How to Use This Page
1. Solve each exercise problem on your own first — attempt every step before looking here. 2. Compare your method with the step-by-step solution shown. 3. If your answer differs, rework from the step where you diverged. 4. Pay special attention to the Common Mistakes section — these are the errors that cost marks in exams. 5. Use the Exam Checklist before your test to ensure you have covered every topic.
Key Concepts (Quick Recap)
Binomial Distribution
- Conditions: Fixed n trials, two outcomes per trial, constant p, independent trials
- PMF: P(X = r) = ⁿCᵣ pʳ qⁿ⁻ʳ, where r = 0, 1, 2, ..., n and q = 1 − p
- Mean: μ = np
- Variance: σ² = npq
- Standard Deviation: σ = √(npq)
- ⁿCᵣ = n! / (r! × (n − r)!)
Poisson Distribution
- When to use: Rare events occurring at a constant average rate
- PMF: P(X = x) = (e⁻ᵛ λˣ) / x!, where x = 0, 1, 2, ...
- Mean: μ = λ
- Variance: σ² = λ (mean equals variance!)
- Additive property: If X ~ Poisson(λ₁) and Y ~ Poisson(λ₂) are independent, then X + Y ~ Poisson(λ₁ + λ₂)
- Poisson as limit of binomial: When n → ∞, p → 0, np = λ (finite), binomial approaches Poisson
Normal Distribution
- PDF: f(x) = (1/(σ√(2π))) × e^(−(x−μ)²/(2σ²))
- Bell-shaped, symmetric about μ
- Mean = μ, Variance = σ², Standard Deviation = σ
- Standard normal: Z = (X − μ)/σ, where Z ~ N(0, 1)
- Area properties: P(μ − σ < X < μ + σ) ≈ 68.27%, P(μ − 2σ < X < μ + 2σ) ≈ 95.45%, P(μ − 3σ < X < μ + 3σ) ≈ 99.73%
- Total area under curve = 1
Relationship Between Distributions
- Binomial → Poisson when n is large, p is small, np = λ
- Normal approximation to Binomial when np > 5 and nq > 5
- Normal approximation to Poisson when λ is large
I. Choose the Best Answer (MCQs)
1. If X ~ B(n, p) with n = 10, p = 0.3, then E(X) and Var(X) are: (a) 3, 2.1 (b) 3, 3 (c) 2.1, 3 (d) 3, 7
Answer(a) 3, 2.1 Explanation: E(X) = np = 10 × 0.3 = 3. Var(X) = npq = 10 × 0.3 × 0.7 = 2.1.
2. In a Poisson distribution, if P(X = 1) = P(X = 2), then the mean λ equals: (a) 1 (b) 2 (c) 3 (d) 4
Answer(b) 2 Explanation: P(X = 1) = λe⁻ᵏ/1! = λe⁻ˡ. P(X = 2) = λ²e⁻ˡ/2! = λ²e⁻ˡ/2. Setting equal: λ = λ²/2 → 2λ = λ² → λ² − 2λ = 0 → λ(λ − 2) = 0. Since λ ≠ 0, λ = 2.
3. The normal curve is symmetric about: (a) σ (b) σ² (c) μ (d) Z
Answer(c) μ Explanation: The normal distribution is symmetric about its mean μ. This means P(X < μ) = P(X > μ) = 0.5.
4. If X ~ B(6, 1/2), then P(X = 3) equals: (a) 15/64 (b) 20/64 (c) 5/16 (d) 1/2
Answer(b) 20/64 Explanation: P(X = 3) = ⁶C₃ (1/2)³ (1/2)³ = 20 × (1/64) = 20/64 = 5/16. Both (b) and (c) represent the same value; the standard form is ⁶C₃/64 = 20/64.
5. If X ~ Poisson(3), then P(X = 0) equals: (a) e⁻³ (b) 3e⁻³ (c) 9e⁻³/2 (d) 1
Answer(a) e⁻³ Explanation: P(X = 0) = e⁻³ × 3⁰/0! = e⁻³ × 1/1 = e⁻³ ≈ 0.0498.
6. In a binomial distribution, if n = 8 and p = 1/4, the mode is: (a) 1 (b) 2 (c) 3 (d) 0
Answer(b) 2 Explanation: Mode = floor((n + 1)p) = floor(9 × 1/4) = floor(2.25) = 2. Since (n+1)p is not an integer, the mode is the single value 2.
7. The standard normal variable Z has: (a) μ = 0, σ = 0 (b) μ = 1, σ = 0 (c) μ = 0, σ = 1 (d) μ = 1, σ = 1
Answer(c) μ = 0, σ = 1 Explanation: The standard normal distribution N(0, 1) has mean 0 and standard deviation 1.
