🎓 MDM4U Final Exam & Culminating Investigation

Comprehensive Final Evaluation — Chapters 1 through 8 plus Culminating Investigation (Strand D)

✅ Counts for 30% of Final Course Mark

⏱️ Duration: 3 hours (180 minutes) for the written exam | Total: /100 marks
Materials allowed: Scientific calculator, ruler, standard normal table (provided). No graphing technology during the exam.
Show all work. Answers without supporting work receive partial credit at best.

K/U

/25

Thinking

/25

Comm.

/25

Applic.

/25

Coverage: Ch 1 Counting & Sample Spaces · Ch 2 Permutations · Ch 3 Combinations · Ch 4 Probability · Ch 5 Discrete Distributions (Binomial, Geometric, Hypergeometric) · Ch 6 Normal Distribution · Ch 7 Statistics & Sampling · Ch 8 Correlation, Regression. The Culminating Investigation rubric is provided at the end of this document.

Part A: Knowledge & Understanding [25 marks]

Question 1 [2 marks]

Compute $ P(8, 3) - C(8, 3) $.

Solution:

$ P(8,3) = 336 $, $ C(8,3) = 56 $. Difference $ = 336 - 56 = 280 $.

Question 2 [2 marks]

A coin is flipped 6 times. $ P(\text{exactly 4 heads}) = ? $ (decimal to 4 places)

Solution:

$ P = \binom{6}{4}(0.5)^6 = 15 \cdot \dfrac{1}{64} = 0.234375 \approx 0.2344 $.

Question 3 [2 marks]

For $ X \sim N(70, 8) $, find the z-score corresponding to $ x = 82 $.

Solution:

$ z = \dfrac{82-70}{8} = 1.5 $.

Question 4 [2 marks]

From a deck of 52 cards, 5 cards are drawn. The number of possible 5-card hands is:

Solution:

$ \binom{52}{5} = 2\,598\,960 $.

Question 5 [2 marks]

For $ X \sim \mathrm{Binomial}(40, 0.25) $, find $ E(X) $.

Solution:

$ E(X) = np = 40(0.25) = 10 $.

Question 6 [2 marks]

A die is rolled until a 6 appears. The expected number of rolls is:

Solution:

Geometric: $ \mu = \dfrac{1}{p} = \dfrac{1}{1/6} = 6 $.

Question 7 [2 marks]

A correlation $ r = -0.7 $. The proportion of variation in $ y $ explained by $ x $ is:

Solution:

$ r^2 = (-0.7)^2 = 0.49 $.

Question 8 [2 marks]

How many distinct arrangements of the letters of "STATISTICS"?

Solution:

10 letters: 3 S, 3 T, 1 A, 2 I, 1 C. $ \dfrac{10!}{3!\,3!\,1!\,2!\,1!} = \dfrac{3\,628\,800}{6 \cdot 6 \cdot 2} = 50\,400 $.

Question 9 [2 marks]

In a stratified random sample, a researcher divides a population by:

Numbering everyone and using random digits
Subgroups, then samples within each
Selecting every $ k $th individual
Choosing whoever volunteers

Solution:

Stratified = subgroups (strata), then sample within each.

Question 10 [2 marks]

A bag has 4 red, 6 blue, 2 green marbles. Two are drawn without replacement. $ P(\text{both red}) = ? $ (decimal to 4 places)

Solution:

$ P = \dfrac{4}{12}\cdot\dfrac{3}{11} = \dfrac{12}{132} = \dfrac{1}{11} \approx 0.0909 $.

Question 11 [3 marks]

Find the median of: 8, 12, 15, 19, 22, 25, 30.

Solution:

For 7 values, median = 4th = 19.

Question 12 [2 marks]

The standard normal table gives $ P(Z < 1.0) \approx $ (4 places):

Solution:

From the table, $ P(Z < 1.0) = 0.8413 $.

Part B: Thinking [25 marks]

Question 13 [5 marks]

A committee of 5 is to be selected from 8 men and 6 women. a) How many committees of 5 are possible? b) How many have exactly 3 women? c) How many have at least 3 women?

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Question 14 [5 marks]

In a class of 30, the probability of a student passing the test is 0.8. Use the binomial distribution to find:
a) $ P(\text{exactly 25 pass}) $.
b) $ E(X), \mathrm{Var}(X) $ for $ X $ = number who pass.
c) Use the normal approximation to estimate $ P(X \le 22) $. Apply the continuity correction.

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Question 15 [5 marks]

A box contains 100 widgets, of which 8 are defective. A sample of 10 is drawn without replacement. a) Identify the distribution. b) Find $ P(\text{exactly 1 defective in sample}) $. c) Find $ P(\text{at least 1 defective}) $. d) State the expected number of defectives.

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Question 16 [5 marks]

For data $ x: 1, 2, 3, 4, 5 $ and $ y: 3, 5, 6, 8, 11 $: a) Compute $ \bar{x}, \bar{y}, s_x, s_y $. b) Compute $ r $. c) State the regression line. d) Predict $ y $ when $ x = 3.5 $. e) Compute $ r^2 $ and interpret.

