JC and Polytechnic Mathematics Material Compilation - Statistics

Title JC and Polytechnic Mathematics Material Compilation – Statistics Editor Lee Jian Lian Date 12/4/2019 Topic [Category: Discrete Probability Distribution] Page Discrete Probability Distribution 2 Binomial Distribution 3 Poisson Distribution – Fundamentals 8 Poisson Distribution as Approximation to Binomial Distribution with large 𝑛 and small 𝑝 11 Topic [Category: Continuous Probability Distribution] Page Continuous Probability Distribution 13 Normal Distribution – Fundamentals 15 Normal Distribution – Distribution of Sample Means 20 Normal Distribution – Central Limit Theorem 22 Confidence Intervals with Normal and 𝑡-distribution: Basics 23 Confidence Intervals with Normal and 𝑡-distribution: Calculation 24 Topic [Category: Hypothesis Testing] Page Hypothesis Testing on One Sample with Normal and 𝑡-distribution 27 Chi-Square-Test for Goodness-of-Fit 32 Chi-Square-Test for Independence 34 Miscellaneous Matters Page Utilizing a Standard Normal Table – Probability from Far-Left (or Negative Infinity) of the Normal Distribution to the 𝑧-score 37 Applicable to Courses of the Following Nature - Information Technology - Engineering - Applied Science/Science - Business/Business Management - GCE ‘A’ Level H1 Mathematics - GCE ‘A’ Level H2 Mathematics & ‘A’ Level H2 Further Mathematics

Title Discrete Probability Distribution Author Liu Hui Ling, Ngee Ann Polytechnic Date 17/10/2018 Discrete Probability Distribution has the following properties. • Takes in discrete variables (Whole number values 𝑘 , where 𝑘 ≥ 0) • Countable number of values involved • Takes in random variables (Sum of all probabilities must be equal to 1) Example Number of Events 0 1 2 3 Probability 0.5 0.25 0.10 0.15 In the case of the Binomial Distribution, as represented by the formula below, 𝑃(𝑋 = 𝑘) = (𝑛 𝑘) 𝑝𝑘(1 − 𝑝)𝑛−𝑘 The following limitation is imposed, as any values that doesn’t comply to the following limitation is undefined. 0 ≤ 𝑘 ≤ 𝑛 In the case of Poisson Distribution, as represented by the formula below, 𝑃(𝑋 = 𝑘) = 𝑒−𝜇 (𝜇𝑘 𝑘!) Where 0 ≤ 𝑘 < ∞ The inequality 0 ≤ 𝑘 < ∞, implies the number of events you are performing the probability calculations for can be any finite whole number greater than or equal to 0. While there is no upper limit to the value of 𝑘, a theorem guarantees the value of all probability will sum up to 1: As the probability within a Binomial Distribution approaches 0 and the number of trials approaches infinity. The Binomial Distribution will converge to the Poisson Distribution. This implies that the Poisson Distribution is just a special case of Binomial Distribution, which means the probability will still sum up to 1 anyway.

Title Polytechnic and A Level H2 Mathematics (Statistics) Binomial Distribution Author Lim Wang Sheng, School of Information Technology, Nanyang Polytechnic [CCA: NYP Mentoring Club] Date 9/6/2018 Applicable to the following levels ✓ School of Information Technology Students (Computing Mathematics) ✓ School of Engineering (Engineering Mathematics – Statistical Analysis) ✓ School of Business Management (Statistics – Business Statistics) ✓ School of Chemical and Life Sciences – Biostatistics ✓ JC/MI Students – H2 Mathematics – Statistics Due to my school’s syllabus, it may or may not cover everything required for H2 Mathematics. JC/MI students should see referring to this guide as a last resort if you still don’t know the basics. To use the binomial distribution, the following requirements must be met. ➢ There will only be 2 possible outcomes (Success/Failure, Yes/No, etc.) ➢ Each trial is an independent event (that is, will not affect the subsequent trial or be affected by past trial) You must also know the following information or able to derive the following details ➢ You know the probability of each trial ➢ You are given the total number of trials and the number of trials the probability is being calculated for, which will be shown in notation form in the next few pages. Formula for Binomial Distribution Probability Given as Follows 𝑃(𝑋 = 𝑘) = (𝑛 𝑘) 𝑝𝑘 (1 − 𝑝)𝑛−𝑘 [It may be written slightly differently in other textbooks. But they should mean the same thing.]

