Introduction to Statistics · Chapter 4 · Random Variables
A lab, not a reading: draw ten cards yourself, run the same experiment as a simulation, then hold both against the binomial table the formula predicts.
bookSHelf · Introduction to Statistics · §4.4 · a self-paced section
Learning objectives — by the end of this section you will be able to
§4.4.1 — Stats Lab: Discrete Distribution (Playing Card Experiment)
In §4.3 you learned to write down a binomial distribution and compute its probabilities on paper. Here you run the experiment those probabilities describe — draw a card, look at the suit, put it back, do it ten times — and hold your own results up against the numbers the formula predicted.
You will end up with three pictures of the same experiment: the theoretical distribution the formula gives you, the empirical distribution your class actually got with real cards, and a simulated distribution a calculator produces in a few seconds. The whole point of the lab is that these three will not match — and that the way they fail to match is itself the lesson.
Context Pause — why nobody's ten draws matches the formula
Ten draws is a small number
A fair coin lands heads half the time, and yet ten flips landing 7–3 surprises nobody. The same slack applies here: the formula says you should average 2.5 diamonds out of ten draws, and a group that gets 5 has not done anything wrong. Watch what happens when the class pools its results — the pooled tally sits much closer to the prediction than any single group's ten draws did.
Your turn — which picture is allowed to change?
Try It Now 4.4.1 — three distributions, one question
Of the three distributions this lab produces — theoretical, empirical, and simulated — which ones would come out differently if your class ran the whole lab again tomorrow with a fresh deck? Explain what that tells you about which one you are allowed to call "the right answer".
Step 1 — ask what each one is built from. The theoretical distribution is computed from the deck's structure: 13 diamonds in 52 cards, ten replaced draws. Nothing in that computation looks at data.
Step 2 — check which ones depend on data. The empirical distribution is a tally of what the cards actually did today, and the simulated distribution is a tally of what the random number generator actually did today. Both are measurements.
Answer: the empirical and simulated distributions would both change; the theoretical one would not. That is why the theoretical distribution is the standard the other two get compared against — not because it is more accurate about your class, but because it is the only one of the three that is not a sample. You are not checking whether the formula got your class right; you are checking whether your class's ten-draw sample landed where a sample from that formula should land.
§4.4.2 — what you need
§4.4.3 — Procedure
Definition 4.4.1 — Theoretical Probability
The probability of an event computed from the structure of the experiment itself, before any data is collected. For a well-described chance experiment with equally likely outcomes,
P(event)=total number of outcomesnumber of outcomes in the eventIt is a statement about the long run, not a prediction about any particular short run.
Insight Note — why replacement keeps every draw independent
The deck does not remember
After you replace the card and reshuffle, the deck is in exactly the state it was in before the draw. It has no record of having just handed you the ace of diamonds, and so it has no tendency to "make up for it" on the next draw. Every one of the ten draws is the first draw all over again.
§4.4.3 — run it once, on paper
Eight steps, one card at a time
§4.4.3 — what Step 5 is buying you
Step 5 is the one students skip, and it is the one that makes the arithmetic work. Because the card goes back before the next draw, every one of the ten draws faces the same full 52-card deck, so every draw has exactly the same chance of being a diamond and no draw tells you anything about the next one. That is sampling with replacement, and it is what lets you treat the ten draws as ten independent trials — the condition a binomial distribution requires.
Once you know the theoretical probability of a diamond, you know the whole distribution. The number of diamonds in ten independent draws, each with the same success probability, is exactly the setup §4.3 called binomial — so X is a binomial random variable and you can name its two parameters without collecting a single card.
§4.4.3 — a thousand repetitions in a few seconds
After you have run the ten real draws by hand, run the same experiment again with technology. Have your calculator or statistics app generate ten random whole numbers from 1 to 4, count how many of them come up as a chosen value (say, a 1), and treat that count as one simulated repetition of the experiment. Repeat until you have as many simulated repetitions as the class produced with real cards, so the two data sets are comparable.
The point is not that the machine is more accurate — it is that the machine will happily give you a thousand repetitions while the deck gives you ten, and the difference between ten and a thousand is exactly what "long-term probability" means.
