4.6 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. Be clear about the direction of that comparison: you are not checking whether the formula got your run right, you are checking whether your 50-value sample landed where a sample from that formula should land.

Step 1 — find the median. There are twelve values, so the median sits between the 6th and 7th.

$$ \text{median} = \frac{0.4508 + 0.5624}{2} = \frac{1.0132}{2} = 0.5066 $$

Step 2 — take the first quartile from the lower half. The lowest six values are 0.0412 through 0.4508, and their median is the average of the 3rd and 4th.

$$ Q_1 = \frac{0.2043 + 0.2765}{2} = \frac{0.4808}{2} = 0.2404 $$

Step 3 — take the third quartile from the upper half. The highest six values are 0.5624 through 0.9564, and their median is the average of the 9th and 10th values overall.

$$ Q_3 = \frac{0.7042 + 0.8219}{2} = \frac{1.5261}{2} \approx 0.7631 $$

Answer: Hannah's median \(= 0.5066\), \(Q_1 = 0.2404\), \(Q_3 = 0.7631\). Hold those against the theoretical 0.5000, 0.2500, and 0.7500 you will compute later — each one is off by less than 0.015, from a sample of only twelve. That is a preview of the whole lab: 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: \(\bar{x} \approx 0.4980\). The theoretical mean is 0.5000, so this sample landed within 0.002 of it. 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 \(\bar{x} \approx 0.4980\),

$$ s^2 = \frac{\sum x^2 - n\bar{x}^2}{n - 1} = \frac{4.0176696 - 12(0.49798)^2}{11} \approx \frac{4.0176696 - 2.9758}{11} \approx \frac{1.0419}{11} \approx 0.0947 $$

Step 2 — take the square root.

$$ s = \sqrt{0.0947} \approx 0.3078 $$

Answer: Cole gets \(s \approx 0.3078\). Hold his number beside the theoretical \(\sigma \approx 0.2887\) you will compute in the Theoretical Distribution section — the sample came out 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 notice that \(s\) drifted further from \(\sigma\) than \(\bar{x}\) did from \(\mu\). A standard deviation is built from squared distances, so a single value out near an end moves it more than it moves the mean.

Step 1 — get the 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 — turn a width into an expected count. Because every value between 0 and 1 is equally likely, the chance of landing in a bar is exactly that bar's width, so 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, and five bars of width 0.2 with 10 expected in each. The 6.25 is worth staring at — 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 that the generator is broken.

Step 1 — compare the 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 that comes out 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. Imani's five bars each absorb the ups and downs of what would have been more than one of Aiko's bars.

Answer: both students are right, because the fewer and wider bars Imani chose smooth the picture while the more and narrower bars Aiko chose expose the run-to-run variation. Neither histogram is more truthful than the other — they trade detail against steadiness. The reason the lab has you draw both is so you notice that "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. Here \(a = 0\) and \(b = 1\).

Step 2 — apply the mean formula.

$$ \mu = \frac{a + b}{2} = \frac{0 + 1}{2} = 0.5 $$

Step 3 — apply the standard deviation formula.

$$ \sigma = \sqrt{\frac{(b - a)^2}{12}} = \sqrt{\frac{1}{12}} = \frac{1}{\sqrt{12}} \approx 0.2887 $$

Answer: \(\mu = 0.5\) and \(\sigma \approx 0.2887\). Notice that \(\sigma\) is a good deal smaller than the 0.5 half-width of the interval. That is the uniform distribution's shape showing up in a number: the values are spread evenly rather than piled at the two ends, so a typical value sits about 0.29 away from the center rather than the 0.5 you would get if every value were at 0 or 1.

Step 1 — write down the cdf. With \(a = 0\) and \(b = 1\),

$$ P(X \leq x) = \frac{x - a}{b - a} = \frac{x - 0}{1 - 0} = x $$

Step 2 — set it equal to each target proportion and solve. The first quartile is the value with 0.25 of the distribution below it, so solve \(x = 0.25\). The median has 0.50 below it, so \(x = 0.50\). The third quartile has 0.75 below it, so \(x = 0.75\).

