18.05 Introduction to Probability and Statistics, Class 6b: Problem Solutions

Class 6b in-class problems, 18.05 Concept questions Concept question 1. Normal distributions 𝑋 has normal distribution, standard deviation 𝜎. within 1 ⋅ 𝜎 ≈ 68% within 2 ⋅ 𝜎 ≈ 95% within 3 ⋅ 𝜎 ≈ 99% Normal PDF 68% 95% 99% 𝜇 − 3𝜎 𝜇 − 2𝜎 𝜇−𝜎 𝜇+𝜎 𝜇 𝜇 + 2𝜎 𝜇 + 3𝜎 𝑧 (a) 𝑃 (−𝜎 < 𝑋 − 𝜇 < 𝜎) is approximately (i) 0.025 (ii) 0.16 (iii) 0.68 (iv) 0.84 (v) 0.95 (iv) 0.84 (v) 0.95 (b) 𝑃 (𝑋 > 𝜇 + 2𝜎) is approximately (i) 0.025 (ii) 0.16 (iii) 0.68 Solution: (a) Correct answer is (iii). The rule of thumb says the probability that 𝑋 is within one standard deviation of the mean is 0.68. (b) Correct answer is (i). This question for the probability in the right tail, beyond 2 standard deviations above the mean. The rule of thumb is that about 95% of the probability is within 2𝜎 of the mean. So about 5% is outside of that. Since this is split symmetrically between two tails, the probability in the right tail is approximately 0.025. Board questions Problem 1. Standardization Suppose 𝑋 is a random variable with mean 𝜇 and standard deviation 𝜎. Let 𝑍 be the standardization of 𝑋. (a) Give the formula for 𝑍 in terms of 𝑋, 𝜇 and 𝜎. (b) Use the algebraic properties of mean and variance to show 𝑍 has mean 0 and standard deviation 1. 𝑋−𝜇 Solution: (a) 𝑍 = . 𝜎 (b) The problem asks us to verify that 𝐸[𝑍] = 0 and Var(𝑍) = 1. We use the properties 𝐸[𝑎𝑋 + 𝑏] = 𝑎𝐸[𝑋] + 𝑏 = 𝑎𝜇 + 𝑏 Var(𝑎𝑋 + 𝑏) = 𝑎2 Var(𝑋) = 𝑎2 𝜎2 . 1 2 In the following, don’t forget that 𝐸[𝑋] = 𝜇 and Var(𝑋) = 𝜎2 . 𝑋−𝜇 1 1 ] = 𝐸 [𝑋 − 𝜇] = (𝐸[𝑋] − 𝜇) = 0. 𝜎 𝜎 𝜎 𝑋−𝜇 1 1 1 Var(𝑍) = Var ( ) = 2 Var(𝑋 − 𝜇) = 2 Var(𝑋) = 2 ⋅ 𝜎2 = 1. 𝜎 𝜎 𝜎 𝜎 𝐸[𝑍] = 𝐸 [ Problem 2. CLT (a) Carefully write the statement of the central limit theorem. (b) To head the newly formed US Dept. of Statistics, suppose that 50% of the population supports the team of Alessandre, Gabriel, Sarah and So Hee, 25% support Jen and 25% support Jerry. A poll asks 400 random people who they support. What is the probability that at least 55% of those polled prefer the team? (c) What is the probability that less than 20% of those polled prefer Jen? Solution: (b) Let 𝑋 be the fraction polled who support the team. So 𝑋 is the average of 400 Bernoulli(0.5) random variables. That is, let 𝑋𝑖 = 1 if the ith person polled prefers the team and 0 if not, so 𝑋 = average of the 𝑋𝑖 . The question asks for the probability 𝑋 > 0.55. 2 Each 𝑋𝑖 has 𝜇 = 0.5 and 𝜎2 = 0.25. So, 𝐸[𝑋] = 0.5 and 𝜎𝑋 = 0.25/400 or 𝜎𝑋 = 1/40 = 0.025. Because 𝑋 is the average of 400 Bernoulli(0.5) variables, the CLT says it is approximately normal and standardizing gives 𝑋 − 0.5 ≈𝑍 0.025 So, 𝑃 (𝑋 > 0.55) ≈ 𝑃 (𝑍 > 2) ≈ 0.025. (c) Let 𝐽 ̄ be the fraction polled who support Jen. The question asks for the probability that 𝐽 ̄ < 0.2. Similar to part (b), 𝐽 ̄ is the average of 400 Bernoulli(0.25) random variables. So, √ 𝐸[𝐽 ]̄ = 0.25 and 𝜎𝑆2 = (0.25)(0.75)/400 ⇒ 𝜎𝑆 = 3/80. 𝐽 ̄ − 0.25 So, √ ≈ 𝑍. Thus, 3/80 √ 𝑃 (𝐽 ̄ < 0.2) ≈ 𝑃 (𝑍 < −4/ 3) = pnorm(−4/sqrt(3), 0, 1) ≈ 0.0105 Problem 3. Sampling from the standard normal distribution How would you approximate a single random sample from a standard normal distribution using 9 rolls of a ten-sided die? Note: 𝜇 = 5.5 and 𝜎2 = 8.25 for a single roll of a 10-sided die. Hint: CLT is about averages. 3 Solution: The average of 9 rolls is a sample from the average of 9 independent random variables. The CLT says this average is approximately normal with 𝜇 = 5.5 and 𝜎 = √8.25/9 = 0.957 If 𝑥 is the average of 9 rolls then standardizing we get 𝑧= 𝑥 − 5.5 0.957 is (approximately) one sample from N(0, 1). Histogram of standardized 9 roll simulation 0.2 0.1 0.0 Density 0.3 0.4 So, to approximate a standard normal, we would roll 9 times and compute 𝑧. −6 −4 −2 0 2 4 6 Standard normal is shown in orange. 𝑋 = average of nine rolls: 𝜇 = 5.5, 𝜎 = √8.25/9. Standarized statistic: 𝑍 = 𝑋−𝜇 ≈ 𝑁 (0, 1). 𝜎 Extra problems Bonus problem An accountant rounds to the nearest dollar. We’ll assume the error in rounding is uniform on [-0.5, 0.5]. Estimate the probability that the total error in 300 entries is more than $5. Solution: Let 𝑋𝑗 be the error in the 𝑗th entry, so, 𝑋𝑗 ∼ 𝑈 (−0.5, 0.5). We have 𝐸[𝑋𝑗 ] = 0 and Var(𝑋𝑗 ) = 1/12. The total error 𝑆 = 𝑋1 + … + 𝑋300 has 𝐸[𝑆] = 0, Var(𝑆) = 300/12 = 25, and 𝜎𝑆 = 5. Standardizing we get, by the CLT, 𝑆/5 is approximately standard normal. That is, 𝑆/5 ≈ 𝑍. So, 𝑃 (𝑆 < −5 or 𝑆 > 5) ≈ 𝑃 (𝑍 < −1 or 𝑍 > 1) ≈ 0.32 .