Your turn — fill in the two blanks
Try It Now 4.4.2 — naming X
Fill in the two blanks from the Procedure section: what is the theoretical probability of drawing a diamond, and what are the two parameters of X?
Step 1 — count the diamonds. A standard deck has 52 cards in four suits of 13 each, so 13 of the 52 cards are diamonds.
P(diamond)=5213=41=0.25Step 2 — check the binomial conditions. There are ten draws, fixed in advance. Each draw is either a diamond (success) or not (failure). Because the card is replaced and the deck reshuffled, the draws are independent and the success probability stays at 0.25 for every one of them. All four conditions hold.
Step 3 — name the parameters. The number of trials is n=10 and the success probability is p=0.25.
Answer: the theoretical probability is 0.25, and X∼B(10, 0.25). Without the replacement step, the second draw would face a 51-card deck whose composition depends on the first draw, p would drift from draw to draw, and X would be hypergeometric instead of binomial.
Your turn — put a number on Step 5
Try It Now 4.4.3 — with vs. without replacement
Find the probability that the first two draws are both diamonds, once with replacement and once without, and say which of the two the binomial formula assumes.
With replacement — the card goes back, so the second draw faces the same 52-card deck as the first and the two draws are independent.
P(both diamonds)=(5213)(5213)=(0.25)(0.25)=0.0625Without replacement — the first diamond is gone, so the second draw faces 51 cards of which only 12 are diamonds.
P(both diamonds)=(5213)(5112)=(0.25)(0.2353)≈0.0588Answer: 0.0625 with replacement and about 0.0588 without — and the binomial formula assumes the first one. Without replacement, the second draw's probability depends on the first draw's outcome, and dependence is the one thing the binomial model is not allowed to have.
Try it in rāSHio — running the simulation without a programming calculator
Open rāSHio and choose File → Random Numbers…, then set Min 1, Max 4, and How many 10; count how many 1s come back and that is one simulated repetition. One setting matters more than all the others here: leave No repeats unticked. Your ten draws put the card back every time, so repeats are not just allowed, they are the whole point — ticking that box would break the independence the binomial model needs.
Figure 4.4.1
File → Random Numbers…
Set the minimum, maximum, and how many values to draw, leave No repeats unticked, and read off the generated whole numbers — that is one simulated repetition.
Figure 4.4.1: Drawing the simulation's random values in rāSHio: File → Random Numbers…
§4.4.4 — organize the data
Turn your ten draws — and your class's — into two columns: how often each count of diamonds happened, and what share of the class that represents.
§4.4.4 — a name for what you counted
Definition 4.4.2 — Relative Frequency
The proportion of times an outcome actually occurred in a collected data set:
RF(x)=total number of repetitionsfrequency of xContext Pause — a formula's column and your afternoon's column
Two columns, two different kinds of number
The P(x) column you are about to build comes out of a formula and will be the same for every class in every school forever. The RF(x) column comes out of your deck this afternoon and belongs to your class alone. When the lab asks you to compare them, it is asking how close a measurement got to the thing it was measuring.
§4.4.4 — measured, not derived
Relative frequency is an empirical quantity — it is measured, not derived. Unlike a theoretical probability, its value changes every time you rerun the experiment.
The two columns are built to be compared, so build them the same way. Round every relative frequency to four decimal places, the same precision the theoretical table will use, and check that your relative frequencies add to 1.0000 before you go any further. If they do not, you have either lost a repetition or divided by the wrong total — finding that now is much cheaper than finding it in the discussion questions.
§4.4.4 — recording what you collect
| X | Frequency | Relative Frequency |
|---|---|---|
| 0 | ||
| 1 | ||
| 2 | ||
| 3 | ||
| 4 | ||
| 5 | ||
| 6 | ||
| 7 | ||
| 8 | ||
| 9 | ||
| 10 |
Table 4.4.1: Your class's empirical results — how often each number of diamonds came up.
Try it in rāSHio — building Table 4.4.1 without tallying by hand
Paste your class's ten-draw counts into rāSHio with File → Delimited List…, then choose Graph → Frequency Table and tick Discrete values so every whole number from 0 to 10 gets its own row — the relative frequency column is built for you.
