Normal Distribution

1 Normal Distribution Exercise 1.1. (From Section 2 Notes, Question 2a, 2b, 2c, In a population, X ~ N(50, 100). Calculate: (a) P(X ≤ 60) P(X ≤ 60) = P(Z ≤ (60-50)/√100) = P(Z ≤ 1) = 0.8413 (from the normal table (Appendix Table 1 of textbook)) (b) P(X > 90) P(X > 90) = P(Z > (90-50)/√100) = P(Z > 4) = 1 - P(Z ≤ 4) ≈ 1 (c) P(45 ≤ X ≤ 55) P(45 ≤ X ≤ 55) = P((45-50)/√100 ≤ Z ≤ (55-50)/√100) = P(-0.5 ≤ Z ≤ 0.5) = P(Z ≤ 0.5) - P(Z ≤ -0.5) = 0.6915 - 0.3085 = 0.383 2 Sample Average (a.k.a. Sample Mean) Exercise 2.1. (Stock & Watson, Review the Concepts, Question 2.4) An econometrics class has 80 students, and the mean student weight is 145   lb. A random sample of 4 students is selected from the class, and their average weight is calculated. Will the average weight of the students in the sample equal   145 lb? Why or why not? Use this example to explain why the sample average,   Ȳ, is a random variable. The average weight of students in the sample will not necessarily be equal to 145 lbs, because of sampling variation. Different groups of 4 students may have different sample averages, hence Ȳ is a random variable. 3 Law of Large Numbers (LLN) Exercise 3.1. (Stock & Watson, Review the Concepts, Question 2.5) Suppose that Y₁, ..., Yn are i.i.d. random variables, with a N(1, 4) distribution. Sketch the probability density of Ȳ when n = 2, n = 10, and n = 100. In words, describe how the densities differ. What is the relationship between your answer and the LLN? When n = 2: E(Ȳ) = 1, SD(Ȳ) ≈ 0.63 When n = 10: E(Ȳ) = 1, SD(Ȳ) ≈ 0.2 When n = 100: E(Ȳ) = 1, SD(Ȳ) ≈ 0.02 Ȳ is normally distributed because each Yi is normally distributed. * Essentially, the difference in the densities is that it is tighter around the mean 1 when n = 100, since SD(Ȳ) is smaller when n gets larger. * As n gets large, SD(Ȳ) gets closer to zero, so the density collapses around the mean 1. * Another way of saying this is that the distribution of Ȳ gets more and more concentrated around μ as n gets large, so the probability that Ȳ is close to μ goes to 1, which is what LLN states. 4 Central Limit Theorem (CLT) Exercise 4.1. (Stock & Watson, Exercise 2.17a) Yi, i = 1, ..., n are i.i.d. Bernoulli random variables with p = 0.4. Let Ȳ denote the sample mean. Use the limit theorem to compute approximations for (i) P(Ȳ ≥ 0.43) when n = 100, (ii) P(Ȳ ≤ 0.37) when n = 400. * Solve for E(Yi) and Var(Yi). E(Yi) = 1*0.4 + 0*0.6 = 0.4 Var(Yi) = E(Yi²) - E(Yi)² = 0.4 - 0.4² = 0.24 * Since E(Yi) and Var(Yi) are finite and n is large, CLT holds. (i) By CLT (n = 100 is sufficiently large), Ȳ ~ N(0.4, 0.24/100) P(Ȳ ≥ 0.43) = 1 - P(Ȳ < 0.43) = 1 - P(Z < 0.6124) ≈ 0.271 (ii) By CLT (n = 400 is sufficiently large), Ȳ ~ N(0.4, 0.24/400) P(Ȳ ≤ 0.37) = P(Z ≤ -1.22) ≈ 0.111 Exercise 4.2. (Stock & Watson, Exercise 2.18) In any year, the weather can inflict damage to a home. From year to year, the damage is random. Let Y denote the dollar value of damage in any given year. Suppose that in 95% of the years, Y = $0, but in 5% of the years, Y = $20,000. (a) What are the mean and standard deviation of the damage in any year? E(Y) = 0.95 * 0 + 0.05 * 20,000 = $1,000 E(Y²) = 0.95 * 0² + 0.05 * (20,000)² = $20M Var(Y) = E(Y²) - E(Y)² = $20M - ($1,000)² = $19M SD(Y) = √$19M = $4359 (b) Consider an "insurance pool" of 100 people whose homes are sufficiently dispersed so that, in any year, the damage to different homes can be viewed as independent identically distributed random variables. Let Ȳ denote the average damage to these 100 homes in a year. (i) What is the expected value of the average damage Ȳ? (ii) What is the probability that Ȳ exceeds $2000? (i) E(Ȳ) = E(ΣYi/n) = (1/n)ΣE(Yi) = E(Y) = $1,000 (from part (a)) Also note: SD(Ȳ) = σ/√n = $4359/√100 = $435.9 (ii) CLT assumptions hold b/c Yi are iid, var(Yi) = $19M, and n = 100 is sufficiently large. Hence, by CLT, Ȳ ~ N(1000, 19M/100) ⇒ P(Ȳ > 2000) = 1 - P(Ȳ ≤ 2000) = 1 - P(Z ≤ (2000-1000)/435.9) = 1 - P(Z ≤ 2.29) = 1 - 0.9890 = 0.011