The branch manager for United Savings and Loan in Seaside, Virginia, has worked with her employees in an effort to reduce the waiting time for customers at the bank. Recently, she and the team concluded that average waiting time is now down to 3.5 minutes with a standard deviation equal to 1.0 minute. However, before making a statement at a managers' meeting, this branch manager wanted to double-check that the process was working as thought. To make this check, she randomly sampled 25 customers and recorded the time they had to wait. She discovered that mean wait time for this sample of customers was 4.2 minutes. Based on the team's claims about waiting time, what is the probability that a sample mean for n = 25 people would be as large or larger than 4.2 minutes?

Question

BlackTom AI · Accepted Answer

The scenario involves testing whether the average waiting time has improved from a population mean of 3.5 minutes with a known standard deviation of 1.0 minute, using a sample of 25 customers.

First, compute the standard error of the mean: SE = sigma / sqrt(n) = 1.0 / sqrt(25) = 1.0 / 5 = 0.2.

Next, translate the observed sample mean into a z-score relative to the claimed process performance: z = (xbar - mu) / SE = (4.2 - 3.5) / 0.2 = 0.7 / 0.2 = 3.5.

We want the probability that a sample mean would be as large or larger than 4.2 minutes if the true mean is 3.5. This is P(Xbar >= 4.2) = P(Z >= 3.5).

Consulting the standard normal distribution, Phi(3.5) ≈ 0.999767, so the tail probability is 1 - Phi(3.5) ≈ 0.000233.

Now evaluate the answer choices:
- 0.0231: This corresponds to a z-value around 2.0 (much smaller tail). It is far too large for a z-score of 3.5, so this is not correct for the given computation.
- 0.0214: Similar to the previous, this is another tail probability near z ≈ 2.0; again it does not match the z = 3.5 result and is therefore not correct for this scenario.
- 0.0000: This is a rounded representation of the true tail probability 0.000233 when expressed with limited precision, effectively capturing the extremely small probability of observing a sample mean as large as 4.2 under the 3.5 baseline.
- 0.0512: This corresponds to a z around 1.66 or so, which would imply a much larger tail probability than the computed 3.5 standard errors away; thus it does not align with the calculation.

In summary, the large z-score (3.5) yields a very small tail probability, on the order of a few ten-thousandths, which is consistent with the rounded option 0.0000. The other options reflect tail probabilities associated with far smaller z-scores and therefore do not match the calculation for z = 3.5.

OSC 2030-009 Fall 2025 SUBMIT: Graded Homework Module 6 (Chap 6&7)

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