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Question 17 [5 marks]

A bag contains 4 white and 6 black balls. Three balls are drawn without replacement. Construct the probability distribution for $ X $ = number of white balls drawn. Compute $ E(X) $ and $ \mathrm{Var}(X) $.

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Part C: Communication [25 marks]

Question 18 [5 marks]

Explain the difference between the binomial and hypergeometric distributions. Use a specific example involving cards, marbles, or quality-control to demonstrate when each applies. Discuss the role of "with vs. without replacement."

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Question 19 [5 marks]

A magazine reports: "Studies show that students who eat breakfast score higher on tests. Therefore, breakfast causes higher test scores." Critique this conclusion. Identify lurking variables and the difference between observational studies and controlled experiments.

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Question 20 [5 marks]

Describe the empirical (68-95-99.7) rule. Sketch a normal curve labelled with $ \mu \pm \sigma, \mu \pm 2\sigma, \mu \pm 3\sigma $. Verify the 68% figure using the standard normal table.

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Question 21 [5 marks]

A student claims, "If we use the binomial distribution to model 100 voters in a national poll, we are assuming the voters are independent." Is this assumption reasonable? When does it break down? Discuss when the hypergeometric distribution should be used instead.

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Question 22 [5 marks]

Describe a step-by-step procedure for designing an unbiased survey, including: (a) defining the population, (b) selecting a sampling method, (c) writing neutral questions, (d) ensuring high response rate, (e) interpreting and reporting results responsibly.

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Part D: Application [25 marks]

Question 23 [5 marks]

In Lotto 6/49, a player chooses 6 distinct numbers from 1 to 49. a) How many possible tickets exist? b) Find $ P(\text{matching all 6}) $. c) Find $ P(\text{matching exactly 5 of 6}) $ (assuming the winning combination is fixed). d) If a ticket costs $3 and the jackpot averages $5 million (assume no shared prize), is the bet "fair" in terms of expected value? Compute and interpret.

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Question 24 [5 marks]

A coffee chain claims its mean wait time is 4.0 minutes with $ \sigma = 0.5 $ min, normally distributed. a) What proportion of customers wait less than 3 minutes? b) The chain promises 95% of customers will wait less than $ T $ minutes. Find $ T $. c) If the actual wait time follows the same distribution, what is $ P(\text{wait} > 5\text{ min}) $?

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Question 25 [5 marks]

A medical screening test has sensitivity 0.98 (true positive rate) and specificity 0.95 (true negative rate). Disease prevalence is 0.5%. a) Find $ P(\text{disease} \mid \text{positive test}) $ using Bayes' theorem and a tree diagram. b) Comment on the implication for medical decision-making.

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Question 26 [5 marks]

A car dealership records monthly sales ($ y $, in units) vs advertising spending ($ x $, in $1000) over 6 months: $ (5, 30), (8, 45), (10, 50), (12, 60), (15, 70), (18, 85) $.
a) Plot the data.
b) Compute $ r $ and interpret strength/direction.
c) State the regression line.
d) Predict sales when $ x = 13 $.
e) Discuss validity of extrapolating to $ x = 30 $.

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Question 27 [5 marks]

A survey of 1000 Canadians asks if they favour a new tax. 540 say yes. The pollster claims this is "evidence of majority support." a) Use the normal approximation to find the probability of getting at least 540 "yes" responses if the true proportion is 0.50. b) Comment on whether the result is statistically meaningful (use a 5% threshold).

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Culminating Data-Management Investigation Rubric

📊 Strand D — The Culminating Investigation (15% of Final Evaluation)

The Culminating Investigation is a separate, multi-week independent inquiry submitted before the final exam window. It contributes 15% of the 30% Final Evaluation, with the remaining 15% from this written exam.

Required Components

Research question (significant, personally meaningful, requires substantial data analysis)
Data plan (variables, source: primary or secondary, sampling method, ethics)
Data collection or acquisition (organized table, properly cited if secondary)
Two-variable analysis (scatter plot, correlation, regression, residuals)
Univariate descriptive statistics (mean, median, spread, distribution shape)
Probability or distribution model (where appropriate, e.g., normal, binomial fit)
Interpretation (causation discussion, lurking variables, limitations)
Written report (1500–2500 words) with proper citations and visualizations
Oral presentation (5–8 minutes) with Q&A
Critique of one peer's investigation

📋 Achievement-Chart Rubric (Total 100 marks)

Level

K/U (25)

Thinking (25)

Comm (25)

Application (25)

4 (80–100%)

Demonstrates thorough understanding of statistical/probability concepts; uses formulas and conventions accurately; quantifies all key statistics correctly.

Shows insightful planning, sophisticated reasoning, and strong critical evaluation of conclusions, lurking variables, and study design.

Communicates findings with high clarity, precise vocabulary, well-labelled tables/graphs, and articulate written/oral presentation.

Makes meaningful connections among investigation, real-world context, and other course strands; interprets results with depth.

3 (70–79%)

Considerable understanding; minor formula or computational gaps; most key statistics correct.

Considerable evidence of planning and reasoning; identifies major confounders and limitations; some critical evaluation.

