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

What proportion of paper produced will fail specifications, if the paper must have strength of at least 225kPa?

Question

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

What proportion of paper produced will fail specifications, if the paper must have strength of at least 225kPa?

BlackTom AI · Accepted Answer

Since tensile strength is normally distributed with mean μ = 240 kPa and standard deviation σ = 15 kPa, we use the normal distribution and standardization (Z-score) to find the probability that strength is below 225 kPa.

1. Relevant knowledge points and concepts
- For a normally distributed variable X ~ N(μ, σ²), probabilities are found by converting X to the standard normal Z = (X − μ)/σ.
- Standard normal tables or a calculator give Φ(z) = P(Z ≤ z).
- The proportion failing specifications is P(X < 225), because failing means strength is less than the minimum acceptable 225 kPa.

2. Step-by-step reasoning and calculations
- Given μ = 240, σ = 15, threshold x = 225.
- Compute Z: z = (x − μ)/σ = (225 − 240)/15 = (−15)/15 = −1.0.
- We want P(X < 225) = P(Z < −1.0).
- Use the standard normal cumulative distribution: Φ(1.0) = 0.8413, and by symmetry P(Z < −1.0) = 1 − Φ(1.0) = 1 − 0.8413 = 0.1587.
  (Alternatively many tables give Φ(−1.0) directly as 0.1587.)

3. Intermediate steps shown
- z calculation: (225 − 240) / 15 = −15 / 15 = −1
- lookup: Φ(1.0) = 0.8413 → Φ(−1.0) = 0.1587

4. Why the correct answer is correct
- The Z-score of −1.0 exactly locates 225 kPa one standard deviation below the mean. For a normal distribution, about 15.87% of the data lie more than one standard deviation below the mean, so 15.87% of papers will be below the 225 kPa specification.

5. Why other answers would be incorrect (brief)
- A value like 0.8413 would correspond to P(X ≥ 225) instead of P(X < 225). Values such as 0.0228 or 0.3085 correspond to z-scores of −2 or −0.5 respectively and do not match the computed z = −1.0, so they are incorrect.

Conclusion: The proportion failing specifications is 0.1587, which matches the provided correct answer.

The tensile strength of paper can be modelled by a normal distribution with a mean of 240kPa and standard deviation 15kPa. What proportion of paper produced will fail specifications, if the paper must have strength of at least 225kPa?单项选择题

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