Lesson Notes By Weeks and Term v3 - Senior Secondary 3
Probability distributions and approximations
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Probability distributions and approximations
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Subject: Further Mathematics
Class: Senior Secondary 3
Term: 1st Term
Week: 3
Theme: Statistics And Probarbilty
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A probability distribution describes how probabilities are distributed over the values of a random variable. Random variables can be discrete (countable values) or continuous (any value within a range).
2.2.1 Binomial Distribution The Binomial distribution models the number of successes in a fixed number of independent Bernoulli trials.
Conditions for Binomial Distribution: Fixed number of trials, n.
Each trial has only two possible outcomes: "success" or "failure." The trials are independent. The probability of success, p, is constant for each trial. The probability of failure is q = 1 - p.
Notation: X ~ B(n, p)
Probability Mass Function (PMF): P(X = k) = C(n, k) p^k q^(n-k) where: X is the random variable representing the number of successes. k is the specific number of successes (k = 0, 1, 2, ..., n). C(n, k) = n! / (k! (n-k)!) is the binomial coefficient. p is the probability of success. q is the probability of failure (1-p). n is the number of trials.
Mean (Expected Value): E(X) = np Variance: Var(X) = npq Worked Example 2.2.1 (Binomial): A quality control inspector at a Nigerian textile factory randomly selects 10 fabric rolls. Past data shows that 5% of fabric rolls produced are defective. a) What is the probability that exactly 2 of the selected rolls are defective? b) What is the expected number of defective rolls in the sample?
Solution: Here, n = 10 (number of trials/rolls), p = 0.05 (probability of a defective roll), q = 1 - 0.05 = 0.95. a) For exactly 2 defective rolls, k =
2. P(X = 2) = C(10, 2) (0.05)^2 (0.95)^(10-2) = [10! / (2! 8!)] (0.05)^2 * (0.95)^8 = (10 9 / 2) 0.0025 * 0.6634204 = 45 0.0025 0.6634204 = 0.0746 (approx. 3 s.f.) The probability that exactly 2 rolls are defective is approximately 0.0746. b)
Expected number of defective rolls: E(X) = np = 10 * 0.05 = 0.5 The expected number of defective rolls is 0.5. 2.2.2 Poisson Distribution The Poisson distribution models the number of events occurring in a fixed interval of time or space, given that these events occur with a known constant mean rate and independently of the time since the last event. It is often used for rare events.
Conditions for Poisson Distribution: Events occur independently. The average rate of events (λ) is constant over the interval. Events cannot occur simultaneously.
Notation: X ~ Po(λ)
Probability Mass Function (PMF): P(X = k) = (e^-λ * λ^k) / k! where: X is the random variable representing the number of events. k is the specific number of events (k = 0, 1, 2, ...). λ (lambda) is the average rate of events in the given interval (mean number of occurrences). e is Euler's number (approximately 2.71828).
Mean (Expected Value): E(X) = λ Variance: Var(X) = λ Worked Example 2.2.2 (Poisson): The average number of power outages in a particular community in Port Harcourt during a month is
3. Assuming power outages follow a Poisson distribution: a) What is the probability that there are exactly 2 power outages in a given month? b) What is the probability that there are more than 1 power outage in a given month?
Solution: Here, λ = 3 (average number of outages per month). a) For exactly 2 outages, k =
2. P(X = 2) = (e^-3 * 3^2) / 2! = (0.049787 * 9) / 2 = 0.448083 / 2 = 0.2240 (approx. 4 s.f.) The probability of exactly 2 power outages is approximately 0.2240. b) For more than 1 outage, P(X > 1) = 1 - [P(X = 0) + P(X = 1)] P(X = 0) = (e^-3 3^0) / 0! = e^-3 1 / 1 = 0.049787 P(X = 1) = (e^-3 3^1) / 1! = e^-3 3 / 1 = 0.049787 * 3 = 0.149361 P(X > 1) = 1 - (0.049787 + 0.149361) = 1 - 0.199148 = 0.8009 (approx. 4 s.f.) The probability of more than 1 power outage is approximately 0.8009. 2.3.1 Normal Distribution The Normal distribution (or Gaussian distribution) is a symmetrical, bell-shaped continuous probability distribution. It is arguably the most important distribution in statistics due to its wide applicability.
