4.1 Sampling Distributions for Sample Means — Answers and Rubrics

$\mu_{\bar x}=12$ minutes and $\sigma_{\bar x}\approx 6/\sqrt{64}=0.75$ minute. These use the known population parameters; 11.9 and 5.7 are statistics from one pilot sample, not newly known true parameters. Individual waiting times retain population SD 6 minutes. Only the size-64 mean’s sampling SD is about 0.75 minute, using the small-fraction approximation.

The SRS is random, and $64\leq 0.10(100{,}000)=10{,}000$, supporting the usual independence approximation without a finite-population correction. The population is not normal, but n=64≥30 and the supplied moderate skew/no-extreme-outlier description supports a CLT-based approximately normal mean model. This is an approximation, not a claim that the individual waiting times are normal.

(i) A normal population gives a normal sampling-mean distribution at n=9 under independent sampling. (ii) Randomness and 40≤1,000 pass; n=40≥30 and moderate skew support an approximate normal mean. (iii) Randomness and 30≤1,000 pass, but extreme skew may require a sample much larger than 30; the threshold alone does not establish an adequate normal approximation. It does not prove every such n=30 distribution fails. (iv) Voluntary selection is not random sampling. (v) 2,000>1,000 fails the usual 10% approximation for the SD formula without a finite-population correction.

$z=(13.5-12)/0.75=2$, giving upper-tail probability about 0.02275. Across repeated SRSs of 64 records from the service’s defined-year population, sample means average 12 minutes and typically deviate from 12 by about 0.75 minute. About 2.28% of these size-64 samples would have an average above 13.5 minutes under the stated approximation. This concerns entire sample means, not the percentage of individual customers waiting above 13.5 minutes.