provided answer.

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