Continuous Distribution

Step 1 — ask what each one is built from. The theoretical distribution is computed from the generator's design: it is set to produce values between 0 and 1, with no value favored over any other. Nothing in that computation looks at data.

Step 2 — check which one depends on data. The empirical distribution is a tally of the 50 numbers the generator actually handed you this afternoon. It is a measurement.

Answer: the empirical distribution would change and the theoretical one would not. That is why the theoretical distribution is the standard the other one gets compared against — not because it is more accurate about your 50 values, but because it is the only one of the two that is not a sample. You are checking whether your 50-value sample landed where a sample from that formula should land, not whether the formula got your run right.

Step 1 — median. Twelve values, so the median sits between the 6th and 7th: $$\text{median} = \frac{0.4508 + 0.5624}{2} = 0.5066$$

Step 2 — $Q_1$ from the lower half. The lowest six values run 0.0412–0.4508; their median is the average of the 3rd and 4th: $$Q_1 = \frac{0.2043 + 0.2765}{2} = 0.2404$$

Step 3 — $Q_3$ from the upper half. The highest six run 0.5624–0.9564; their median is the average of the 9th and 10th overall: $$Q_3 = \frac{0.7042 + 0.8219}{2} \approx 0.7631$$

Answer: median $= 0.5066$, $Q_1 = 0.2404$, $Q_3 = 0.7631$. Hold those against the theoretical 0.5000, 0.2500, 0.7500 — each is off by less than 0.015 from a sample of only twelve. The sample does not land on the theoretical value, it lands near it.

Step 1 — add the twelve values. $$0.0412+0.1187+0.2043+0.2765+0.3391+0.4508+0.5624+0.6130+0.7042+0.8219+0.8873+0.9564 = 5.9758$$

Step 2 — divide by the count. $$\bar{x} = \frac{5.9758}{12} \approx 0.4980$$

Answer: $\bar{x} \approx 0.4980$, within 0.002 of the theoretical mean 0.5000. Do not read that as "twelve values is plenty" — it is one run, and a different twelve values could easily have come out at 0.44 or 0.56. What the mean does reliably is sit near the center; how near depends on how many values you averaged.

Step 1 — subtract the part explained by the mean. With $n=12$ and $\bar{x}\approx0.4980$, $$s^2 = \frac{\sum x^2 - n\bar{x}^2}{n-1} = \frac{4.0176696 - 12(0.49798)^2}{11} \approx \frac{1.0419}{11} \approx 0.0947$$

Step 2 — take the square root. $$s = \sqrt{0.0947} \approx 0.3078$$

Answer: $s \approx 0.3078$, against a theoretical $\sigma \approx 0.2887$ — about 7% more spread out than the model predicts. Divide by $n-1$, not $n$: these twelve values are a sample, not every value the generator could ever produce. And $s$ drifted further from $\sigma$ than $\bar{x}$ did from $\mu$ — built from squared distances, a single value out near an end moves it more than it moves the mean.

Step 1 — bar widths. The interval from 0 to 1 has length 1, split into equal pieces: $$\text{eight bars: } \frac{1}{8}=0.125 \qquad \text{five bars: } \frac{1}{5}=0.2$$

Step 2 — width to expected count. Every value between 0 and 1 is equally likely, so the chance of landing in a bar is exactly that bar's width, and the expected count is 50 times the width: $$50(0.125)=6.25 \qquad 50(0.2)=10$$

Answer: eight bars of width 0.125 with 6.25 values expected in each, five bars of width 0.2 with 10 expected. No bar can hold a quarter of a value — an expected count is an average over many reruns of the lab, not a prediction of what any one histogram will show, which is exactly why a bar holding 4 or 9 is not evidence the generator is broken.

Step 1 — compare expected counts. Aiko's eight bars expect 6.25 values each; Imani's five bars expect 10 each.

Step 2 — ask how much a bar can wobble. A bar 3 above expectation is off by about half its height at 6.25, but only about a third of its height at 10 — the same absolute wobble looks bigger on a shorter bar.

