SOLUTION: Traveling between two campuses of a university in a city via shuttle bus takes, on average, 28 minutes with a standard deviation of 5 minutes. In a given week, a bus transported pa

Question 1178032: Traveling between two campuses of a university in a city via shuttle bus takes, on average, 28 minutes with a standard deviation of 5 minutes. In a given week, a bus transported passengers 40 times. What is the probability that the average transport time was more than 30 minutes? Assume the mean time is measured to the nearest minute.
Answer by CPhill(1959) (Show Source): You can put this solution on YOUR website!
Let's solve this problem step-by-step.
**Understanding the Problem**
* **Population Mean (μ):** 28 minutes
* **Population Standard Deviation (σ):** 5 minutes
* **Sample Size (n):** 40
* **Goal:** Find the probability that the sample mean (X̄) is more than 30 minutes, i.e., P(X̄ > 30).
**Applying the Central Limit Theorem**
Since the sample size (n = 40) is large (n > 30), we can apply the Central Limit Theorem. The sample mean (X̄) will be approximately normally distributed with:
* **Mean of X̄ (μ_X̄):** μ_X̄ = μ = 28 minutes
* **Standard Deviation of X̄ (σ_X̄):** σ_X̄ = σ / √n = 5 / √40 ≈ 5 / 6.3246 ≈ 0.7906 minutes
**Calculating the Z-score**
We need to find the z-score for X̄ = 30 minutes.
* Z = (X̄ - μ_X̄) / σ_X̄
* Z = (30 - 28) / 0.7906
* Z = 2 / 0.7906 ≈ 2.53
**Finding the Probability**
We want to find P(X̄ > 30), which is equivalent to P(Z > 2.53).
Using a standard normal distribution table or calculator:
* P(Z ≤ 2.53) ≈ 0.9943
Therefore:
* P(Z > 2.53) = 1 - P(Z ≤ 2.53) = 1 - 0.9943 = 0.0057
**Continuity Correction**
Since the mean time is measured to the nearest minute, we need to apply a continuity correction. We want to find P(X̄ > 30), which we will approximate as P(X̄ > 29.5).
* Z = (29.5 - 28) / 0.7906 = 1.5 / 0.7906 ≈ 1.8973
Using a standard normal distribution table or calculator:
* P(Z ≤ 1.8973) ≈ 0.9711
Therefore:
* P(Z > 1.8973) = 1 - P(Z ≤ 1.8973) = 1 - 0.9711 = 0.0289
**Therefore, the probability that the average transport time was more than 30 minutes is approximately 0.0057 without continuity correction, and approximately 0.0289 with continuity correction.**
Given the problem statement, it is likely that the continuity correction is desired, so the answer should be 0.0289.

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