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

What is the probability that the tensile strength will be between 226kPa and 243kPa?

Question

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

What is the probability that the tensile strength will be between 226kPa and 243kPa?

BlackTom AI · Accepted Answer

Standardize the tensile strength using Z = (X − μ)/σ where μ = 240 kPa and σ = 15 kPa.

1. Relevant concepts
- For a normal distribution, probabilities for X are found by converting to the standard normal Z and using the standard normal cumulative distribution function (Φ).
- P(a < X < b) = Φ((b−μ)/σ) − Φ((a−μ)/σ).

2. Step-by-step standardization
- Lower bound: z1 = (226 − 240)/15 = −14/15 = −0.933333...
- Upper bound: z2 = (243 − 240)/15 = 3/15 = 0.2

3. Use standard normal CDF values (from a table or calculator)
- Φ(0.2) ≈ 0.5793
- Φ(−0.933333) = 1 − Φ(0.933333) ≈ 1 − 0.824 (more precisely ≈ 0.1754)

4. Compute the probability
- P(226 < X < 243) = Φ(0.2) − Φ(−0.933333)
- ≈ 0.5793 − 0.1754 = 0.4039

5. Why this is correct
- Converting to Z accounts for the given mean and spread so the standard normal table or calculator applies correctly. The subtraction of Φ values gives the probability between the two z-scores.

6. Why other answers would be wrong (common errors)
- Forgetting to standardize (using raw values in the standard normal table).
- Using σ = 15 incorrectly (e.g., dividing by 15 twice or using 15 as variance).
- Using the wrong tail (taking absolute values or computing one-sided probabilities instead of the difference).

Conclusion: The probability that tensile strength is between 226 kPa and 243 kPa is 0.4039, matching the provided answer.

The tensile strength of paper can be modelled by a normal distribution with a mean of 240kPa and standard deviation 15kPa. What is the probability that the tensile strength will be between 226kPa and 243kPa?单项选择题

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