8. If X ~ N(100, 25), then P(X > 110) equals approximately: (a) 0.0228 (b) 0.9772 (c) 0.1587 (d) 0.8413
Answer(a) 0.0228 Explanation: σ² = 25, so σ = 5. Z = (110 − 100)/5 = 2. P(X > 110) = P(Z > 2) = 0.5 − P(0 < Z < 2) = 0.5 − 0.4772 = 0.0228.
9. For a binomial distribution with n = 5, p = 0.4, P(X ≥ 4) equals: (a) 0.0768 (b) 0.0870 (c) 0.0768 (d) 0.2304
Answer(b) 0.0870 approximately Explanation: P(X ≥ 4) = P(X = 4) + P(X = 5) = ⁵C₄(0.4)⁴(0.6)¹ + ⁵C₅(0.4)⁵(0.6)⁰ = 5(0.0256)(0.6) + 1(0.01024)(1) = 0.0768 + 0.01024 = 0.08704.
10. If λ = 2 for a Poisson distribution, then P(X ≤ 1) equals: (a) e⁻² (b) 3e⁻² (c) 2e⁻² (d) 5e⁻²
Answer(b) 3e⁻² Explanation: P(X ≤ 1) = P(X = 0) + P(X = 1) = e⁻² + 2e⁻² = 3e⁻² ≈ 0.406.
II. Fill in the Blanks
1. In a binomial distribution, the mean is np and the variance is npq.
2. For a Poisson distribution, both mean and variance equal λ.
3. The standard normal variable is defined as Z = (X − μ)/σ.
4. In a binomial distribution with n = 20 and p = 0.05, the Poisson approximation uses λ = 1 (since λ = np = 20 × 0.05 = 1).
5. The total area under the normal curve equals 1.
6. The binomial distribution is symmetric when p = 0.5.
7. In N(μ, σ²), the notation means μ is the mean and σ² is the variance.
8. P(Z > 0) in a standard normal distribution equals 0.5.
9. For X ~ B(n, p), the maximum value of variance occurs when p = 0.5 (since Var = npq = np(1−p) is maximised at p = 1/2).
10. If X ~ Poisson(λ), then P(X = 0) = e⁻ᵏ.
III. True or False
1. Statement: A binomial distribution is always symmetric.
False. Correction: A binomial distribution is symmetric only when p = 0.5. When p ≠ 0.5, the distribution is skewed — right-skewed when p < 0.5 and left-skewed when p > 0.5.
2. Statement: In a Poisson distribution, mean equals variance.
True. For Poisson(λ), mean = λ and variance = λ, so they are equal.
3. Statement: The normal distribution is defined for all real values of the variable.
True. The normal distribution has support on (−∞, +∞), though the probability of extreme values is very small.
4. Statement: As n increases, the binomial distribution approaches the normal distribution when p is not too close to 0 or 1.
True. By the Central Limit Theorem, for large n, the binomial distribution is well approximated by a normal distribution with mean np and variance npq.
5. Statement: In a Poisson distribution, P(X = x) can be greater than 1.
False. Correction: P(X = x) is always between 0 and 1 since it is a probability. The PMF gives P(X = x) = e⁻ᵏλˣ/x!, which is always in [0, 1].
6. Statement: The area between μ and μ + σ under the normal curve is approximately 0.3413.
True. P(0 < Z < 1) = 0.3413, so P(μ < X < μ + σ) ≈ 34.13%.
7. Statement: If X ~ B(n, p) and Y ~ B(m, p) are independent, then X + Y ~ B(n + m, p).
True. The sum of independent binomial random variables with the same probability p is also binomial with parameters (n + m, p).
IV. Match the Following
| Column A | Column B |
|---|
| 1. Binomial mean | (a) λ |
| 2. Poisson mean | (b) np |
| 3. Normal curve symmetry point | (c) e⁻ᵏλˣ/x! |
| 4. Poisson PMF | (d) μ |
| 5. Standard normal variance | (e) 1 |
Answer1 → (b): Binomial mean = np 2 → (a): Poisson mean = λ 3 → (d): Normal curve is symmetric about μ 4 → (c): Poisson PMF = e⁻ᵏλˣ/x! 5 → (e): Standard normal N(0,1) has variance = 1
V. Answer Briefly
1. State the conditions under which a binomial distribution is applicable.
Answer
A binomial distribution applies when: 1. There are a fixed number of trials n 2. Each trial has exactly two outcomes — success or failure 3. The probability of success p is constant for each trial 4. All trials are independent 5. The random variable X counts the number of successes in n trials
2. Write the PMF of the Poisson distribution and state its mean and variance.
Answer
PMF: P(X = x) = e⁻ᵏλˣ/x!, for x = 0, 1, 2, ... where λ > 0 is the parameter (average rate of occurrence). Mean = λ, Variance = λ. The key property of Poisson is that mean equals variance.
