Week 7 — Review & Exam 1 (Ch 1–3) - Statistics in the Modern World (MATH 1209)

This week in one line: there is no new content — you’ll review Weeks 1–6, work six fully worked exam-style problems, and take Exam 1.

Think of this week as practice: no new moves to learn, just the moves you already know, practiced until you can do them without thinking. Over Weeks 1–6 you learned to size up a study, describe a batch of numbers or categories honestly, and reason about chance with probability, expected value, and the Normal curve. Exam 1 asks you to do all three, on new scenarios, under time pressure, with no notes but the formula sheet you build yourself.

This unit is your self-study companion: a checklist of exactly what “ready” looks like, a compact reminder of what belongs on that formula sheet, six fully worked exam-style problems so you can check your own reasoning line by line, and a short list of exam-day tips. There is no new content this week — if a term below feels unfamiliar, that’s your cue to go back to the week that introduced it before exam day.

Exam 1 logistics

Coverage: Chapters 1–3 (everything from Weeks 1–6). Roughly Ch 1 ≈ 20%, Ch 2 ≈ 25%, Ch 3 ≈ 55% of the exam, by teaching time — Chapter 3 (probability, random variables, the Normal curve) is the biggest single piece, so weight your review time accordingly.
Length: given during your normal 75-minute class period; ~22 items total, mixing a small number of multiple-choice questions (at most ~18% of the exam) with short-answer and free-response problems where you must show your work to earn credit.
Closed book, closed notes. The only reference you may use is one handwritten 8.5″×11″ formula sheet, front and back — no worked examples or example problems on it, formulas and brief notes only — and it is collected with your exam.
Calculator: TI-83/84 (or your instructor’s stated equivalent). No phone calculators.
No makeup exams. If you miss Exam 1 for any reason, Exam 2 counts double toward your grade — showing up and doing your best on Exam 1 is always the better option.
Exams are printed in several equivalent randomized versions; you’ll simply work the version in front of you. Content and difficulty are the same across versions.

1Review checklist: what “ready for Exam 1” looks like¶

Work down this list chapter by chapter. For each bullet, ask yourself: could I do this without any notes or preparation, on paper, right now? If not, that’s your next review target — go back to the week named and redo its worked example without looking at the solution first.

1.1Chapter 1 — Data & Study Design (Week 1)¶

Classify a variable as numerical (discrete or continuous) or categorical (nominal or ordinal) — and catch the trap where a column of digits (like a birth month) is really categorical because averaging it would mean nothing.
Define population, sample, parameter (a fixed, usually unknown number describing the population, Greek letter like $\mu$ ), and statistic (a number you compute from a sample, Roman letter like $\bar{x}$ ) — and correctly label which is which in a description.
Identify a sampling method (simple random, stratified, cluster) and name a plausible source of bias in a described sampling plan.
Tell an observational study apart from an experiment, and explain in one sentence why only a randomized experiment supports a cause-and-effect claim (the idea of confounding).

1.2Chapter 2 — Summarizing Data (Weeks 2–3)¶

Compute the mean, median, mode, sample standard deviation $s$ , and the five-number summary ( $\min, Q_1, \text{median}, Q_3, \max$ ) for a list of numbers, by hand or by reading calculator/R output.
Compute the IQR ( $Q_3 - Q_1$ ) and apply the 1.5·IQR outlier rule to flag unusual values.
Read a histogram or boxplot and describe a distribution’s shape (symmetric, left/right-skewed, unimodal/bimodal).
Explain why skew pulls the mean away from the median, and why the median and IQR are more resistant to outliers than the mean and SD.
Build a frequency / relative-frequency table, and explain why a bar chart’s y-axis must start at 0 to be read honestly.
Read a two-way (contingency) table and correctly compute a marginal, joint, or conditional proportion — and know which one a question is actually asking for.
Judge from conditional proportions (or a segmented bar chart) whether two categorical variables appear associated.

