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. The value 0.356 is P(X = 1) and 0.132 is P(X = 3), so both answer for the wrong number of successes.

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. Taking P(X = 1) alone gives 0.259 and P(X = 0) alone gives 0.078; 'at most 1' requires both terms to be added.

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. The value 2.55 is the variance np(1−p) = 20 × 0.15 × 0.85, 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. The value 0.273 is P(X = 4) = 70/256, the most likely single outcome rather than the one asked for.

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. The value 0.1587 is P(Z > 1), the upper tail, which answers the opposite inequality.

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. Quoting 0.9772 gives P(Z < 2), the lower tail, rather than the probability of exceeding 130.

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.

X ~ B(10, 1/2). Find P(X = 3), to 3 s.f.

• A5.00
• B15.0
• C0.125
• D0.117✓

Answer: P(X = k) = nCk p^k (1 − p)^(n − k) = 120 × (1/2)³ × (1/2)⁷ = 0.1172, so 0.117 to 3 s.f. The mean np is offered as well, and it is not a probability.

X ~ B(8, 1/4). Find P(X = 2), to 3 s.f.

• A0.311✓
• B1.75
• C0.0625
• D2.00

Answer: P(X = k) = nCk p^k (1 − p)^(n − k) = 28 × (1/4)² × (3/4)⁶ = 0.3115, so 0.311 to 3 s.f. The mean np is offered as well, and it is not a probability.

X ~ B(6, 1/3). Find P(X = 2), to 3 s.f.

• A0.111
• B1.67
• C0.329✓
• D2.00

Answer: P(X = k) = nCk p^k (1 − p)^(n − k) = 15 × (1/3)² × (2/3)⁴ = 0.3292, so 0.329 to 3 s.f. The mean np is offered as well, and it is not a probability.

X ~ B(20, 1/4). State the mean and variance of X.

• AMean 15/4, variance 5
• BMean 5, variance 15/4✓
• CMean 5, variance 5
• DMean 5, variance √(15)/2

Answer: For a binomial distribution the mean is np = 5 and the variance is np(1 − p) = 15/4. One distractor quotes the standard deviation in place of the variance — the variance is the square of it.

X ~ B(50, 1/5). State the mean and variance of X.

• AMean 10, variance 10
• BMean 8, variance 10
• CMean 10, variance 8✓
• DMean 10, variance 2√(2)

Answer: For a binomial distribution the mean is np = 10 and the variance is np(1 − p) = 8. One distractor quotes the standard deviation in place of the variance — the variance is the square of it.

X ~ B(12, 1/2). State the mean and variance of X.

• AMean 3, variance 6
• BMean 6, variance 3✓
• CMean 6, variance 6
• DMean 6, variance √(3)

Answer: For a binomial distribution the mean is np = 6 and the variance is np(1 − p) = 3. One distractor quotes the standard deviation in place of the variance — the variance is the square of it.

For a normal distribution, approximately what percentage of data lies within one standard deviation of the mean?

• A99.7%
• B95%
• C68%✓
• D50%

Answer: The empirical rule: about 68% within one standard deviation, 95% within two and 99.7% within three. The 95% figure is the one quoted for confidence intervals, which is why it is so often given here by mistake.

X ~ N(50, 4²). What is P(X > 50)?

• A0.5✓
• B0.68
• C0.95
• D0.32

Answer: The normal distribution is symmetric about its mean, so exactly half the probability lies above 50. No tables are needed — symmetry alone settles it, and that symmetry is worth spotting before reaching for a calculator.

For X ~ B(n, p), what condition must the trials satisfy?

• AIndependent, with a fixed probability of success✓
• BEach trial must produce a result that is drawn from a continuous range of values
• CThe number of trials must be large enough to exceed thirty in total
• DThe probability of success must be exactly one half for the model to apply

Answer: The binomial model needs a fixed number of independent trials, two outcomes, and a constant p. Continuous outcomes rule the model out entirely, and n = 30 or p = 0.5 are not requirements.

Reaction times, X ms, of drivers in a simulator are modelled by X ~ N(250, 30²). Find P(220 < X < 300), to 4 decimal places.

• A0.1109
• B0.7936✓
• C0.4522
• D0.2064

Answer: Standardise: z = (220 − 250)/30 = −1 and z = (300 − 250)/30 = 5/3. P = Φ(1.667) − Φ(−1) = 0.9522 − 0.1587 = 0.7936. The value 0.1109 subtracts Φ(+1) = 0.8413 instead of Φ(−1).

The time, X minutes, that customers wait for a delivery is normally distributed with mean 460 and standard deviation σ. Given that P(X > 500) = 0.15, find σ to 3 significant figures.

• Aσ = 38.6✓
• Bσ = 27.8
• Cσ = 24.3
• Dσ = 41.5

Answer: P(X < 500) = 0.85, so (500 − 460)/σ = Φ⁻¹(0.85) = 1.0364. Then σ = 40/1.0364 = 38.6. Using 1.645 (the 5% point) gives 24.3, and multiplying 40 by 1.0364 instead of dividing gives 41.5.

The masses of bags of sugar are normally distributed with mean 1.02 kg and standard deviation 0.015 kg. A bag is underweight if its mass is less than 1 kg. Bags are packed in boxes of 12, independently. Find the probability that a box contains at most one underweight bag, to 3 significant figures.

• A0.382
• B0.317
• C0.300
• D0.700✓

Answer: P(underweight) = P(Z < (1 − 1.02)/0.015) = Φ(−1.333) = 0.0912. The number underweight in a box is Y ~ B(12, 0.0912). P(Y ≤ 1) = 0.9088¹² + 12(0.0912)(0.9088¹¹) = 0.317 + 0.382 = 0.700. The value 0.382 is P(Y = 1) only, and 0.317 is P(Y = 0) only.

X ~ B(200, 0.4). Use a normal approximation, with a continuity correction, to estimate P(X ≤ 70), to 4 decimal places.

• A0.0745
• B0.0852✓
• C0.0648
• D0.9148

Answer: Mean np = 80 and variance np(1 − p) = 48, so Y ~ N(80, 48). With the continuity correction, P(X ≤ 70) ≈ P(Y < 70.5) = Φ((70.5 − 80)/√48) = Φ(−1.371) = 0.0852. Without the correction (using 70) the estimate is 0.0745; correcting the wrong way (69.5) gives 0.0648.

The graph of the probability density of X ~ N(45, σ²) has points of inflection at x = 41 and x = 49. Find σ.

• Aσ = 8
• Bσ = 16
• Cσ = 4✓
• Dσ = 2

Answer: The points of inflection of a normal curve are at μ ± σ, one standard deviation either side of the mean. So 45 ± σ = 41 and 49, giving σ = 4. The value 8 is the distance between the two points, and 16 is the variance σ².