Step 3 — see what wider bars do. Widening the bars pools more values into each one, and pooling averages the wobble out.

Answer: both are right — Imani's fewer, wider bars smooth the picture, while Aiko's more, narrower bars expose the run-to-run variation. Neither histogram is more truthful; they trade detail against steadiness. "The data looks uniform" is partly a statement about the data and partly a statement about how you chose to draw it.

Step 1 — identify the endpoints. $a=0$, $b=1$.

Step 2 — the mean. $$\mu = \frac{a+b}{2} = \frac{0+1}{2} = 0.5$$

Step 3 — the standard deviation. $$\sigma = \sqrt{\frac{(b-a)^2}{12}} = \sqrt{\frac{1}{12}} \approx 0.2887$$

Answer: $\mu = 0.5$, $\sigma \approx 0.2887$ — a good deal smaller than the 0.5 half-width of the interval. The uniform distribution's shape shows up in that number: values are spread evenly rather than piled at the two ends, so a typical value sits about 0.29 from the center rather than the 0.5 you'd get if every value sat at 0 or 1.

Step 1 — write the cdf. $$P(X \le x) = \frac{x-a}{b-a} = \frac{x-0}{1-0} = x$$

Step 2 — set it to each target proportion. $Q_1$ has 0.25 below it, so solve $x=0.25$; the median has 0.50, so $x=0.50$; $Q_3$ has 0.75, so $x=0.75$.

Answer: $Q_1 = 0.2500$, median $= 0.5000$, $Q_3 = 0.7500$ — the four-decimal values your sorted 50 values get held up against. The median equals the mean here because the uniform distribution is symmetric, a fact that will stop being true the moment you meet a skewed distribution.

Step 1 — line the pairs up. Mean 0.4980 vs. 0.5000; median 0.5066 vs. 0.5000; $Q_1$ 0.2404 vs. 0.2500; $Q_3$ 0.7631 vs. 0.7500.

Step 2 — measure each gap. The gaps are 0.0020, 0.0066, 0.0096, and 0.0131 — all under 0.014, on a scale where the whole distribution is only 1 wide.

Answer: yes, close — every statistic is within about 1.3% of the width of the distribution. They are not exact because the empirical values are measurements of a random sample while the theoretical values are computed from the generator's design; the quartiles drifted more than the mean, which is normal — each quartile is pinned by the position of just two of the twelve values, so it is the shakier estimate of the pair.

Step 1 — the theoretical $IQR$. $$IQR = Q_3 - Q_1 = 0.75 - 0.25 = 0.50$$

Step 2 — the fences. $$Q_1 - 1.5(IQR) = 0.25 - 0.75 = -0.50 \qquad Q_3 + 1.5(IQR) = 0.75 + 0.75 = 1.50$$

Answer: the fences sit at $-0.50$ and $1.50$ — outside the 0-to-1 interval the generator can ever produce — so no value it produces could ever be flagged as a potential outlier. That is a real property of the uniform distribution: with the data spread evenly and no tails, there is nothing far from the middle for the rule to catch. If your own sample's fences do flag a value, look at it — a typo or a mis-entered decimal is the likelier explanation, not a surprising draw.

Step 1. The empirical column holds $\bar{x}$, $s$, the quartiles, and the median of numbers the generator actually produced. The theoretical column holds values computed from $a=0$ and $b=1$ — it never looks at the generator.

Step 2. With 500 values, every empirical statistic is built from ten times as much information, so each settles closer to its theoretical partner: the mean tightens toward 0.5000, the quartiles toward 0.2500 and 0.7500, $s$ toward 0.2887.

Step 3. With eight bars, the expected count per bar rises from 6.25 to 62.5, so a bar a few values off is now a small fraction of its height instead of a large one — the histogram flattens toward the rectangle the density function actually is.

Answer: the empirical column changes and tightens around the theoretical one, which does not move. The 0.5 was never a promise about your 50 values — it was a promise about what the average of generated values does as you keep generating. Fifty values cannot show you that. Five hundred can start to.