JC H2 Mathematics Material Compilation - Statistics

Title Junior College H2 Mathematics Materials Compilation – Statistics Author AprilDolphin Date 19/5/2025 Page Title 2 Binomial Distribution 18 Normal Distribution – Basic Operations 37 Normal Distribution – Operation Involving Linear Combinations of Normal Random Variables and Sum of Multiple Independent Identically Distributed Normal Random Variables

Title Junior College ‘A’ Levels H1/H2 – Binomial Distribution Author AprilDolphin Date 25/3/2025 Conditions for a random variable to be modelled by a Binomial Distribution includes the following: The experiment must consist of Bernoulli trials (Where there are two possible outcomes in the experiment, which we can call as “outcome” and “complement outcome”.) All trials in the experiment have to be independent (Where the probability of obtaining “outcome” of each trial isn’t affected by a previous trial or will affect a future trial within the experiment). All trials within the experiment have to be identically distributed. (Such that each Bernoulli trial has constant probability of obtaining “outcome” and “complement outcome”.) Example of common outcomes and complement outcomes as follows: Outcome Complement Outcome Yes No No Yes Success Failure Failure Success Picking a red ball Not picking a red ball If the experiment in question satisfies the above requirements, it is said the follow a Binomial Distribution with parameters 𝑛 and 𝑝,where, 𝑛 refers to the total number of trials. 𝑝 refers to the probability of obtaining the “outcome” of each trial. When it is written in standard Binomial Notation, it looks like the following, where 𝑋 refers to the random variable. 𝑋~𝐵(𝑛, 𝑝)

The formula for Binomial Probability distribution of a specific number of trials to be calculated for is given below: 𝑃(𝑋 = 𝑥) = (𝑛 𝑥) 𝑝𝑥(1 − 𝑝)𝑛−𝑥 The formula for Binomial Probability distribution from 0 up till 𝑥 number of trials can be calculated as follows: 𝑃(𝑋 ≤ 𝑥) = (𝑛 0) 𝑝0(1 − 𝑝)𝑛−0 + (𝑛 1) 𝑝1(1 − 𝑝)𝑛−1 + (𝑛 2) 𝑝2(1 − 𝑝)𝑛−2 + ⋯ + (𝑛 𝑥) 𝑝𝑥(1 − 𝑝)𝑛−𝑥 Since A Levels permit the use of Texas Instrument Graphing Calculators in exam condition, I would also have to demonstrate the two rather commonly used functionality in TI-84 Plus CE, namely BinomialPDF and BinomialCDF that is equivalent the above two respectively. BinomialPDF can be used when you are tasked to find 𝑃(𝑋 = 𝑥) given parameters 𝑛 and 𝑝. Example 1. Given the random variable 𝑋~𝐵 (3, 1 6), find 𝑃(𝑋 = 2). This case requires the use of BinomialPDF functionality, which can be accessed by pressing the following buttons on the TI-84 PLUS CE in the following order: Press [2ND] then [VARS] in exact order as mentioned and press the down arrow key repeatedly until you see your calculator cursor reaching an option called “binompdf”. Press [ENTER] key on the calculator and you should see something similar to the below example on the screen trials: p: x value: Paste Key in value of 𝑛 into the number of trials, and press [ENTER] Key in value of p into the field “p” and press [ENTER] Key in number of trials being computed for into the field “x value” and press [ENTER] Once the cursor is on the “Paste”, you should have the following,

trials: 3 p:1/6 x value: 2 Paste Press [ENTER] after checking if the values are correct and you should see the following on your graphing calculator screen binompdf(3, 1/6, 2) Press [ENTER] and the answer should appear as follows (If the calculator isn’t in fraction mode): binompdf(3, 1/6, 2) .0694444444 Example 2. Given the random variable 𝑋~𝐵 (3, 1 6), find the probability that 𝑃(𝑋 ≤ 2). This case requires the use of BinomialCDF functionality, which can be accessed by pressing the following buttons on the TI-84 PLUS CE in the following order. Press [2ND] key, then Press [VARS] in exact order as mentioned and press the down arrow key repeatedly until you see your calculator cursor reaching an option called “binomcdf”. Press [ENTER] key on the calculator and you should see something similar to the below example on the screen trials: p: x value: Paste Key in value of 𝑛 into the number of trials, and press [ENTER] Key in value of p into the field “p” and press [ENTER] Key in number of trials being computed for into the field “x value” and press [ENTER] Once the cursor is on the “Paste”, you should have the following,