Figure 4.4.2
Graph → Frequency Table
Tick Discrete values and Calculate — every whole number from 0 to 10 gets its own row, frequency and relative frequency both filled.
Figure 4.4.2: Building a discrete-value frequency table in rāSHio: Graph → Frequency Table.
§4.4.4 — two numbers and a picture
Two blanks, then a histogram
Try it in rāSHio — drawing both histograms on the same scale
Keep your counts in the rāSHio spreadsheet and choose Graph → Histogram; set the bin start to 0 and the bin width to 1 so each bar is one value of X. Re-run the same steps on the P(x) column to get the theoretical histogram on a matching scale.
Your turn — thirty groups, pooled
Try It Now 4.4.4 — Ms. Delgado's class
Ms. Delgado's class splits into 30 groups, and each group runs the experiment once. She records: 2 groups got 0 diamonds, 5 got 1, 9 got 2, 7 got 3, 4 got 4, 2 got 5, and 1 got 6. Find the relative frequency of x=2, and find the sample mean xˉ.
Step 1 — check the total. 2+5+9+7+4+2+1=30 repetitions, matching the 30 groups.
RF(x=2)=309=0.3000Step 2 — weight each value by how often it happened.
xˉ=300(2)+1(5)+2(9)+3(7)+4(4)+5(2)+6(1)=3076≈2.5333Answer: RF(x=2)=0.3000 and xˉ≈2.5333. Any one of those 30 groups, looking only at its own ten draws, saw a whole number — none of them saw 2.5 at all. The average is not a value the experiment can produce; it is a value the experiment's results cluster around.
Your turn — the spread this time
Try It Now 4.4.5 — same 30 groups, s this time
Using the same 30-group data from Ms. Delgado's class, find the sample standard deviation s.
Step 1 — sum of squared values, weighted by frequency.
∑x2f=02(2)+12(5)+22(9)+32(7)+42(4)+52(2)+62(1)=254Step 2 — subtract the part explained by the mean (n=30, xˉ≈2.5333).
s2=29254−30(2.5333)2=2961.4667≈2.1195Answer: s=2.1195≈1.4559. Hold that beside the theoretical σ≈1.3693 two slides ahead. The sample came out slightly more spread out than the model predicts — ordinary behavior for 30 repetitions. Divide by n−1, not n; this is a sample, not the whole population of every ten-draw run that could ever happen.
Try it in rāSHio — reading xˉ and s in one step
Once your counts are in the spreadsheet, rāSHio's Stats → Summary Statistics returns xˉ and s for the number-of-diamonds column in one step, so you can check your two blanks against the μ and σ you compute from the formula next.
Figure 4.4.3
Stats → Summary Statistics
Point it at the number-of-diamonds column and read xˉ and s straight off the output — no manual weighting required.
Figure 4.4.3: Reading the mean and standard deviation in rāSHio: Stats → Summary Statistics.
§4.4.5 — the same lab, from the formula alone
Nothing here comes from the cards
§4.4.5 — the binomial table
| X | P(x) |
|---|---|
| 0 | |
| 1 | |
| 2 | |
| 3 | |
| 4 | |
| 5 | |
| 6 | |
| 7 | |
| 8 | |
| 9 | |
| 10 |
Table 4.4.2: The theoretical probability distribution of X, the number of diamonds in ten replaced draws.
§4.4.5 — one formula, eleven rows
Every entry in the P(x) column comes from the binomial probability formula of §4.3, with n=10 and p=0.25:
P(X=x)=(x10)(0.25)x(0.75)10−xThe mean and standard deviation come from the binomial shortcuts μ=np and σ=npq, where q=1−p. You do not have to compute all eleven probabilities by hand — a calculator's binomial PDF command fills the column in one pass — but compute two or three of them the long way first, so you know what the command is doing for you.
Insight Note — a millionth chance is not a zero chance
The tail is not empty, it is just thin
Getting all ten draws to come up diamonds has probability 0.2510, which is about one chance in a million. That is not zero. A probability table that rounds it to 0.0000 is telling you the outcome is rare, not that it is impossible — and the difference matters the day somebody's class actually rolls one.