1.3Chapter 3 — Probability, Random Variables & the Normal Curve (Weeks 3–6)¶

State probability as long-run relative frequency, and give the sample space of a simple process.
Apply the complement rule $P(A^c) = 1 - P(A)$ and the addition rule ( $P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B)$ , or just $P(A)+P(B)$ for disjoint events).
Apply the multiplication rule for independent events ( $P(A \text{ and } B) = P(A) \times P(B)$ ), and explain in words what independence means.
Build a discrete random variable’s probability distribution and check that the probabilities sum to 1.
Compute and interpret the expected value $E(X) = \sum x \cdot P(x)$ as a long-run average.
Compute the standard deviation of a random variable, $SD(X) = \sqrt{\sum (x-\mu)^2 P(x)}$ .
Explain why, for a continuous variable, probability is area under a density curve (and total area always equals 1), and name the key features of a Normal curve (bell shape, symmetric, centered at $\mu$ , spread controlled by $\sigma$ ).
Compute a z-score, $z = (x-\mu)/\sigma$ , and interpret it as “how many standard deviations from the mean.”
Apply the 68–95–99.7 empirical rule to estimate what proportion of values fall within 1, 2, or 3 standard deviations of the mean.
Compare two values from different Normal distributions by standardizing both to z-scores.
Sketch and shade a Normal curve to represent a stated probability.

2What belongs on your formula sheet¶

Your formula sheet may hold formulas, symbol definitions, and short notes — not worked examples. Here is the complete cumulative list for Ch 1–3; copying it by hand (rather than printing it) is itself good review, since writing each formula out forces you to notice which ones you’ve already memorized.

Concept	Formula	Symbols
Sample mean	$\bar{x} = \dfrac{\sum x_i}{n}$	$x_i$ = each value, $n$ = sample size
Sample standard deviation	$s = \sqrt{\dfrac{\sum (x_i-\bar{x})^2}{n-1}}$	$s$ = sample SD
Interquartile range	$IQR = Q_3 - Q_1$	$Q_1, Q_3$ = 25th/75th percentiles
Outlier fences	below $Q_1 - 1.5\cdot IQR$ or above $Q_3 + 1.5\cdot IQR$	—
Complement rule	$P(A^c) = 1 - P(A)$	$A^c$ = “ $A$ does not happen”
Addition rule (general)	$P(A \text{ or } B) = P(A)+P(B)-P(A \text{ and } B)$	disjoint events: drop the last term
Multiplication rule (independent)	$P(A \text{ and } B) = P(A) \times P(B)$	requires independence
Expected value	$E(X) = \sum x \cdot P(x)$	$\mu = E(X)$
SD of a random variable	$SD(X) = \sqrt{\sum (x-\mu)^2 P(x)}$	—
z-score	$z = \dfrac{x-\mu}{\sigma}$	$\mu$ = population mean, $\sigma$ = population SD
Empirical rule	68% within $\pm 1\sigma$ , 95% within $\pm 2\sigma$ , 99.7% within $\pm 3\sigma$	of the mean, for an approximately Normal distribution

Symbol legend to have memorized (not written on the sheet, since these are notation, not formulas): $\mu$ (population mean) vs. $\bar{x}$ (sample mean); $\sigma$ (population SD) vs. $s$ (sample SD); $n$ (sample size); $P(A)$ (probability of event $A$ ).

3Exam-style worked examples¶

Work each one yourself first — cover the solution with your hand or a sheet of paper — then check your steps against what’s shown. Every number below was computed in R (mosaic/BSDA, R 4.5.2) so you can trust it to the decimal place.

Exam-style Example 1 (Ch 1) — Reading a study and its vocabulary

Part A — design. A campus wellness center wants to know whether a 10-minute guided morning walk lowers self-reported stress. They recruit 60 student volunteers and use a random-number generator to assign 30 of them to take the morning walk every day for two weeks and the other 30 to change nothing (control). At the end of two weeks, every volunteer rates their stress on a continuous 0–10 scale.

Explanatory variable: whether a student walked or not — categorical, nominal (two labels, “walk” / “no walk,” with no natural order).
Response variable: the stress rating — numerical, continuous (it can fall anywhere on the 0–10 scale, and averaging it is meaningful).
Observational study or experiment? Experiment — the researchers themselves assigned the treatment (who walks), rather than just observing an existing group of walkers.
Why does this design support a cause-and-effect claim? Because the 60 volunteers were randomly assigned to the two groups, the walk and no-walk groups should be similar, on average, in every other way (sleep habits, coursework, personality) before the study starts. If stress ends up lower in the walking group, that difference isn’t explained by some other systematic difference between the groups (a confounding variable) — it’s attributable to the walk itself. An observational study (just asking students whether they already walk in the morning) couldn’t rule out confounding this cleanly, because morning walkers might already differ from non-walkers in ways connected to stress (schedule, motivation, fitness).