trials: 3 p:1/6 x value: 2 Paste Press [ENTER] after checking if the values are correct and you should see the following on your graphing calculator screen binomcdf(3,1/6,2) Press [ENTER] and the answer should appear as follows (If calculator isn’t in fraction mode) binomcdf(3,1/6,2) .9953703704 Using graphing calculator manipulation to solve problems involving binomial distribution: (Questions mostly taken from Power Math H2 Second Edition Volume 2 by PK Lim and slightly changed) Q1. In XYZ Junior College, 65% of the student population are male. 12 students are randomly selected from this school. Find the probability that (i) Exactly 3 of them are male. (ii) At most 3 of them are male. (iii) Not less than 3 of them are male. (iv) More than 5 of them are male. (v) Between 4 to 8 of them inclusively are male. Written working Graphing Calculator Actions Performed (i) 𝑋: Number of male students selected, out of 12 students 𝑋~𝐵(12,0.65) 𝑃(𝑋 = 3) = 0.00476 Go to “binompdf” option and press enter Key in trials as 12, Key in p as 0.65 Key in x value as 3 Answer obtained = 0.00476

(ii) 𝑃(𝑋 ≤ 3) = 0.00561 Go to “binomcdf” option and press enter. Key in trials as 12 Key in p as 0.65 Key in x value as 3 Answer obtained = 0.00561 Final Answer: 0.00561 (iii) 𝑃(𝑋 ≥ 3) = 1 − 𝑃(𝑋 ≤ 2) = 0.999 Go to “binomcdf” option and press enter. Key in trials as 12 Key in p as 0.65 Key in x value as 2 Answer obtained = 8.479084920E-4 Press 1 – 8.479084920E-4 to get 0.9991520915 Final Answer: 0.999 (iv) 𝑃(𝑋 > 5) = 1 − 𝑃(𝑋 ≤ 5) = 0.915 Go to “binomcdf” option and press enter. Key in trials as 12 Key in p as 0.65 Key in x value as 5 Answer obtained = 0.0846320652 Press 1 – 0.0846320652 to get 0.9153679348 Final Answer: 0.915 (v) 𝑃(4 ≤ 𝑋 ≤ 8) = 𝑃(𝑋 ≤ 8) − 𝑃(𝑋 ≤ 3) = 0.648 Go to “binomcdf” option and press enter. Key in trials as 12 Key in p as 0.65 Key in x value as 8 Answer obtained = 0.6533473038 Go to “binomcdf” options and press enter.

Key in trials as 12 Key in p as 0.65 Key in x value as 3 Answer obtained = 0.0056097523 Press 0.6533473038- 0.0056097523=0.6477375515 Final answer: 0.648 Q2. A survey shows that only 60% of all drivers in a town uses their seatbelts. (i) 5 drivers in the town are randomly selected. Let 𝑋 be the number of drivers who use their seat belts, out of 5 drivers from this town. Find the exact standard deviation of 𝑋. (ii) If a sample of 500 drivers are taken, what is the expected number of drivers who use their seat belts? (i) 𝑋: Number of drivers who uses their seat belts, out of 5 drivers selected from this town. 𝑋~𝐵(5, 0.6) Variance of Binomial Distribution 𝜎2 = 𝑛𝑝(1 − 𝑝) Standard Deviation of Binomial Distribution 𝜎 = √𝑛𝑝(1 − 𝑝) = √5(0.6)(1 − 0.6) = √1.2 (ii) 𝑌: Number of drivers who uses their seat belts, out of 500 drivers selected from this town 𝑌~𝐵(500,0.6) Expected number of drivers who uses their seat belts 𝐸(𝑌) = 𝑛𝑝 = 500(0.6) = 300 Q3. A random variable 𝑋~𝐵(𝑛, 𝑝) has mean of 8 and variance of 6. (i) Find the value of both 𝑛 and 𝑝. (ii) Find the probability that 𝑋 lies within 1 standard deviation of the mean. (i) Equation 1: 𝑛𝑝 = 8 Equation 2: 𝑛𝑝(1 − 𝑝) = 6

Sub Equation 1 into Equation 2. 8(1 − 𝑝) = 6 8 − 8𝑝 = 6 −8𝑝 = 6 − 8 = −2 Equation 3: 𝑝 = 2 8 = 0.25 Substitute Equation 3 into Equation 1. 𝑛(0.25) = 8 𝑛 = 8 0.25 =