Try these eight questions on paper first. They cover the probability distributions chapter and rise in difficulty.
Questions
Question 1. X ~ B(5, 0.4). Find P(X = 2).
Answer
P(X = 2) = 5C2 × 0.4² × 0.6³ = 10 × 0.16 × 0.216 = 0.3456.
Question 2. X ~ B(8, 0.1). Find P(X ≥ 1).
Answer
P(X ≥ 1) = 1 − P(X = 0) = 1 − 0.9⁸ = 1 − 0.43047 = 0.5695 (4 d.p.).
Question 3. A binomial variable has mean 24 and variance 9.6. Find n and p.
Answer
q = 9.6 ÷ 24 = 0.4, so p = 0.6.
n = 24 ÷ 0.6 = 40.
Check: 40 × 0.6 = 24 and 40 × 0.6 × 0.4 = 9.6. So n = 40 and p = 0.6.
Question 4. X ~ N(60, 16). Find P(X > 66).
Answer
The variance is 16, so σ = 4.
z = (66 − 60) ÷ 4 = 1.5. The upper-tail probability for z = 1.5 is 0.0668.
Question 5. For the same X ~ N(60, 16), find P(56 < X < 64).
Answer
The limits standardise to z = −1 and z = 1, since (56 − 60) ÷ 4 = −1 and (64 − 60) ÷ 4 = 1.
Each tail beyond ±1 has probability 0.1587. So P = 1 − 2(0.1587) = 0.6826.
Question 6. X ~ N(100, σ²) and P(X < 88) = 0.1587. Find σ.
Answer
The lower tail is 0.1587, so z = −1 (88 is below the mean).
(88 − 100) ÷ σ = −1, so σ = 12.
Question 7. Decide whether each is binomial or normal: (a) the number of heads in 20 coin tosses, (b) the heights of Form 5 students in a school.
Answer
(a) Binomial. There are 20 fixed, independent yes-or-no trials, and the variable counts heads.
(b) Normal. Height is measured on a continuous scale, and the values cluster around a mean.
Question 8. X ~ B(6, 0.2). Find P(X ≤ 1).
Answer
P(X = 0) = 0.8⁶ = 0.262144.
P(X = 1) = 6 × 0.2 × 0.8⁵ = 1.2 × 0.32768 = 0.393216.
The sum is 0.6554 (4 d.p.).
If you got these wrong
- Questions 1, 2 and 8 test calculating binomial probabilities.
- Question 3 tests finding mean and variance of a binomial variable.
- Questions 4 and 5 test standardising a normal random variable.
- Question 6 tests finding unknown normal-distribution parameters.
- Question 7 tests recognising binomial conditions.
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