Part B — parameter vs. statistic. Separately, to practice the vocabulary: suppose (a luxury real research never actually has) the true mean commute time for all 8,500 CSUB commuter students happens to be known: $\mu = 19.4$ minutes — this is a parameter. A random sample of $n=6$ students self-reports commute times, in minutes: 12, 25, 18, 31, 15, 22.

suppressMessages({library(mosaic); library(BSDA)})
commute <- c(12, 25, 18, 31, 15, 22)
favstats(commute)
#>  min    Q1 median    Q3 max mean      sd n missing
#>   12 15.75     20 24.25  31 20.5 6.94982 6       0

The sample mean $\bar{x} = 20.5$ minutes is a statistic. Notice $\bar{x} = 20.5$ is close to, but not exactly equal to, $\mu = 19.4$ — a gap of 1.1 minutes. That gap isn’t an error; it’s ordinary sampling variability, the fact that a sample almost never lands exactly on the population value. (We build precise tools for that gap starting in Week 9 — for now, just know the vocabulary that separates a known-in-theory population number from a computed-from-data sample number.)

Exam-style Example 2 (Ch 2, numerical) — mean, median, spread, and an outlier

Eleven classmates share how many songs are on their current music playlist: 18, 22, 19, 25, 21, 20, 23, 19, 24, 17, 52.

songs <- c(18, 22, 19, 25, 21, 20, 23, 19, 24, 17, 52)
favstats(songs)
#>  min Q1 median   Q3 max     mean      sd  n missing
#>   17 19     21 23.5  52 23.63636 9.73933 11       0

Five-number summary: min = 17, $Q_1 = 19$ , median = 21, $Q_3 = 23.5$ , max = 52.

Center: the median (21 songs) is a better “typical” value here than the mean (23.6 songs) — the mean is pulled upward by that one long playlist.

Spread: $IQR = Q_3 - Q_1 = 23.5 - 19 = 4.5$ songs.

Outlier check: the fences are

Q_1 - 1.5 \cdot IQR = 19 - 1.5(4.5) = 12.25 \qquad Q_3 + 1.5 \cdot IQR = 23.5 + 1.5(4.5) = 30.25

The 52-song playlist is well above 30.25, so it’s a formally flagged outlier; every other value falls inside the fences.

Read-back: because of that outlier, the distribution is right-skewed (a long tail toward larger values), which is exactly why the mean (23.6) sits above the median (21) — the classic mean-gets-pulled-toward-the-tail pattern from Week 2.

Exam-style Example 3 (Ch 2, categorical) — a two-way table

A campus dining survey cross-tabulates 180 students’ meal-plan tier (Basic or Premium) with whether they’re satisfied with dining options:

	Basic	Premium	Total
Satisfied	54	72	126
Not satisfied	36	18	54
Total	90	90	180

dining <- matrix(c(54, 36, 72, 18), nrow = 2,
                  dimnames = list(Outcome = c("Satisfied", "Not satisfied"),
                                   Plan = c("Basic", "Premium")))
prop.table(dining, margin = 2)   # P(Outcome | Plan) -- conditional on each column
#>                Plan
#> Outcome         Basic Premium
#>   Satisfied       0.6     0.8
#>   Not satisfied   0.4     0.2

Marginal: overall, $126/180 = 0.700$ , so 70.0% of all 180 students are satisfied, ignoring plan tier entirely.
Joint: $54/180 = 0.300$ , so 30.0% of all 180 students are both on the Basic plan and satisfied.
Conditional: among Basic-plan students, $54/90 = 0.600$ (60.0% satisfied); among Premium-plan students, $72/90 = 0.800$ (80.0% satisfied).

Read-back: the conditional satisfaction rate jumps from 60.0% (Basic) to 80.0% (Premium) — a 20-point gap — so plan tier and satisfaction appear associated. If they weren’t associated, both conditional rates would sit close to the 70.0% marginal rate.

Exam-style Example 4 (Ch 3, probability rules) — complement, addition, multiplication

On a large campus, 62% of students drink coffee regularly ( $P(C) = 0.62$ ) and 35% report pulling at least one all-nighter this semester ( $P(N) = 0.35$ ). Assume, for this problem, that coffee-drinking and all-nighters are independent events.

pC <- 0.62; pN <- 0.35
p_and <- pC * pN            # multiplication rule (independent)
p_or  <- pC + pN - p_and    # general addition rule
c(p_and = p_and, p_or = p_or, `1-pC` = 1 - pC, `1-p_or` = 1 - p_or)
#>  p_and   p_or   1-pC 1-p_or
#>  0.217  0.753  0.380  0.247

Multiplication rule (independence lets you multiply): $P(C \text{ and } N) = 0.62 \times 0.35 = 0.217$ — 21.7% of students both drink coffee and pulled an all-nighter.
Addition rule (general form, since the events can overlap): $P(C \text{ or } N) = 0.62 + 0.35 - 0.217 = 0.753$ — 75.3% of students did at least one of the two.
Complement: $P(\text{not } C) = 1 - 0.62 = 0.38$ (38% don’t drink coffee regularly), and $P(\text{neither } C \text{ nor } N) = 1 - 0.753 = 0.247$ — 24.7% did neither.