3. What are the properties of the normal distribution?
Answer
1. The normal curve is bell-shaped and symmetric about the mean μ 2. Mean = Median = Mode = μ 3. The curve is asymptotic to the x-axis (tails extend infinitely) 4. The total area under the curve = 1 5. About 68.27% of values fall within μ ± σ, 95.45% within μ ± 2σ, 99.73% within μ ± 3σ 6. It is completely determined by two parameters: μ (mean) and σ (standard deviation)
4. Define the standard normal variable. How is any normal variable converted to standard normal?
Answer
The standard normal variable Z has distribution N(0, 1) with mean 0 and standard deviation 1. Any normal variable X ~ N(μ, σ²) is converted to Z using the formula: Z = (X − μ)/σ. This standardisation allows us to use the standard normal table to find probabilities for any normal distribution.
5. When does the Poisson distribution approximate the binomial distribution?
Answer
The Poisson distribution approximates the binomial when: 1. n is large (n > 20 typically) 2. p is small (p < 0.05) 3. np = λ is moderate (λ ≤ 10) Under these conditions, B(n, p) ≈ Poisson(λ) where λ = np.
6. If X ~ B(5, 0.6), find P(X = 3).
Answer
P(X = 3) = ⁵C₃ (0.6)³ (0.4)² = 10 × 0.216 × 0.16 = 10 × 0.03456 = 0.3456
VI. Answer in Detail
1. If X ~ B(6, 0.5), find the complete probability distribution, mean, and variance.
Answer
Step 1: Complete probability distribution P(X = r) = ⁶Cᵣ (0.5)ʳ (0.5)⁶⁻ʳ = ⁶Cᵣ / 64 | r | ⁶Cᵣ | P(X = r) | |---|---|---| | 0 | 1 | 1/64 ≈ 0.0156 | | 1 | 6 | 6/64 ≈ 0.0938 | | 2 | 15 | 15/64 ≈ 0.2344 | | 3 | 20 | 20/64 ≈ 0.3125 | | 4 | 15 | 15/64 ≈ 0.2344 | | 5 | 6 | 6/64 ≈ 0.0938 | | 6 | 1 | 1/64 ≈ 0.0156 | Verification: Sum = 64/64 = 1 ✓ Step 2: Mean E(X) = np = 6 × 0.5 = 3 Step 3: Variance Var(X) = npq = 6 × 0.5 × 0.5 = 1.5
2. A factory produces items of which 10% are defective. A sample of 5 items is inspected. Find: (i) P(all are good), (ii) P(exactly 2 are defective), (iii) P(at most 1 is defective), (iv) mean and variance.
Answer
Here n = 5, p = 0.1 (probability of defective), q = 0.9 X ~ B(5, 0.1) (i) P(all are good) = P(X = 0) = ⁵C₀ (0.1)⁰ (0.9)⁵ = 1 × 1 × 0.59049 = 0.59049 (ii) P(exactly 2 are defective) = P(X = 2) = ⁵C₂ (0.1)² (0.9)³ = 10 × 0.01 × 0.729 = 10 × 0.00729 = 0.0729 (iii) P(at most 1 defective) = P(X = 0) + P(X = 1) = ⁵C₀(0.1)⁰(0.9)⁵ + ⁵C₁(0.1)¹(0.9)⁴ = 0.59049 + 5(0.1)(0.6561) = 0.59049 + 0.32805 = 0.91854 (iv) Mean = np = 5 × 0.1 = 0.5; Variance = npq = 5 × 0.1 × 0.9 = 0.45
3. On average, 3 customers arrive at a service counter per minute. Using Poisson distribution, find: (i) P(no customer arrives in a minute), (ii) P(exactly 2 customers arrive), (iii) P(at most 2 customers arrive), (iv) P(more than 2 customers arrive).
Answer
λ = 3 (average arrivals per minute), X ~ Poisson(3) (i) P(X = 0) = e⁻³ × 3⁰/0! = e⁻³ ≈ 0.0498 (ii) P(X = 2) = e⁻³ × 3²/2! = e⁻³ × 9/2 = 4.5e⁻³ ≈ 4.5 × 0.0498 ≈ 0.2240 (iii) P(X ≤ 2) = P(X = 0) + P(X = 1) + P(X = 2) = e⁻³(1 + 3 + 4.5) = 8.5e⁻³ ≈ 8.5 × 0.0498 ≈ 0.4232 (iv) P(X > 2) = 1 − P(X ≤ 2) = 1 − 0.4232 = 0.5768 Verification: Σ P(X = x) from x = 0 to ∞ = 1 (since e³ = e³ × Σ(3ˣ/x!))