Notation Meaning 𝑃(𝑋 = 𝑘) Probability of obtaining an outcome, given the variable or the number of trials being calculated for will be exactly equal to 𝑘 Or simply put, the number of trials the outcome is being calculated for (𝑛 𝑘) Total number of combinations the 2 outcomes can be rearranged 𝑛 refers to the total number trials 𝑘 refers to the number of trials the outcome is being is being calculated for 𝑝𝑘 The probability the outcome you are finding for after 𝑘 number of independent trials. (Example, the outcome can be Yes or Success) (1 − 𝑝)𝑛−𝑘 The probability of obtaining the alternate outcome after 𝑛 − 𝑘 number of independent trials. (Example, if your outcome is Yes or Success, then the alternate corresponding outcome are No or Failure respectively.) 𝑋~𝐵(𝑛, 𝑝) The random variable 𝑋 is to follow a binomial distribution, over 𝑛 number of independent trials, which trial shall have a 𝑝 probability of obtain the outcome mentioned in question. Formula List for Analyzing a Binomial Distribution Formula for Mean (𝜇) (Also called Expected Value) 𝜇 = 𝑛𝑝 Formula for Variance (𝜎2) 𝜎2 = 𝑛𝑝(1 − 𝑝) Formula for Standard Deviation (𝜎) 𝜎 = √𝑛𝑝(1 − 𝑝)

Binomial Distribution Questions and Example [Section I]: Basic Calculation Q1: Given the following binomial distribution and information. 𝑋~𝐵(5,0.3) Evaluate the following (a) 𝑃(𝑋 = 2) (b) 𝑃(𝑋 < 2) (c) 𝑃(𝑋 < 3) (d) 𝑃(𝑋 ≥ 2) Q1(a) 𝑃(𝑋 = 2) = (5 2) 0.32(1 − 0.3)5−2 = 10(0.09)(0.343) = 0.3087 Q1(b) 𝑃(𝑋 < 2) = 𝑃(𝑋 = 0) + 𝑃(𝑋 = 1) 𝑃(𝑋 = 0) = (5 0) 0.30(1 − 0.3)5−0 = 1(0.3)0(0.7)5−0 = 0.16807 𝑃(𝑋 = 1) = (5 1) 0.31(1 − 0.3)5−1 = 5(0.3)1(0.7)5−1 = 0.36015 𝑃(𝑋 < 2) = 0.16807 + 0.36015 = 0.52822 Q1(c) 𝑃(𝑋 < 3) = 1 − [𝑃(𝑋 = 3) + 𝑃(𝑋 = 4) + 𝑃(𝑋 = 5)] [Values of all probabilities in binomial distribution must sum up to 1] **Use the method that require the least number of calculation. 𝑃(𝑋 = 3) = (5 3) 0.33(1 − 0.3)5−3 = 10(0.027)(0.7)2 = 0.1323 𝑃(𝑋 = 4) = (5 4) 0.34(1 − 0.3)5−4 = 5(0.0081)(0.7) = 0.02835 𝑃(𝑋 = 5) = (5 5) 0.35(1 − 0.3)5−5 = 1(0.00243)(1) = 0.00243

𝑃(𝑋 = 3) + 𝑃(𝑋 = 4) + 𝑃(𝑋 = 5) = 0.16308 𝑃(𝑋 < 3) = 1 − 0.16308 = 0.83692 Q1(d) [From Answers Derived in Q1(b)] 𝑃(𝑋 ≥ 2) = 1 − [𝑃(𝑋 = 0) + 𝑃(𝑋 = 1)] = 1 − (0.16807 + 0.3015) = 1 − 0.46458 = 0.53542 Section II (Application of Binomial Distribution) Q2 A survey indicates that 60% of the school’s student population is interested to participate in an event. You randomly selected 7 students who had participated in the survey. (a) Is binomial distribution suitable for this question, please justify your answer. (b) Find the probability that exactly 4 students are interested in the event. (c) Find the probability that at most 3 students are interested in the event. (d) Find the expected value, standard deviation and variance of the distribution. (a) Yes. Every student’s interest in the event can be regarded as independent. There are only two possible outcomes, either a “YES” or a “NO”. (b) 𝑃(𝑋 = 4) = (7 4) (0.6)4(1 − 0.6)7−4 = 35(0.1296)(0.064) = 0.290304 (c) 𝑃(𝑋 ≤ 3) = 𝑃(𝑋 = 0) + 𝑃(𝑋 = 1) + 𝑃(𝑋 = 2) + 𝑃(𝑋 = 3) 𝑃(𝑋 = 0) = (7 0) (0.6)0(1 − 0.6)7−0 = 0.00164 𝑃(𝑋 = 1) = (7 1) (0.6)1(1 − 0.6)7−1 = 0.01720 𝑃(𝑋 = 2) = (7 2) (0.6)2(1 − 0.6)7−2 = 0.00741 𝑃(𝑋 = 3) = (7 3) (0.6)3(1 − 0.6)7−3 = 0.19354

𝑃(𝑋 ≤ 3) = 0.00164 + 0.01720 + 0.00741 + 0.19354 = 0.21979 (d) 𝜇 = 𝑛𝑝 𝜇 = 0.6(7) = 4.2 𝜎2 = 𝑛𝑝(1 − 𝑝) 𝜎2 = 4.2(1 − 0.6) = 1.62 𝜎 = √1.62 = 1.2728 Q3 [Question Taken from Nanyang Polytechnic Computing Mathematics 2 Exam Paper] Given that the mean and variance of a B