Considerable clarity in communication; good use of conventions; minor gaps in labels or terminology.

Considerable transfer of concepts to context; reasonable interpretation; minor gaps in connecting strands.

2 (60–69%)

Some understanding; recurring errors in formulas or computations; gaps in core ideas.

Some planning and reasoning; limited critical evaluation; identifies obvious confounders only.

Some clarity but with notable inconsistencies in conventions, vocabulary, or visuals; report needs revision.

Some transfer; basic interpretation; weak connections to context or other strands.

1 (50–59%)

Limited understanding; many basic errors; unclear use of vocabulary.

Limited planning; lacks reasoning depth; conclusions not justified by data.

Limited clarity; conventions and vocabulary frequently incorrect; weak presentation.

Limited transfer; superficial interpretation; little connection to context.

R (<50%)

Insufficient understanding shown.

Insufficient planning or reasoning shown.

Insufficient communication of findings.

Insufficient connection to context shown.

Per Growing Success (2010): Final mark = 70% Term Work + 30% Final Evaluation. The Final Evaluation comprises 15% Culminating Investigation + 15% written Final Exam.

📑 Brief Final Exam Answer Key (open after attempting)

Q1

280

Q2

0.2344 (15/64)

Q3

z = 1.5

Q4

2 598 960

Q5

Q6

Q7

0.49 (49%)

Q8

50 400

Q9

(b) Stratified random

Q10

1/11 ≈ 0.0909

Q11

median = 19

Q12

0.8413

Q13

(a) C(14,5) = 2002 (b) C(6,3)·C(8,2) = 20·28 = 560 (c) C(6,3)·C(8,2)+C(6,4)·C(8,1)+C(6,5)·C(8,0) = 560+120+6 = 686

Q14

(a) C(30,25)·(0.8)^25·(0.2)^5 ≈ 0.1023 (b) E = 24, Var = 4.8 (c) μ = 24, σ ≈ 2.19; z = (22.5−24)/2.19 ≈ −0.685; P(Z < −0.685) ≈ 0.247

Q15

(a) Hypergeometric (b) C(8,1)·C(92,9)/C(100,10) ≈ 0.396 (c) 1 − C(92,10)/C(100,10) ≈ 0.572 (d) E = n·r/N = 10·8/100 = 0.8

Q16

x̄ = 3, ȳ = 6.6; s_x ≈ 1.581, s_y ≈ 2.966; r ≈ 0.992; ŷ = 0.6 + 2x; ŷ(3.5) = 7.6; r² ≈ 0.985 → 98.5% of variation explained.

Q17

X ∈ {0,1,2,3}; using hypergeometric: P(0)=20/120, P(1)=60/120, P(2)=36/120, P(3)=4/120. E(X) = 1.2; E(X²) = 1.8; Var = 1.8 − 1.44 = 0.36 (matches n·r/N · (N−r)/N · (N−n)/(N−1) = 3·0.4·0.6·7/9 = 0.56? — see note: the simple formula gives Var = n·(r/N)·((N−r)/N)·((N−n)/(N−1)) ≈ 3·0.4·0.6·7/9 ≈ 0.56. Direct calculation from the distribution above gives 0.56. Recompute: P(0)=20/120, P(1)=60/120, P(2)=36/120, P(3)=4/120. Sums must total 1: 20+60+36+4 = 120 ✓. E = (0·20+1·60+2·36+3·4)/120 = 144/120 = 1.2. E(X²)=(0+60+144+36)/120 = 240/120=2. Var = 2 − 1.44 = 0.56.

Q23

(a) C(49,6) = 13 983 816 (b) 1/13 983 816 ≈ 7.15×10^-8 (c) C(6,5)·C(43,1)/C(49,6) = 6·43/13 983 816 ≈ 1.84×10^-5 (d) E ≈ $5 000 000/13 983 816 − $3 ≈ $0.36 − $3 = −$2.64. Unfair (negative expected return).

Q24

(a) z = (3−4)/0.5 = −2; P(Z < −2) = 0.0228 (b) z = 1.645 → T = 4 + 1.645·0.5 ≈ 4.82 min (c) z = 2 → P(Z > 2) = 0.0228

Q25

P(D)=0.005, P(¬D)=0.995. P(+|D)=0.98, P(+|¬D)=0.05. P(+) = 0.005·0.98 + 0.995·0.05 = 0.0049 + 0.04975 = 0.05465. P(D|+) = 0.0049/0.05465 ≈ 0.0897 ≈ 9.0%. Even after a positive test, only ~9% have the disease — illustrates the base-rate fallacy.

Q26

r ≈ 0.998; ŷ ≈ 13.5 + 4.0x (approx); ŷ(13) ≈ 65.5. Extrapolation to $30k may not be valid (outside observed range, market saturation may flatten the relationship).

Q27

μ = 500, σ = √(1000·0.5·0.5) ≈ 15.81. z = (539.5−500)/15.81 ≈ 2.50. P(Z ≥ 2.50) = 0.0062. p < 0.05, so result is statistically significant — there is meaningful evidence against the 50% null hypothesis.