4. The marks in a subject are normally distributed with mean 60 and standard deviation 10. Find: (i) P(mark > 70), (ii) P(mark between 50 and 80), (iii) P(mark < 40), (iv) the marks above which 10% of students score.
Answer
μ = 60, σ = 10, X ~ N(60, 100) (i) P(X > 70) Z = (70 − 60)/10 = 1 P(X > 70) = P(Z > 1) = 0.5 − P(0 < Z < 1) = 0.5 − 0.3413 = 0.1587 (ii) P(50 < X < 80) Z₁ = (50 − 60)/10 = −1, Z₂ = (80 − 60)/10 = 2 P(−1 < Z < 2) = P(−1 < Z < 0) + P(0 < Z < 2) = 0.3413 + 0.4772 = 0.8185 (iii) P(X < 40) Z = (40 − 60)/10 = −2 P(Z < −2) = 0.5 − P(−2 < Z < 0) = 0.5 − 0.4772 = 0.0228 (iv) Find x such that P(X > x) = 0.10 P(Z > z) = 0.10 → P(0 < Z < z) = 0.40 → z ≈ 1.28 (from table) x = μ + zσ = 60 + 1.28(10) = 60 + 12.8 = 72.8
5. A machine produces screws, 2% of which are defective. In a pack of 100 screws, find using Poisson approximation: (i) P(no defective screw), (ii) P(2 defective screws), (iii) P(at least 2 defective screws).
Answer
n = 100, p = 0.02, λ = np = 100 × 0.02 = 2 Using Poisson approximation: X ~ Poisson(2) (i) P(X = 0) = e⁻² × 2⁰/0! = e⁻² ≈ 0.1353 (ii) P(X = 2) = e⁻² × 2²/2! = 2e⁻² ≈ 2 × 0.1353 = 0.2707 (iii) P(X ≥ 2) = 1 − P(X = 0) − P(X = 1) = 1 − e⁻² − 2e⁻² = 1 − 3e⁻² ≈ 1 − 3 × 0.1353 = 1 − 0.4060 = 0.5940
6. If X ~ B(100, 0.3), use normal approximation to find P(25 < X < 35).
Answer
Step 1: Find parameters Mean = np = 100 × 0.3 = 30 Variance = npq = 100 × 0.3 × 0.7 = 21 σ = √21 ≈ 4.583 Step 2: Apply continuity correction P(25 < X < 35) = P(25.5 ≤ X ≤ 34.5) (continuity correction) Step 3: Convert to Z Z₁ = (25.5 − 30)/4.583 = −4.5/4.583 ≈ −0.98 Z₂ = (34.5 − 30)/4.583 = 4.5/4.583 ≈ 0.98 Step 4: Find probability P(−0.98 < Z < 0.98) = 2 × P(0 < Z < 0.98) = 2 × 0.3365 = 0.6730
Common Mistakes to Avoid
1. Using p where q is needed — In binomial variance, Var(X) = npq, not np. Always compute q = 1 − p first. For example, if p = 0.3, then q = 0.7, not 0.3.
2. Forgetting factorial in Poisson PMF — P(X = x) = e⁻ᵏλˣ/x!, NOT e⁻ᵏλˣ. The x! in the denominator is essential.
3. Not applying continuity correction for normal approximation to binomial — When approximating a discrete distribution (binomial) with a continuous one (normal), use continuity correction: P(X = r) becomes P(r − 0.5 < X < r + 0.5).
4. Confusing σ and σ² in normal distribution — N(μ, σ²) means the second parameter is variance, not standard deviation. If given N(100, 25), then σ = 5 (not σ = 25).
5. Wrong Z transformation — Use Z = (X − μ)/σ, not Z = (X − σ)/μ or any other arrangement. Always subtract the mean and divide by standard deviation.
6. Incorrect Poisson approximation conditions — Use Poisson approximation only when n is large (n > 20) and p is small (p < 0.05). If these conditions are not met, use exact binomial.
Exam Checklist
- [ ] I can identify when to use binomial, Poisson, or normal distribution based on problem context
- [ ] I can compute binomial probabilities using P(X = r) = ⁿCᵣ pʳ qⁿ⁻ʳ
- [ ] I can compute Poisson probabilities using P(X = x) = e⁻ᵏλˣ/x!
- [ ] I can convert a normal variable to standard normal using Z = (X − μ)/σ
- [ ] I can use the standard normal table to find probabilities
- [ ] I know when to apply continuity correction (binomial to normal approximation)
- [ ] I can compute mean and variance for all three distributions from given parameters
- [ ] I can handle cumulative probabilities like P(X ≤ r), P(X ≥ r) correctly