Characteristics: Bell-shaped and symmetrical about its mean. The mean, median, and mode are all equal. Total area under the curve is
1. Asymptotic to the x-axis (tails extend indefinitely without touching).
Defined by two parameters: mean (μ) and standard deviation (σ).
Notation: X ~ N(μ, σ2) (where σ2 is variance)
Probability Density Function (PDF): f(x) = (1 / (σ sqrt(2π))) e^(-(x - μ)2 / (2σ2)) (Students typically do not need to use this formula directly but should be aware of it.) Standard Normal Distribution (Z-distribution): A special case of the Normal distribution with mean μ = 0 and standard deviation σ =
1. Any Normal random variable X can be transformed into a standard normal variable Z using the formula: Z = (X - μ) / σ This transformation allows us to use standard normal tables (Z-tables) to find probabilities.
Key Properties using Z-tables: P(Z a) = 1 - P(Z 50). The probability of success p is very small (typically p 50). p = 0.01 (small, k - 0.5) in Normal. P(X ≤ k) in Binomial becomes P(Y k) in Binomial becomes P(Y > k + 0.5) in Normal. P(X 85) in Binomial becomes P(Y > 85.5) in Normal (with continuity correction). For Y = 85.5: Z = (85.5 - 90) / 6 = -4.5 / 6 = -0.75 P(Y > 85.5) = P(Z > -0.75) = 1 - P(Z 5) = 1 - P(X ≤ 5) P(X ≤ 5) = P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4) + P(X=5) P(X=0) = 0.018316 P(X=1) = (e^-4 4^1) / 1! = 0.018316 4 = 0.073264 P(X=2) = (e^-4 4^2) / 2! = (0.018316 16) / 2 = 0.146528 P(X=3) = (e^-4 4^3) / 3! = (0.018316 64) / 6 = 0.195371 P(X=4) = (e^-4 4^4) / 4! = (0.018316 256) / 24 = 0.195371 P(X=5) = (e^-4 4^5) / 5! = (0.018316 1024) / 120 = 0.156297 P(X ≤ 5) = 0.018316 + 0.073264 + 0.146528 + 0.195371 + 0.195371 + 0.156297 = 0.785147 P(X > 5) = 1 - 0.785147 = 0.214853 = 0.2149 (approx. 4 s.f.)
Commentary: This problem reinforces the application of Poisson PMF and cumulative probabilities. Students must be careful with calculations involving e. Question 3 (Normal Approximation to Binomial): A local government council in Kaduna is conducting a survey. It is known that 70% of households in the area use borehole water. If a random sample of 100 households is selected, use the Normal approximation with continuity correction to find the probability that between 65 and 75 (inclusive) households use borehole water.
Solution 3: This is a Binomial distribution: X ~ B(n=100, p=0.7).
Check conditions for Normal approximation: np = 100 * 0.7 = 70 (≥ 5) n(1-p) = 100 * 0.3 = 30 (≥ 5) Conditions met. Approximate with N(μ, σ2): μ = np = 70 σ2 = np(1-p) = 70 * 0.3 = 21 σ = sqrt(21) ≈ 4.5826 We want P(65 ≤ X ≤ 75) in Binomial. With continuity correction, this becomes P(64.5 35) in Poisson. With continuity correction, this becomes P(Y > 35.5) in Normal. For Y = 35.5: Z = (35.5 - 30) / 5.4772 = 5.5 / 5.4772 = 1.004 ≈ 1.00 P(Y > 35.5) = P(Z > 1.00) = 1 - P(Z < 1.00) (Assuming P(Z < 1.00) = 0.8413 from Z-tables) = 1 - 0.8413 = 0.1587 The probability of more than 35 connectivity issues is approximately 0.
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7. Commentary: This problem checks the application of Poisson to Normal approximation, including calculations of mean, variance, and the continuity correction for "greater than" probabilities. --- This section provides a detailed exposition of the three main probability distributions and their approximations. Differentiation (for varied learning styles and paces): Visual Learners: Utilize diagrams for distribution shapes (bell curve, bar charts for discrete distributions), flowcharts for deciding which distribution to use or which approximation applies, and visual step-by-step problem solutions.