Your turn — one row, plus the center and spread
Try It Now 4.4.6 — P(X=2), μ, σ
Compute P(X=2) from the formula, then find μ and σ for this distribution.
Step 1 — count the arrangements. (210)=210×9=45 different orders.
P(X=2)=45(0.25)2(0.75)8=45(0.0625)(0.1001)≈0.2816Step 2 — use the binomial shortcuts.
μ=np=10(0.25)=2.5σ=npq=10(0.25)(0.75)=1.875≈1.3693Answer: P(X=2)≈0.2816, μ=2.5, σ≈1.3693. Compare that μ to your class's xˉ and that σ to your s — same question, two completely different directions: one from the formula, one from the cards.
Try it in rāSHio — filling Table 4.4.2 without the by-hand arithmetic
Open rāSHio and choose Calculators → Binomial, set Trials 10 and Probability 0.25, then pick the Exact (P(X = k)) tab and step k from 0 to 10 to read each row of the P(x) column. The walkthrough below demonstrates the calculator in its At Least mode on different numbers, so use it for where the controls live rather than for this table's values — it reports probabilities only, μ and σ still come from np and npq.
Figure 4.4.4
Calculators → Binomial
Enter the number of trials and the success probability, then read the probability column for whichever k you need.
Figure 4.4.4: Building a binomial probability table in rāSHio: Calculators → Binomial.
Your turn — the check that catches a dropped row
Try It Now 4.4.7 — P(X=0) and what the column must add to
Compute P(X=0), and explain what the eleven entries of the P(x) column must add to and why.
Step 1 — apply the formula at x=0. Exactly one way to get no diamonds at all, and (0.25)0=1, so the whole expression collapses to the probability that all ten draws miss.
P(X=0)=(010)(0.25)0(0.75)10=(0.75)10≈0.0563Answer: P(X=0)≈0.0563, and the column must add to exactly 1 — §4.1's requirement that the probabilities of a discrete random variable are each between 0 and 1 and sum to 1. Use it as your check: totaling 0.9987 or 1.0021 is a rounding artifact you can move past, but totaling 0.87 means a row was dropped or mis-entered, and the histogram you build next will be wrong in a way that is hard to see by eye.
Your turn — expected count vs. observed count
Try It Now 4.4.8 — how many groups should get 2?
The theoretical distribution says P(X=2)≈0.2816. If 30 groups each run the experiment once, how many should get exactly 2 diamonds? Compare that to the 9 groups in Ms. Delgado's class who actually did.
Step 1 — turn a probability into an expected count.
30×0.2816=8.448Answer: about 8.4 groups were expected and 9 were observed — a difference of roughly half a group. The expected count is not a whole number even though the observed count has to be, which is the same "you cannot draw 2.5 diamonds" point from a different angle. This is the arithmetic that turns "the graphs look similar" into something you can defend: compare each bar's observed count against 30×P(x) and say which bars are off, and by how much.
§4.4.6 — using the data
Six blanks, in three matched pairs: read each one first off the formula, then off your own tally.
Context Pause — matched pairs, formula then data
RF means relative frequency
From here on, P(x) always means a probability read off the theoretical table, and RF(x) always means a relative frequency read off your class's empirical table. The questions ahead come in matched pairs on purpose: the same three events, answered first from the formula and then from the deck.
§4.4.6 — the same three events, twice
From Table 4.4.2
P(x=3)= _________
P(1<x<4)= _________
P(x≥8)= _________
From Table 4.4.1
RF(x=3)= _________
RF(1<x<4)= _________
RF(x≥8)= _________
Read the inequalities carefully before you add anything up. 1<x<4 is a strict inequality on both ends, so it covers x=2 and x=3 only — 1 and 4 are outside it. x≥8 does include 8, so it covers 8, 9, and 10. Getting an endpoint wrong here changes the answer far more than any rounding decision will.
Your turn — the theoretical half
Try It Now 4.4.9 — X∼B(10, 0.25)
Compute P(1<x<4) and P(x≥8) to four decimal places.
