Drawing repeated samples from a known year of ambulance response times, the HS345 Unit 5 sampling distribution exercise example shows sample means tightening and losing their skew as n grows. Searches like "hs 345 unit 5 assignment example", "hs345 unit 5 sample" and "hs345 unit 5 example" land here.
What a finished HS345 Unit 5 sampling distribution exercise looks like
The exercise opens with the population described completely: 5,200 response times, mean 9.37 minutes, standard deviation 4.82, median 8.59, and a right tail reaching past 37 minutes, with 12.3 percent of calls over 15. A histogram of the population sits beside three histograms of sample means, each built from 2,000 random samples, for sizes 4, 16 and 64. A table under the four charts compares the observed spread of the sample means, 2.30, 1.24 and 0.59 minutes, with the values the standard error formula predicts, 2.41, 1.20 and 0.60. A skewness column shows the tail fading, from 1.0 in the population to 0.19 at n of 64. The final page works a probability question about a sample of 36 calls.
How a HS345 Unit 5 example is structured
The order separates what is known from what is shown. The population comes first and is described as fully as the course allows, because the exercise depends on the reader accepting that its mean and spread are facts, not estimates. Simulated sample means come next, one size at a time, so the narrowing is visible before any formula names it. Only then does the standard error formula appear, presented as a prediction the simulation can check; the close agreement in the table is the exercise's main evidence. Shape is treated separately from spread, since the central limit theorem makes a claim about each. The closing problem applies both: a supervisor auditing 36 random calls would see an average above 10.5 minutes only about 7.9 percent of the time, while a single call exceeds 10.5 about 35 percent of the time.
A population with nothing estimated
All 5,200 calls, their mean, standard deviation, median and tail. The exercise notes that a complete dispatch log makes this a rare case where the population is genuinely known.
Means of four, sixteen and sixty-four
Three histograms on one horizontal scale. At n of 4 the means still lean right; by 64 they form a narrow, nearly symmetric hump centered on 9.4.
Prediction against simulation
Sigma over the square root of n predicts 2.41, 1.20 and 0.60 minutes. The simulated spreads land within a few hundredths of the last two and about a tenth of the first.
Share within a minute of the truth
About 36 percent of means from samples of 4 fall within one minute of 9.4, against 58 percent at 16 and 91 percent at 64.
One call against thirty-six
A standard error of about 0.80 minutes and a z of 1.41 give 7.9 percent for a sample mean above 10.5. The single-call figure, near 35 percent, is set beside it for contrast.
Where marks go in HS345 Unit 5
Three distinctions and one calculation carry most of the credit on an HS345 sampling distribution exercise. The first separates the population from the distribution of sample means; papers labeling the n of 64 histogram as response times rather than mean response times usually lose it. The second separates standard deviation from standard error, which the table makes concrete. The third separates the claim about spread, which holds at any n, from the claim about shape, which improves with n and is only approximate for a skewed population at n of 4. The calculation is the closing probability, where credit requires the standard error, not the population standard deviation, in the denominator of z. Graders also deduct for stating that samples of 30 are always normal, and for simulations reported without the number of repetitions.
Get a HS345 Unit 5 example written to your instructions
Your HS345 Unit 5 exercise may supply a population, a simulation applet or only parameters, and a custom sample works from whichever it provides. Forward the prompt with the rubric and any file. Returned in 24-48 hours at no charge the first time, the sample keeps population, sample and sampling distribution visibly separate and checks every standard error against the data.
HS345 Unit 5 questions, answered
Why do the simulated spreads not match the formula exactly?
Because 2,000 samples is a large but finite number, so the simulated standard deviation of the means carries its own sampling error. At n of 4 the skewed population adds a small further gap. The example reports both figures and explains that close agreement, not perfect agreement, is what a simulation is expected to show.
Is it realistic to know a population mean?
Occasionally. A full year of dispatch records, every discharge in a hospital's database or a complete registry can be treated as a population for a defined question. Most research samples because the population is unavailable. The exercise uses a known population so that the behavior of sample means can be checked against a true value rather than assumed.
What if my course uses an applet instead of a dataset?
The same structure works. Report the population settings the applet uses, the sample sizes chosen, the number of repetitions and the resulting spread and shape of the means. A custom sample can be built around the applet your section names, with screenshots described in words where the rubric asks for them.