Flashcards · 9709 · Topic 5

Probability & Statistics 1 flashcards

28 cards on Topic 5, subtopics 5.1 to 5.5. Pick a subtopic or study them all.

Study mode needs JavaScript. Every card is listed below, so you can still cover the answers and test yourself.

All 28 cards

For a quick read-through or printing. Cover the right-hand side to test yourself.

Histogram: what does the height of each bar show?: Frequency density = frequency ÷ class width. The area of each bar is proportional to the frequency.
Mean and standard deviation of raw data: x̄ = Σx/n σ = √(Σx²/n − x̄²) Grouped: use mid-points, x̄ = Σxf/Σf and σ = √(Σx²f/Σf − x̄²).
Coded data: if y = x − a, how are the mean and standard deviation of x found from y?: x̄ = a + ȳ. The standard deviation of x equals the standard deviation of y (subtracting a constant does not change the spread).
Cumulative frequency graph: where do you plot points, and how do you read the median and quartiles for n values?: Plot at upper class boundaries. Median at n/2, Q₁ at n/4, Q₃ at 3n/4 on the cumulative frequency axis.
How do you compare two data sets?: One comment on centre (median or mean) and one on spread (IQR or standard deviation), both in the context of the data.
How do you find the mean and standard deviation of two data sets combined?: Add n, Σx and Σx² across the two sets, then use the formulas once. Do not average the two means unless the sets are the same size.

Permutation vs combination: Permutation: order matters, ⁿPᵣ = n!/(n − r)!. Combination: order does not matter, ⁿCᵣ = n!/(r!(n − r)!).
How many arrangements are there of the letters of LETTER?: 6 letters with E twice and T twice: 6!/(2! × 2!) = 720/4 = 180.
Arrangements where certain items must be together: Glue the items into one block, arrange the blocks, then multiply by the number of arrangements inside the block.
7 people stand in a line. In how many ways can A and B not be next to each other?: Total − together = 7! − 2 × 6! = 5040 − 1440 = 3600.
How do you handle "at least one" or "at most" in a selection?: List the exact cases (for example exactly 1, exactly 2, …), count each with combinations, and add. A single product usually double-counts.

Mutually exclusive events: They cannot happen together: P(A ∩ B) = 0, so P(A or B) = P(A) + P(B).
Independent events: definition and test: One does not affect the probability of the other. Test: P(A ∩ B) = P(A) × P(B). Compare the numbers to justify it.
Conditional probability formula: P(A | B) = P(A ∩ B)/P(B).
Can two events with non-zero probabilities be both mutually exclusive and independent?: No. Exclusive gives P(A ∩ B) = 0, but independence needs P(A ∩ B) = P(A)P(B) > 0.
How do you find P(A | B) from a tree diagram?: Numerator: the probability of the branches where both A and B happen. Denominator: the total probability of all branches where B happens. Divide.

Discrete random variable: the probability rule, E(X) and Var(X): Σp = 1. E(X) = Σxp. Var(X) = Σx²p − {E(X)}².
Conditions for a binomial distribution B(n, p): A fixed number n of independent trials, each with two outcomes (success or failure), and a constant probability p of success. X is the number of successes.
Binomial B(n, p): P(X = r), mean and variance: P(X = r) = ⁿCᵣ pʳ(1 − p)ⁿ⁻ʳ. Mean np; variance np(1 − p).
Geometric Geo(p): P(X = r), mean, and P(X > r): P(X = r) = p(1 − p)ʳ⁻¹, r = 1, 2, 3, … Mean 1/p. P(X > r) = (1 − p)ʳ (the first r trials all fail).
Binomial or geometric?: Binomial: fixed number of trials; X counts the successes (0 to n). Geometric: trials continue until the first success; X is the trial on which it occurs (1, 2, 3, …).
X takes the values 1, 2, 3, 4 with probabilities 0.1, a, b, 0.3, and E(X) = 2.7. Find a and b.: Σp = 1: a + b = 0.6. E(X): 0.1 + 2a + 3b + 1.2 = 2.7, so 2a + 3b = 1.4. b = 0.2, a = 0.4.

Standardising X ~ N(μ, σ²): Z = (X − μ)/σ, where Z ~ N(0, 1). Then P(X < x) = Φ((x − μ)/σ).
How do you find Φ(−z), and P(Z > z)?: Φ(−z) = 1 − Φ(z). P(Z > z) = 1 − Φ(z). Between two values: Φ(z₂) − Φ(z₁).
Normal approximation to B(n, p): conditions and parameters: Use when np > 5 and nq > 5 (q = 1 − p). Approximate by N(np, npq), with a continuity correction.
Continuity corrections for P(X ≤ 15), P(X < 15), P(X ≥ 15) and P(X = 15): P(X ≤ 15) → P(Y < 15.5) P(X < 15) → P(Y < 14.5) P(X ≥ 15) → P(Y > 14.5) P(X = 15) → P(14.5 < Y < 15.5)
What z-value marks the top 10% of a normal distribution?: z = +1.282 (the area to its left is 0.9). The bottom 10% is z = −1.282.
P(X < 40) = 0.05 and P(X > 70) = 0.1 for X ~ N(μ, σ²). Find μ and σ.: (40 − μ)/σ = −1.645 and (70 − μ)/σ = 1.282. Subtract: 30 = 2.927σ, so σ = 10.2 and μ = 40 + 1.645σ = 56.9 (3 s.f.). Keep σ unrounded when finding μ.