Step 1. 1<x<4 means x∈{2,3}; x≥8 means x∈{8,9,10}.
P(1<x<4)=P(2)+P(3)=0.2816+0.2503=0.5319 P(x≥8)=P(8)+P(9)+P(10)=0.0004+0.0000+0.0000=0.0004Answer: P(1<x<4)=0.5319, P(x≥8)=0.0004. More than half of all classes land on 2 or 3 diamonds, while fewer than 1 class in 2,000 lands on 8 or more. If your class did get 8 or more, you have not broken statistics — you have met the tail, and that is a better lab result than the boring one.
Your turn — the empirical half
Try It Now 4.4.10 — the same 30 groups, held beside the formula
Using Ms. Delgado's 30-group data (2 at 0, 5 at 1, 9 at 2, 7 at 3, 4 at 4, 2 at 5, 1 at 6), compute RF(x=3), RF(1<x<4), and RF(x≥8), and hold each beside its theoretical partner.
Step 1. RF(x=3)=307≈0.2333, RF(1<x<4)=309+7≈0.5333.
Step 2. No group got 8, 9, or 10, so RF(x≥8)=300=0.0000.
Answer: theoretical 0.2503 against empirical 0.2333; theoretical 0.5319 against empirical 0.5333; theoretical 0.0004 against empirical 0.0000. An empirical relative frequency of 0 does not mean the event is impossible; it means 30 repetitions were nowhere near enough to see something that happens about four times in ten thousand. That is exactly why you need the theoretical column to know it is there at all.
§4.4.7 — discuss with your group
For questions 1 and 2, think about the shapes of the two graphs, the probabilities, the relative frequencies, the means, and the standard deviations.
Your turn — work question 4 all the way through
Try It Now 4.4.11 — 500 repetitions instead of 30
Which of the two tables changes when the experiment is repeated 500 times instead of 30, in which direction, and what stays fixed?
Step 1. Table 4.4.1 holds frequencies and relative frequencies — counts of what actually happened. Table 4.4.2 holds P(x) values computed from n=10 and p=0.25, and never looks at the deck.
Step 2. With 500 repetitions instead of 30, each relative frequency is an average over many more trials, so the whole column settles closer to the matching P(x) — including RF(x≥8), which would finally start showing something other than 0.
Answer: Table 4.4.1 changes and Table 4.4.2 does not; the empirical histogram takes on the theoretical one's shape. The 0.25 was never a promise about your ten draws — it was a promise about what the fraction of diamonds does as the number of draws grows. Ten draws cannot show you that. Five hundred can start to.
Key Terms — six words this lab pins down
empirical probability — the relative frequency with which an outcome actually occurred in collected data; a measured quantity that changes each time the experiment is rerun.
long-term probability — the value a relative frequency settles toward as the number of repetitions grows large; what a theoretical probability is a statement about.
relative frequency — the frequency of an outcome divided by the total number of repetitions.
sampling with replacement — returning each selected item to the population before the next selection, so every trial faces the same population and the trials stay independent.
simulation — a technology-generated imitation of a chance experiment, used to produce many more repetitions than could be collected by hand.
theoretical probability — the probability of an outcome computed from the structure of the experiment rather than from data.
The headline result
Ten draws is a sample; five hundred is the long run.
Only Table 4.4.1 has data in it. Rerun the lab and the empirical and simulated distributions move — the theoretical table never does.
More repetitions do not change what the formula predicts; they change how closely your measurement of it agrees.
† 0.25 was never a promise about any ten draws in particular — it is a promise about where the relative frequency settles as the number of draws grows. A single class of ten cannot show that. Pooling every class in the building starts to.
§4.4 — conclusions
Sampling with replacement turns ten card draws into ten independent trials, so the count of diamonds is binomial — X∼B(10,0.25) — before a single card is drawn. The theoretical table is the standard; the empirical and simulated tables are measurements of it.
Small samples disagree with the formula and that is not an error — it is what "long-term" means. An empirical relative frequency of 0.0000 does not mean an event is impossible; it means the sample was too small to catch something rare.
Next: §4.5 — Continuous Probability Functions, where the discrete tables you just built give way to a curve.