Assume that the time needed by a worker to perform a maintenance operation is normally distributed, with a mean of 60 minutes and a standard deviation of 6 minutes. What is the probability that the average time needed by a sample of 5 workers to perform the maintenance is between 63 and 68 minutes?

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BlackTom AI · Accepted Answer

We start by recognizing that the maintenance times are normally distributed with a mean μ = 60 minutes and a standard deviation σ = 6 minutes. When taking a sample of n = 5 workers, the sampling distribution of the sample mean X̄ is also normal, with mean μ_X̄ = μ = 60 and standard deviation (standard error) σ_X̄ = σ / √n = 6 / √5.

Compute the standard error: √5 ≈ 2.236, so σ_X̄ = 6 / 2.236 ≈ 2.683.

We want P(63 ≤ X̄ ≤ 68). Convert to a standard normal Z using Z = (X̄ − μ_X̄) / σ_X̄.

Lower bound Z1 = (63 − 60) / 2.683 ≈ 3 / 2.683 ≈ 1.118.
Upper bound Z2 = (68 − 60) / 2.683 ≈ 8 / 2.683 ≈ 2.981.

From standard normal tables or a calculator, find Φ(Z2) and Φ(Z1):
- Φ(2.981) is approximately 0.9985 (very close to Φ(3) ≈ 0.99865).
- Φ(1.118) is approximately 0.869 (since Φ(1.12) is around 0.870, and Φ(1.118) is very close).

The probability is Φ(2.981) − Φ(1.118) ≈ 0.9985 − 0.869 ≈ 0.1295.

In decimal form, this difference is about 0.13 when rounded to two decimal places. The steps involve identifying the sampling distribution of the mean, computing the standard error, converting bounds to Z-scores, and using the standard normal CDF to obtain the probability.

COMM111_S2 Quiz 5: Sampling Distributions

Assume that the time needed by a worker to perform a maintenance operation is normally distributed, with a mean of 60 minutes and a standard deviation of 6 minutes. What is the probability that the average time needed by a sample of 5 workers to perform the maintenance is between 63 and 68 minutes?

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