Auditory Learners: Encourage group discussions, peer teaching sessions, and verbal explanations of concepts and problem-solving steps.
Kinesthetic Learners: Provide opportunities for hands-on activities, such as using physical counters or dice to simulate Binomial experiments, or manipulating values on a spreadsheet to observe changes in distributions. Group work on large whiteboards for problem-solving.
Differentiated Questioning: Ask simpler, recall-based questions to struggling learners and more complex, analytical questions to advanced learners during discussions.
Remediation (for struggling learners): Review of Prerequisites: Conduct a quick review of basic probability, factorials, combinations, and the use of mathematical tables (e, logarithms) if necessary.
Focused Practice: Provide additional, simpler practice problems that isolate specific concepts (e.g., only calculating Binomial probabilities, only applying Poisson formula).
Step-by-Step Guidance: Break down complex problems into smaller, manageable steps. Provide partially worked examples where learners complete the final steps.
Peer Tutoring: Pair struggling learners with high-achieving peers who can explain concepts in a more accessible way.
Visual Aids and Simulators: Use online simulators or physical tools to demonstrate how parameters affect the shape of the distributions.
Remediation Activity: "Distribution ID Card" - Provide cards with scenarios and learners must identify the appropriate distribution, state its parameters, and explain why.
Extension (for high-achieving learners): Advanced Problems: Challenge them with multi-step problems that involve combining concepts (e.g., a scenario requiring a Binomial calculation, then an approximation, and then interpreting the difference).
Investigation Tasks: Assign a mini-research project to investigate other probability distributions (e.g., Geometric, Hypergeometric, Exponential) and their real-world applications in Nigeria.
Real Data Analysis: Provide access to simplified real-world data sets (e.g., from agricultural yields, traffic counts, health statistics) and ask them to determine which distribution best fits the data and to make predictions.
Conceptual Exploration: Ask them to explore the theoretical underpinnings of why these approximations work (e.g., how the Binomial PMF converges to Poisson under specific conditions, or the Central Limit Theorem's role in Normal approximation).
Extension Activity: "Approximation Accuracy Challenge" - Students calculate probabilities using both the exact distribution formula and its approximation, then calculate the percentage error to evaluate the accuracy of the approximation under different parameter values. This encourages a deeper understanding of the conditions for good approximation.
Public Health and Epidemiology (Binomial, Poisson, Normal, and Approximations): Binomial: Estimating the probability of a certain number of individuals contracting a specific non-contagious disease (e.g., malaria) in a community given the general prevalence rate.
Poisson: Modeling the occurrence of rare diseases or health events (e.g., cholera outbreaks) in specific regions or over time intervals, which helps public health officials plan interventions.
Normal: Analyzing the distribution of physiological measurements in the Nigerian population (e.g., blood pressure, cholesterol levels, body mass index) to identify risk factors for non-communicable diseases.
Approximations: Using the Normal approximation to the Binomial distribution for large-scale vaccination campaigns to estimate the number of successful immunizations out of millions of recipients. Agriculture and Food Security (Binomial, Poisson, Normal, and Approximations): Binomial: Predicting the number of successful germinations of seeds, successful crop yields based on specific farming techniques, or the number of healthy livestock in a herd.
Poisson: Modeling the number of pest infestations in a given farm area or the frequency of specific crop diseases over a planting season, aiding in pest control and disease management strategies.
Normal: Analyzing the distribution of crop yields (e.g., tonnes of rice per hectare) across different farms or regions to understand productivity variations and inform agricultural policies.
Approximations: Using the Poisson approximation to the Binomial when dealing with the probability of a very small number of defective fruits (e.g., rotten tomatoes) in a very large harvest from a commercial farm. Manufacturing and Quality Control (Binomial, Poisson, and Normal): Binomial: Assessing the probability of producing a certain number of defective items (e.g., faulty electronics, incorrectly assembled vehicle parts) in a production run.
Poisson: Modeling the number of defects found per square meter of fabric or per hour of machine operation in a textile or processing plant.
Normal: Analyzing variations in product dimensions, weights, or material strengths to ensure quality standards are met within acceptable tolerances.
Approximations: Applying the Normal approximation to the Binomial to quickly estimate the likelihood of a high number of compliant products in a very large production batch, rather than doing complex Binomial calculations. ---