Mathematics Advanced • Year 12 • Module 5 • Lesson 11

Introduction to Random Variables

Practise HSC-style writing on random variables, the CDF, and the discrete/continuous distinction — including an extended analytical response.

Master · Past-Paper Style

1. Short-answer questions

1.1 A discrete random variable X has the probability function:

x	0	1	2	3	4
p(x)	0.05	0.20	k	0.30	0.15

(a) Find k. (b) Find P(X ≤ 2). (c) Find P(X = 2 | X ≥ 1). 3 marks Band 3-4

1.2 The cumulative distribution function of a discrete random variable Y is given by

y	1	2	3	4	5
F(y)	0.10	0.35	0.60	0.85	1.00

(a) Find the probability function p(y). (b) Calculate P(2 ≤ Y ≤ 4) using F. (c) Find the smallest y such that F(y) ≥ 0.5. 3 marks Band 3-4

1.3 A discrete random variable X has p(x) = c (6 − x) for x = 1, 2, 3, 4, 5.
(a) Find c. (b) State P(X ≥ 4). (c) Find P(X > 3 | X > 1). 4 marks Band 4

Stuck on 1.3(c)? P(X > 3 | X > 1) = P(X > 3) / P(X > 1) — use the conditional probability formula.

2. Extended response

2.1 A factory sells DVD storage boxes in packs of 4. From historical inspection, the number X of defective boxes per pack has the probability function

x	0	1	2	3	4
p(x)	0.60	0.25	0.10	0.04	0.01

(a) Verify that p is a valid probability function and build the full CDF F(0), F(1), …, F(4).

(b) A pack is "acceptable" if it contains at most 1 defective box, "marginal" if it contains exactly 2, and "rejected" otherwise. Find the probability of each of these three outcomes.

(c) The quality manager proposes a new policy: "Reject any pack with 2 or more defectives." Calculate the probability that a random pack is rejected under this policy and compare it to the previous "rejected" probability from (b). State, in one sentence, whether the new policy is stricter or more lenient and quantify the change.

(d) A retailer buys 50 packs. Using your CDF from (a), state the expected proportion of "acceptable" packs in the order, and explain in one sentence why the actual count in a real order might still differ from the expected count.

7 marks Band 5-6

Explicit marking criteria

Part (a) — 2 marks

• 1 mark — checks all p(x) ∈ [0, 1] AND Σ p(x) = 0.60 + 0.25 + 0.10 + 0.04 + 0.01 = 1.

• 1 mark — produces correct cumulative table F(0) = 0.60, F(1) = 0.85, F(2) = 0.95, F(3) = 0.99, F(4) = 1.00.

Part (b) — 2 marks

• 1 mark — Acceptable = P(X ≤ 1) = F(1) = 0.85 (using the CDF, not just summing).

• 1 mark — Marginal = p(2) = 0.10; Rejected = P(X ≥ 3) = 1 − F(2) = 0.05.

Part (c) — 2 marks

• 1 mark — Computes new rejection probability P(X ≥ 2) = 1 − F(1) = 0.15.

• 1 mark — Stricter; rejection rate rises by 0.10 (from 0.05 to 0.15), tripling.

Part (d) — 1 mark

• 1 mark — Expected acceptable = 0.85 × 50 = 42.5 packs; observed count varies because each order is a finite random sample (sampling variability) and the probability function is a long-run frequency, not a guaranteed outcome.

Your response:

Stuck on (d)? Expected count = (probability) × (sample size); discuss why real outcomes drift around this expectation.

How did this worksheet feel?

Got it Partly Lost

What I'll revisit before next class:

Answers — sample responses + marking notes

1.1 — Find k, P(X ≤ 2), P(X = 2 | X ≥ 1) (3 marks)

Sample response.
(a) 0.05 + 0.20 + k + 0.30 + 0.15 = 1 ⇒ 0.70 + k = 1 ⇒ k = 0.30.
(b) P(X ≤ 2) = p(0) + p(1) + p(2) = 0.05 + 0.20 + 0.30 = 0.55.
(c) P(X ≥ 1) = 1 − p(0) = 1 − 0.05 = 0.95. P(X = 2 | X ≥ 1) = P(X = 2)/P(X ≥ 1) = 0.30/0.95 = 6/19 ≈ 0.316.

Marking notes. 1 mark — correct k via Σ p(x) = 1. 1 mark — correct P(X ≤ 2). 1 mark — correct conditional probability with the line "P(X = 2)/P(X ≥ 1)" visible. Common errors: forgetting that "X ≥ 1" excludes only X = 0 (not also X = 1).

