Statistical Distributions

The random variable X follows the binomial distribution X ~ B(6, 0.25). Find P(X = 2), giving your answer to 3 decimal places.

• A0.297✓
• B0.356
• C0.132
• D0.063

Answer: P(X = 2) = ⁶C₂ × 0.25² × 0.75⁴ = 15 × 0.0625 × 0.316406… = 0.2966… ≈ 0.297. (Option B is P(X = 1) and option C is P(X = 3).)

The random variable X follows the binomial distribution X ~ B(5, 0.4). Find P(X ≤ 1), giving your answer to 3 decimal places.

• A0.259
• B0.337✓
• C0.078
• D0.087

Answer: P(X ≤ 1) = P(X = 0) + P(X = 1) = 0.6⁵ + ⁵C₁ × 0.4 × 0.6⁴ = 0.07776 + 0.2592 = 0.33696 ≈ 0.337. (Option A is P(X = 1) only; option C is P(X = 0) only.)

The random variable X follows the binomial distribution X ~ B(20, 0.15). Find the mean of X.

• A3✓
• B17
• C2.55
• D0.15

Answer: For X ~ B(n, p) the mean is E(X) = np = 20 × 0.15 = 3. (Option C is the variance np(1−p) = 20 × 0.15 × 0.85 = 2.55, not the mean.)

The random variable X follows the binomial distribution X ~ B(8, 0.5). Find P(X = 3).

• A0.375
• B0.273
• C0.125
• D0.219✓

Answer: P(X = 3) = ⁸C₃ × 0.5³ × 0.5⁵ = 56 × 0.5⁸ = 56/256 = 0.21875 ≈ 0.219. (Option B is P(X = 4).)

The weights of bags of flour are normally distributed with mean 50 g and standard deviation 8 g. Using P(Z < 1) ≈ 0.8413, find P(X < 58).

• A0.1587
• B0.5000
• C0.8413✓
• D0.9772

Answer: Standardise: Z = (58 − 50)/8 = 1. So P(X < 58) = P(Z < 1) ≈ 0.8413. (Option A is P(Z > 1), the upper tail.)

The lifetimes of a type of battery are normally distributed with mean 100 hours and standard deviation 15 hours. Using P(Z < 2) ≈ 0.9772, find P(X > 130).

• A0.9772
• B0.0228✓
• C0.0456
• D0.4772

Answer: Standardise: Z = (130 − 100)/15 = 2. So P(X > 130) = P(Z > 2) = 1 − P(Z < 2) = 1 − 0.9772 = 0.0228. (Option A is P(Z < 2), the wrong tail.)

The marks in a test are normally distributed with mean 60 and standard deviation 5. Using the fact that P(Z < 1.96) ≈ 0.975, find the mark m such that P(X < m) = 0.975.

• A67.0
• B50.2
• C69.8✓
• D62.0

Answer: This is an inverse-normal problem. P(X < m) = 0.975 corresponds to Z = 1.96. Rearranging Z = (m − μ)/σ gives m = μ + Zσ = 60 + 1.96 × 5 = 60 + 9.8 = 69.8.

A multiple-choice quiz has 10 questions, each with 4 options and exactly one correct answer. A student guesses every answer at random and X is the number answered correctly. Which condition required for X to be modelled as a binomial distribution is satisfied here?

• AThe number of trials is not fixed
• BThe probability of success changes from question to question
• CThere are a fixed number of independent trials each with the same probability of success✓
• DThe trials are dependent on previous answers

Answer: A binomial model requires a fixed number of independent trials, each with two outcomes (success/failure) and a constant probability of success. Here there are n = 10 fixed questions, each guess is independent, and each has the same success probability p = 0.25, so X ~ B(10, 0.25) is valid. The other options describe situations that would break the binomial conditions.