University:
York University
Course:
MATH 1131 | Introduction to Statistics I
Academic year:

2022
Views:

218

Pages:

9
Author:

Niklas

Exercise 1: A contractor is interested in the total cost of a project on which she intends to bid. She estimates that materials will cost 25000$ and that labor will cost 900$ per day. If a project takes X days to complete, the total cost of labor will be 900X dollars and the total cost of the project (in dollars) will be C = 25000 + 9000X You are given the following probability distribution of likely completion times for the project: Completion time X (days) Probability 10 0.1 11 0.3 12 0.3 13 0.2 14 0.1 Find the mean and the standard deviation for total cost C. Calculations: F(C) = E(25000 + 900X) = 25000 + 900E(X) E(X) = Σx_i * p(x_i) = (10 * 0.1) + (11 * 0.3) + (12 * 0.3) + (13 * 0.2) + (14 * 0.1) = 11.9 E(C) = 25000 + 900 * 11.9 = 35410$ σ^2 = Var(C) = Var(25000 + 900X) = 0 + (900)^2 * Var(X) = 810000 * Var(X) Var(X) = Σ(x_i - E(X))^2 * p_i = 4.29 Var(C) = 810000 * 4.29 = 3474900$ σ(C) = √Var(C) = 1032.1$ Exercise 2: A discrete random variable X has probability distribution given by: P(X = x) = k(x+2) where X = 0, 1, 2, 3, 4, 5. a. Find the value of the constant k. b. Compute mean, variance, and standard deviation of X. Calculations: ∑ P(X=x) = 1 => P(X=0) + P(X=1) + P(X=2) + P(X=3) + P(X=4) + P(X=5) = 1 => 27k = 1 => k = 1/27 ≈ 0.037 x f(x) = P(X=x) x^2 * f(x) 0 2k 0 1 3k 3k 2 4k 8k 3 5k 15k 4 6k 24k 5 7k 35k E(X) = Σx_i * f(x_i) = 85k = 3.15 Var(X) = Σx_i^2 * f(x_i) - (Σx_i * f(x_i))^2 = 12.395 - 9.99 = 2.475 σ^2 = Var(X) = 2.475 σ = √Var(X) = 4.57 (standard deviation) Exercise 3: Calls arrive at a call center during a certain period of the day at a rate of 300 per hour. Assuming that the calls arrive at random, determine the probability that: a. Exactly one call arrives during a one-minute period. b. No calls arrive in an interval of 1 minute. c. Compute mean and standard deviation of the number of calls per minute. Event: # calls per minute a. # calls in 1 minute = 1: If an event occurs at random at a constant average rate λ per unit/space, then the probability of r occurrences of the event in the unit interval can be obtained from the probability function of the Poisson distribution: P(x = r) = (λ^r / r!) * e^(-λ) X = calls in 1 minute period X ~ Poisson distribution with rate λ = 300/60 = 5 per minute P(X = 1) = (5^1 / 1!) * e^(-5) = 5/e^5 ≈ 0.034 b. # calls in t minutes = 0: Y ~ Poisson with rate λ = dt P(Y = 0) = e^(-dt) c. X ~ Poisson(5): E(X) = Var(X) for Poisson distribution E(X) = Var(X) = λ = 5 per minute Exercise 4: A store manager estimates that the probability that a customer will make a purchase is 0.4. Of the next 3 customers who enter the store, what are the respective probabilities that 0, 1, 2, and 3 of them will make a purchase? What is the mean and standard deviation of the number of customers out of 3 making a purchase? P = 0.4 X = {0, 1, 2, 3} # Customers n = 3 We only have 2 possible outcomes: A customer will make a purchase A customer will make no purchase X ~ Bin(3, 0.4) P(X = x) = (nCx) * p^x * (1-p)^(n-x) P(X = 0) = (3C0) * (0.4)^0 * (0.6)^3 = 0.216 P(X = 1) = (3C1) * (0.4)^1 * (0.6)^2 = 0.432 P(X = 2) = (3C2) * (0.4)^2 * (0.6)^1 = 0.288 P(X = 3) = (3C3) * (0.4)^3 * (0.6)^0 = 0.064 E(X) = n * p = 3 * 0.4 = 1.2 V(X) = σ^2 = n * p * (1-p) = 3 * 0.4 * 0.6 = 0.72 σ = √V(X) = √0.72 ≈ 0.848