Answer: \(Q_1 = 0.2500\), median \(= 0.5000\), \(Q_3 = 0.7500\). These are the four-decimal values to write in the Theoretical Distribution blanks, and they are the numbers 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 four pairs up. Mean 0.4980 against 0.5000; median 0.5066 against 0.5000; \(Q_1\) 0.2404 against 0.2500; \(Q_3\) 0.7631 against 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.

Step 3 — say why they are not exact. Twelve values are a sample. Each one is drawn independently, and there is no mechanism forcing a sample's quartile to equal the distribution's quartile; the only guarantee is that the sample statistics cluster around the theoretical ones and cluster more tightly as the sample grows.

Answer: yes, close — every statistic is within about 1.3% of the width of the distribution — and 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. Notice which gap is biggest: the quartiles drifted more than the mean did. That is normal. The mean uses all twelve values, while each quartile is pinned by the position of just two of them, so a quartile is the shakier estimate of the pair.

Step 1 — get the IQR from the theoretical quartiles.

$$ IQR = Q_3 - Q_1 = 0.75 - 0.25 = 0.50 $$

Step 2 — build the fences.

$$ Q_1 - 1.5(IQR) = 0.25 - 1.5(0.50) = 0.25 - 0.75 = -0.50 $$ $$ Q_3 + 1.5(IQR) = 0.75 + 1.5(0.50) = 0.75 + 0.75 = 1.50 $$

Step 3 — compare the fences to the possible values. The generator only ever produces values between 0 and 1, and both fences sit outside that interval.

Answer: the fences are at \(-0.50\) and \(1.50\), so no value this generator can produce could ever be flagged as a potential outlier. That is a real property of the uniform distribution, not a quirk of the numbers: with the data spread evenly and no tails, there is nothing far from the middle for the rule to catch. If your own fences (computed from your sample's quartiles rather than the theoretical ones) do flag a value, look at it — the likeliest explanations are a typo when you copied a number into the table or a value entered at the wrong number of decimal places, not a surprising draw.

Step 1 — the first quartile (part b). The group's \(Q_1 = 0.2404\) against a theoretical 0.2500, so the sample's first quartile is 0.0096 lower than expected.

Step 2 — the third quartile (part d). The group's \(Q_3 = 0.7631\) against a theoretical 0.7500, so the sample's third quartile is 0.0131 higher than expected.

Step 3 — the maximum (part e). The group's largest value is 0.9564. The distribution runs to 1, so the sample fell 0.0436 short of the top of the interval — expected, since landing in the top 4% of the range takes a lucky draw and there were only twelve.

Step 4 — the width of the IQR (part f).

$$ IQR = 0.7631 - 0.2404 = 0.5227 $$

against a theoretical 0.5000, so the sample's middle half is 0.0227 wider than the model predicts.

Answer: \(Q_1\) came in 0.0096 low, \(Q_3\) came in 0.0131 high, the maximum fell 0.0436 short of 1, and the IQR came out 0.5227 — about 4.5% wider than 0.50. Read those together and the picture is coherent: both quartiles drifted outward, so of course the distance between them grew. Reporting the IQR gap as a separate surprise, when it is just the sum of the two quartile gaps you already reported, is the most common way this part of the lab gets written up wrong.

Step 1 — identify what each column is made of. 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 \(b = 1\).

Step 2 — see which one has data in it. Only the empirical column does. The theoretical column never looks at the generator, so producing ten times as many values cannot move a single entry in it.

Step 3 — say which direction the empirical column moves. With 500 values, every statistic is built from ten times as much information, so each one settles closer to its theoretical partner. The mean tightens toward 0.5000, the quartiles toward 0.2500 and 0.7500, and the sample standard deviation toward 0.2887.

Step 4 — say what happens to the picture. With eight bars the expected count per bar rises from 6.25 to 62.5, so a bar that runs a few values over or under is now a small fraction of its height instead of a large one. The histogram flattens out and starts looking like the rectangle the density function actually is.

Answer: the empirical column changes and the theoretical column does not; the empirical statistics tighten around the theoretical ones and the histogram flattens toward a rectangle. This is the whole meaning of a continuous distribution's parameters, and it is the lab's real payoff. 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.