Exam-style Example 5 (Ch 3, random variable) — expected value and SD of raffle winnings

A $5 fundraiser raffle ticket pays: $0 with probability 0.90, $10 with probability 0.08, and $100 with probability 0.02.

prize <- c(0, 10, 100)
p     <- c(0.90, 0.08, 0.02)
sum(p)                                  # check the distribution is valid
#> [1] 1
mu <- sum(prize * p)                    # E(prize)
sd_prize <- sqrt(sum((prize - mu)^2 * p))
mu
#> [1] 2.8
sd_prize
#> [1] 14.14779

net <- prize - 5                        # net winnings = prize - $5 ticket cost
sum(net * p)                            # E(net winnings)
#> [1] -2.2
sqrt(sum((net - sum(net * p))^2 * p))   # SD(net winnings)
#> [1] 14.14779

Check the distribution first: $0.90 + 0.08 + 0.02 = 1.00$ — a valid probability distribution.

Expected value: $E(\text{prize}) = 0(0.90) + 10(0.08) + 100(0.02) = 0 + 0.8 + 2 = \$2.80$ . Subtracting the $5 ticket cost, the expected net winnings are $E(X) = 2.80 - 5 = -\$2.20$ — on average, a player loses $2.20 per ticket. That’s normal for a fundraiser: the raffle is designed to have negative expected value for the player, since the difference funds the cause.

Standard deviation: $SD(\text{prize}) = \sqrt{(0-2.8)^2(0.90) + (10-2.8)^2(0.08) + (100-2.8)^2(0.02)} = \sqrt{200.16} \approx \$14.15$ . Subtracting a constant (the $5 ticket price) shifts the mean but never changes the spread, so $SD(\text{net winnings}) \approx \$14.15$ too — confirmed by the R output above. That large SD relative to the $2.20 average loss tells you why people buy raffle tickets despite a negative expected value: the small chance of the $100 prize creates a lot of variability around that average.

Exam-style Example 6 (Ch 3, Normal model) — z-scores and the empirical rule

Exam scores in Section A are approximately Normal with $\mu = 76$ , $\sigma = 8$ . A Section A student scores 92. Exam scores in Section B (a different exam) are approximately Normal with $\mu = 70$ , $\sigma = 5$ . A Section B student scores 82. Whose score is more impressive, relative to their own section?

z1 <- (92 - 76) / 8
z2 <- (82 - 70) / 5
c(z1 = z1, z2 = z2)
#>  z1  z2
#> 2.0 2.4

xpnorm(92, mean = 76, sd = 8)   # exact proportion below 92 in Section A
#> If X ~ N(76, 8), then
#>   P(X <= 92) = P(Z <= 2) = 0.9772
#>   P(X >  92) = P(Z >  2) = 0.02275

Standardize both scores: $z_1 = (92-76)/8 = 2.0$ and $z_2 = (82-70)/5 = 2.4$ . Even though 92 is a higher raw score than 82, the Section B student is farther above their own class’s mean (2.4 SDs vs. 2.0 SDs) — so the Section B score is the more impressive one, relative to its own distribution. Raw scores from different distributions can’t be compared directly; z-scores put them on the same scale.

Empirical rule, applied to Section A: about 95% of Section A scores fall within $\mu \pm 2\sigma = 76 \pm 16$ , i.e., between 60 and 92. That leaves about 5% split between the two tails, so roughly 2.5% of students scored above 92 — the Section A student is around the 97.5th percentile. The exact Normal calculation (which you’ll compute directly starting in Week 8 with xpnorm/normalcdf) gives 97.72%, confirming the empirical rule’s estimate was a close approximation, not a coincidence.

