Introduction to Probability and Statistics. Test 3

Name: Nyi Nyi Moe Htet 59/60 MATH 340, Introduction to Probability and Statistics, Test 3 No calculators needed or allowed! 1. [10 Points] Suppose that the probability that an offer sent by a university is accepted by the student is 0.8. Assume that 10,000 offers have been sent in 2023 and each student’s decision of acceptance is independent of others. (a). Find the approximate probability that the new enrollment is larger than 8,050. (b). Find the approximate probability that the new enrollment is smaller than 7,900. Use the table of the normal distribution on the final page. μ = 10,000 x 0.8 = 8000 σ² = 10,000 x 0.8 (1 - 0.8) = 8000 x 0.2 = 1600 σ = 40 Y = (X - 8000) / 40 standard X = 40Y + 8000 P(X > 8050) = P(40Y + 8000 > 8050) P(X < 7900) = P(40Y + 8000 < 7900) = P(Y > 5/4) = P(Y < -100/40) = 1 - P(Y < 5/4) = P(Y < -5/2) = 1 - 0.8944 = P(Y > 5/2) = 0.1056 = 1 - P(Y < 5/2) = 1 - 0.9938 = 0.0062 2. [10 Points] Consider four independent tosses of a fair coin. Let X denote the number of heads that appear in the first two tosses and Y denote the number of tails that appear in the last two tosses. (a). Find the joint probability mass function of X and Y. (b). Find p_{X|Y}(1|2) and p_{Y|X}(2|1). (c). Determine whether X and Y are independent. (a) P_X(0) = 1/16 P_X(1) = 8/16 P_X(2) = 4/16 P_Y(0) = 4/16 P_Y(1) = 8/16 P_Y(2) = 4/16 Y\X 0 1 2 0 1/16 2/16 1/16 1 2/16 4/16 2/16 2 1/16 2/16 1/16 P_{X|Y}(1|0) = P_X(2) * P_Y(0) = 4/16 * 4/16 = 1/16 P_{X|Y}(2|1) = P_X(2) * P_Y(1) = 4/16 * 8/16 = 2/16 P_{X|Y}(2|2) = P_X(2) * P_Y(2) = 4/16 * 4/16 = 1/16 P_{X|Y}(1|0) = P_Y(1) * P_X(0) = 2/16 * 3/16 = 2/16 P_{X|Y}(1|1) = P_Y(1) * P_X(1) = 8/16 * 8/16 = 4/16 P_{X|Y}(1|2) = P_Y(1) * P_X(2) = 2/16 * 8/16 = 2/16 (b) P_{X|Y}(1|2) = 2/16 / 4/16 = 1/2 P_{Y|X}(2|1) = 2/16 / 8/16 = 1/4 (c) p_x,y(0|0) = p_x(0) * p_y(0) 1/16 = 4/16 * 4/16 = 1/16 p_x,y(0|1) = p_x(0) * p_y(1) 2/16 = 4/16 * 8/16 = 2/16 p_x,y(0|2) = p_x(0) * p_y(2) 1/16 = 4/16 * 4/16 = 1/16 Thus, x and y are independent. The tools are independent, and x and y events do not depend on each other. 3. [10 Points] Consider three independent tosses of a fair coin. Let X denote the number of heads that appear and Y denote the number of tails that appear. (a). Find the joint probability mass function of X and Y. (b). Find P_{X|Y}(0|0) and P_{Y|X}(2|1). (c). Find the correlation coefficient ρ(X,Y) of X and Y, in which ρ(X,Y) = Cov(X,Y) / √[Var(X)Var(Y)]. (a) P_X(0) = 1/8 P_X(1) = 3/8 P_X(2) = 3/8 P_X(3) = 1/8 P_Y(0) = 1/8 P_Y(1) = 3/8 P_Y(2) = 3/8 P_Y(3) = 1/8 Y\X 0 1 2 3 0 0 0 0 1/8 1 0 0 3/8 0 2 0 3/8 0 0 3 1/8 0 0 0 (b) P_{X|Y}(0|0) = 0 / (1/8) = 0 P_{Y|X}(2|1) = (3/8) / (3/8) = 1 (c) E[X] = 1 * 3/8 + 2 * 3/8 + 3 * 1/8 = 3/8 + 6/8 + 3/8 = 12/8 = 3/2 E[Y] = 1 * 3/8 + 2 * 3/8 + 3 * 1/8 = 3/2 E[X²] = 1² * 3/8 + 2² * 3/8 + 3² * 1/8 = 3/8 + 12/8 + 9/8 = 24/8 = 3 E[Y²] = 3 Var[X] = E[X²] - (E[X])² = 3 - (3/2)² = 3 - 9/4 = (12-9)/4 = 3/4 Var[Y] = E[Y²] - (E[Y])² = 3 - (3/2)² = 3/4 E[X,Y] = 123/8 + 213/8 = 6/8 + 6/8 = 12/8 = 3/2 Cov(X,Y) = E[X,Y] - E[X]E[Y] = 3/2 - (3/2)(3/2) = 3/2 - 9/4 = (6-9)/2 = -3/4 ρ(X,Y) = (-3/4) / √(3/4 * 3/4) = 1 (crossed out, corrected to -1) 4. [10 Points] Let X be a binomial random variable with parameter (n, p). (a). Show that the expectation E[X] = np. (b). Show that the variance Var[X] = np(1-p). Let X_i be binomial random variable with parameter (1, p) E[X_i] = 1 * p = p Var[X_i] = E[X_i²] - (E[X_i])² = 1² * p(X_i=1) - p² = p - p² = p(1-p) Since X₁, ..., X_n are identical. E[X] = E[X₁] + ... + E[X_n] = p + ... + p = np Var[X] = Var[X₁] + ... + Var[X_n] = p(1-p) + ... + p(1-p) = np(1-p) 5. [10 Points] Let X₁, ..., X₉ be independent Poisson random variables with expectation 1. Let f(x) = X₁ + ... + X₉. (a). Using Markov’s inequality, find an upper bound of P(X > 15). (b). Use the table of the normal distribution on the final page, find an approximation of P(X > 15). E[X] = 9 (a) P(X > 15) ≤ 9/15 = 3/5 (b) μ = 9, σ² = 9, σ = 3 Y = (X-9)/3 X = 3Y + 9 P(X > 15) = P(3Y + 9 > 15) = P(Y > 2) = 1 - P(Y < 2) = 1 - 0.9772 = 0.0228