From long-term observation, it is known that the waiting-time for non-ambulance adult arrivals in a hospital's A&E has a mean of 120 minutes and a standard deviation of 50 minutes. On a given day, there are 200 such arrivals.

What is the probability that the daily mean for arrivals exceeds 125 minutes?

Question

From long-term observation, it is known that the waiting-time for non-ambulance adult arrivals in a hospital's A&E has a mean of 120 minutes and a standard deviation of 50 minutes. On a given day, there are 200 such arrivals.

What is the probability that the daily mean for arrivals exceeds 125 minutes?

BlackTom AI · Accepted Answer

The sampling distribution of the sample mean is approximately normal by the Central Limit Theorem, so we can use the normal distribution to find the probability.

1. Relevant concepts
- Population mean μ = 120 minutes, population standard deviation σ = 50 minutes.
- Sample size n = 200. For the sample mean X̄, the standard error (SE) = σ / √n.
- Use Z = (X̄ - μ) / SE and standard normal tables (or a calculator) to find probabilities.

2. Step-by-step reasoning
- Compute the standard error:
  √n = √200 = 14.1421356237
  SE = σ / √n = 50 / 14.1421356237 = 3.5355339059
- Compute the Z-score for X̄ = 125:
  Z = (125 − 120) / SE = 5 / 3.5355339059 = 1.414213562
- Find P(X̄ > 125) = P(Z > 1.414213562).

3. Intermediate calculation with the normal CDF
- Standard normal CDF values: Φ(1.41) ≈ 0.9207, Φ(1.42) ≈ 0.9222.
- Interpolating for Z = 1.4142 (about 0.42 of the way from 1.41 to 1.42):
  Φ(1.4142) ≈ 0.9207 + 0.42*(0.9222 − 0.9207) ≈ 0.9207 + 0.42*0.0015 ≈ 0.92133
- Tail probability: P(Z > 1.4142) = 1 − Φ(1.4142) ≈ 1 − 0.92133 = 0.07867 ≈ 0.0786.

4. Why this is correct
- With n = 200 (large), the CLT justifies treating the sample mean as normally distributed with mean 120 and SE 3.5355. The Z-score calculation and use of the standard normal CDF give the probability that the sample mean exceeds 125.

5. Why typical incorrect approaches are wrong
- Using σ = 50 directly (not dividing by √n) would give Z = 0.1 and a much larger probability (~0.46), which ignores the reduction in variability of the sample mean.
- Using a t-distribution is unnecessary here because n is large; the normal approximation is appropriate.
- Arithmetic errors in computing √200 or dividing by it will produce wrong z and probability.

6. Conclusion
The probability that the daily mean exceeds 125 minutes is approximately 0.0786, which matches the provided answer 0.0786.

类似问题

What is a sampling distribution?

If you increase the sample size from 25 to 100 while keeping the population standard deviation constant, what happens to the standard deviation of the sampling distribution of the mean?

Which of the following statements is correct about a parameter or a statistic associated with repeated random samples of the same size from the same population?

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