Figure 1. Two bell curves are drawn side by side: Section A’s curve centered at 76 with a shaded right tail starting at 92 (2.0 SDs out), and Section B’s curve centered at 70 with a shaded right tail starting at 82 (2.4 SDs out). Section B’s shaded tail is visibly thinner — farther into its own curve’s tail — even though 82 is a smaller raw number than 92, illustrating why standardizing, not the raw score, decides which result is relatively more impressive.

4Quick self-check before Exam 1¶

Try each one on paper first. Brief answers follow so you can check yourself — no full step-by-step solutions are given here on purpose; if your answer doesn’t match, redo the matching worked example above before you take Exam 1.

In Example 1’s wellness study, if the researchers had instead just surveyed students about their existing morning habits (no random assignment), what type of study would that be, and why would it not support a cause-and-effect conclusion? (Observational; a confounding variable, like existing fitness level or schedule, could explain any stress difference.)
Using Example 2’s playlist data, what would the mean become if you dropped the 52-song outlier and recomputed it on the other 10 values? Is it closer to or farther from the median of 21? (Mean of the remaining 10 values ≈ 20.8 songs — much closer to the median of 21, showing how strongly one point can pull the mean.)
In Example 3’s dining table, is “30.0% of all students are Basic-plan and satisfied” a marginal, joint, or conditional proportion? (Joint.)
A campus flyer plots four majors’ average weekly study hours (Business 12.4, Nursing 14.1, Psychology 13.0, CS 15.8) but starts the y-axis at 10 instead of 0. What’s the true ratio of the tallest bar to the shortest (CS to Business), and roughly how many times more dramatic does that gap look once the axis is truncated to start at 10? (True: 15.8/12.4 ≈ 1.27, about 27% higher. Truncated: apparent bars of 5.8 vs. 2.4, a ratio of about 2.42, i.e., about 142% “higher” — roughly five times more dramatic than the real 27% difference, from the same four numbers.)
If two events $A$ and $B$ are disjoint (they can never both happen), what is $P(A \text{ and } B)$ , and how does the addition rule simplify? ( $P(A \text{ and } B) = 0$ ; addition rule becomes $P(A \text{ or } B) = P(A) + P(B)$ .)
In Example 5’s raffle, would a $5 ticket with the same prizes but probabilities 0.80, 0.15, 0.05 (instead of 0.90, 0.08, 0.02) have a higher or lower expected net winnings than $-\$2.20$ ? You don’t need to compute the exact value — reason from the fact that the $100 prize is now more likely. (Higher — moving probability toward the $100 prize raises $E(\text{prize})$ , which raises the net expected value, even though it’s still probably negative.)
A z-score of $z = -1.5$ means what, in plain language? (The value is 1.5 standard deviations below the mean.)
According to the empirical rule, about what percent of values in an approximately Normal distribution fall more than 3 standard deviations from the mean, in either direction? (About 0.3%, since 99.7% fall within 3 SDs.)

MATH 952 exam clinic

Three arithmetic habits cause most of the point loss on Exam 1 — none of them are new statistics, just care with numbers:

Percents. Converting a count or proportion to a percent is one step: divide, then move the decimal two places right (multiply by 100). $54/90 = 0.600 \rightarrow 60.0\%$ — don’t stop at the decimal and call it a percent.
Order of operations in the SD formula. Square each deviation before you add them up, and take the square root last — squaring after summing gives a wrong, meaningless number.
Reading the right cell of a table. Before you divide, ask “a percent (or proportion) of what group?” A conditional proportion divides by a row or column total; a joint proportion divides by the grand total. Mixing these two up is the single most common two-way-table mistake.

Exam-day tips

Budget your time roughly by chapter weight: on a 75-minute exam, that’s about 15 minutes on Ch 1 material, 19 minutes on Ch 2, and 41 minutes on Ch 3 — don’t let one tough Chapter 1 question eat time you need for the Chapter 3 section.
Read the whole question before you start writing — many points are lost by answering the question you expected instead of the one actually asked (e.g., “conditional” vs. “marginal”).
Show every step. Partial credit exists for a correctly set up formula even if the final arithmetic slips — a bare final number with no work shown often can’t earn partial credit.
Sketch and shade a Normal curve for every z-score / empirical-rule question, even if it’s not required — it catches sign errors (above vs. below the mean) before you commit to an answer.
Bring your TI-83/84 with fresh batteries and your handwritten, front-and-back formula sheet. No phones, no shared calculators once the exam starts.
If you finish early, don’t leave — revisit any question where you weren’t fully confident and double-check your rounding and units (minutes vs. hours, percent vs. proportion).