1.2 — Recover p from F, find P(2 ≤ Y ≤ 4) and median-like cut-off (3 marks)

Sample response.
(a) p(1) = F(1) = 0.10; p(2) = F(2) − F(1) = 0.25; p(3) = F(3) − F(2) = 0.25; p(4) = F(4) − F(3) = 0.25; p(5) = F(5) − F(4) = 0.15.
(b) P(2 ≤ Y ≤ 4) = F(4) − F(1) = 0.85 − 0.10 = 0.75. (Check: 0.25 + 0.25 + 0.25 = 0.75 ✓.)
(c) F(2) = 0.35 < 0.5; F(3) = 0.60 ≥ 0.5. Smallest y is y = 3.

Marking notes. 1 mark — correct p column (all five values). 1 mark — correct interval probability using F (NOT direct sum only — must reference F). 1 mark — correct smallest y identified, with both F(2) and F(3) compared explicitly to 0.5.

1.3 — p(x) = c(6 − x), x = 1, …, 5 (4 marks)

Sample response.
(a) Σ p(x) = c(5 + 4 + 3 + 2 + 1) = 15c = 1, so c = 1/15. Then p(1) = 5/15, p(2) = 4/15, p(3) = 3/15, p(4) = 2/15, p(5) = 1/15.
(b) P(X ≥ 4) = p(4) + p(5) = 2/15 + 1/15 = 3/15 = 1/5.
(c) P(X > 1) = 1 − p(1) = 1 − 5/15 = 10/15. P(X > 3) = p(4) + p(5) = 3/15. So P(X > 3 | X > 1) = (3/15)/(10/15) = 3/10.

Marking notes. (a) 1 mark — correct c with working. (b) 1 mark — correct P(X ≥ 4). (c) 1 mark — correct numerator and denominator from the conditional definition; 1 mark — correct simplified fraction. Common error: students compute P(X > 3) × P(X > 1) (multiplying instead of dividing) — that gives a much smaller (and incorrect) value.

2.1 — DVD storage boxes extended response (7 marks): sample Band-6 response with annotations

Sample Band-6 response.

Part (a) — validity and CDF. Each p(x) lies in [0, 1] and Σ p(x) = 0.60 + 0.25 + 0.10 + 0.04 + 0.01 = 1, so p is a valid probability function. [1 mark.] Building F cumulatively:

F(0) = 0.60
F(1) = 0.60 + 0.25 = 0.85
F(2) = 0.85 + 0.10 = 0.95
F(3) = 0.95 + 0.04 = 0.99
F(4) = 0.99 + 0.01 = 1.00 ✓

[1 mark.]

Part (b) — classify packs.

Acceptable (X ≤ 1): F(1) = 0.85. [1 mark — uses F.]
Marginal (X = 2): p(2) = 0.10.
Rejected (X ≥ 3): 1 − F(2) = 1 − 0.95 = 0.05. [1 mark — marginal and rejected.]
Check: 0.85 + 0.10 + 0.05 = 1.00 ✓.

Part (c) — new policy comparison. Under the proposed policy, a pack is rejected iff X ≥ 2:

P(X ≥ 2) = 1 − F(1) = 1 − 0.85 = 0.15. [1 mark.]

This is stricter than the previous rejection rate of 0.05: the rejection probability has tripled (0.15 / 0.05 = 3), so 3× as many packs would be turned away. [1 mark — comparison and quantification.]

Part (d) — retailer of 50 packs. Expected number of acceptable packs ≈ 0.85 × 50 = 42.5 packs. The actual count will differ from 42.5 because each order is a finite random sample, not the long-run population: the probability function describes the long-run frequency, so individual outcomes fluctuate around the expected value (this is sampling variability). [1 mark — expectation calculation AND explanation of variability.]

Total: 7/7.

Band descriptors for marker.

Band 3: Verifies validity, computes F partially, identifies one of (acceptable/marginal/rejected) correctly. ≈ 2-3 marks.

Band 4: All of part (a) and (b) correct; attempts (c) but does not explicitly compare or quantify ("triples"). ≈ 4-5 marks.

Band 5: Parts (a)-(c) complete with comparison; computes expectation in (d) but does not explain why real counts differ. ≈ 5-6 marks.

Band 6: All parts complete, every probability computed via F (not just direct sums), final answer in (d) names "sampling variability" or equivalent and ties it to "finite sample vs. long-run frequency". 7/7.