Math 340 Homework for Sections 4.1 and 4.2 _____________________________________________________________________ 1. (#3, page 275) a. First we need to determine c so that the total area under the graph of the density function is 1. So, 1 = integral_{0}^{1} x (1 - x) dx => c = 6. b. P( X \leq 1/2) = integral_{0}^{1/2} 6 x (1 - x) dx = 1/2 c. P( X \leq 1/3) = integral_{0}^{1/3} 6 x (1 - x) dx = 7/27 d. P(1/3 \leq X \leq 1/2) = 1/2 - 7/27 e. E(X) = integral_{0}^{1} x f(x) dx E(X^2) = integral_{0}^{1} x^2 f(x) dx var(X) = E(X^2) - E(X)^2 __________________________________________________ 2. (#9, page 275) P(S_4 \geq 3) = P(Z > z) = P(Z > 1.73) = 1 - .9582 = 4.18% where z = (3 - E(S_4)) / sigma(S_4) = (3 - 2) / (1 / sqrt(3)) = sqrt(3) = 1.73 Note that E(S_4) = 4 E(X_1) = (4) (1/2) sigma(S_4) = sqrt(4) sigma(X_1) = (2)(1/sqrt(12)) = 1/sqrt(3) __________________________________________________ 3. (#13, page 275) First we need to find the height of the triangle so that the total area under it is 1. Call the height of the triangle c. Then 1 = (1/2)(c)(0.1)(2) => c = 10. a. proportion of the output scrapped = 2 P(X \leq 0.925) = (2) (1/2) (0.025) (10/4) = (0.25) / 4 = 1/16 = 6.25% b. Let L be the lenght of a randomly selected rod. Then, P(.95 \leq L \leq 1.05) = 1 - (2) (1/2)(5)(.05) = 75%. In other words there is 75% chance that a randomly selected rod satisfies the required length criterion. We would like to choose n so that the no. of rods that satisfy the length criterion in a collection of n rods be at least 100. So, the hueristic way to do this is to say (75%) (n) \geq 100 => n \geq 134. However, there is a problem with the answer in the book when we consider the probability for assurance, 95%. I will need to resolve the problem and update this solution later. __________________________________________________ 4. (#1, page 293) First of all we are told that the half-life tau (Greek letter tau) is 1 year. This means that 1/2 = P(T > tau) = e^(- lambda tau) = e^(-lambda) Therefore, decay rate lambda = ln 2. a. P(T > 5) = e^(-5 lambda) b. 10% = P(T > t) = e^(-lambda t) => t = (ln 10) / (ln 2) c. 1/1024 = 2^(-10) = P(T > t) = e^(-lambda t) => t = (ln 1024) / ln 2 = 10. d. The simplest way to get the answer to this problem is the following approach. P(no survivors after 10 yrs) = P(1024 decays within 10 yrs) = [P(T < 10)]^(1024), assumin decays are indep. = [1 - P(T > 10)]^(1024) = [1 - e^(-10 ln 2)]^(1024) = (1 - 1/1024)^(1024) = 0.3677 (exact result) Note that since lim_{n->\infty} (1 + 1/n)^n = e, we could have made the estimate: (1 - 1/1024)^1024 ~ e^(-1) = 0.3679 (the answer in the book). __________________________________________________ 5. (#5, page 293) We are given that lambda = 1. Below W_i will stand for the i-th arrival time (recall that waiting times have exponential distribution with parameter lambda, namely, P(W_i > t) = e^(-lambda t)). a. P(W_4 \leq 2) = 1 - e^(-2). b. P(T_4 < 5) = 1 - P(T_4 \geq 5) = 1 - P( N(0, 5] \leq 3 ) = 1 - e^(-lambda t) sum_{0}^{3} (lambda t)^k / k! where lambda = 1 and t = 5. c. E(T_4) = r / lambda = 4 / 1. __________________________________________________ 6. (#11, page 293) a. If we can establish that T has the memoryless property, we have showed that it has exponential distribution with parameter lambda. So, take the time interval (0, t] and suppose that it is n microseconds. Namely, 0 < t_1 < t_2 < ... < t_n = t where t_k - t_{k-1} = 10^(-6). According to the given information we have that: P(T > t_n | T > t_{n-1}) = lambda. Hence, P(T > t) = P(T > t_n | T > t_{n-1}) P(T > t_{n-1}) = P(T > t_n | T>t_{n-1})P(T>t_{n-1} | T>t_{n-2})P(T>t_{n-2}) = product_{k=1}^{k=n} P(T > t_{k} | T > t_{k-1}) = lambda^n Thus, we see that the P(T > t) is independent of t which proves the result. b. P(1 < T < 2) = e^(-a lambda) - e^(-b lambda) = e^(-lambda) - e^(-2 lambda) __________________________________________________ 7. (#13, page 293) Let T_i be the operating timeuntil failure of the i-th component. Then, we are given that T_i's have exponential dist with parameter lambda, and n = the no. of components = 10000. a. 20 = [E(T_1) + ... + E(T_n)] / n = n E(T_1) / n = E(T_1) = 1 / lambda => lambda = 1 / 20 = 5%. b. N_d = no. of components, among n, which survive more than d days => consider n trials and success = the event that a component survives more than d days. Then, prob of success = p = P(T > 10) = e^(-10 lambda) Hence, we have: E(N_d) = n p SD(N_d) = sqrt(n p q) where q = 1 - p. __________________________________________________ 8. (#15, page 293) T_tot = T_1 + ... + T_4 => E(T_tot) = 4 E(T_1) = 4/lambda = (4)(20) = 80 => SD(T_tot) = sqrt(4) SD(T_1) = (2)(20) = 40 => P(T_tot \geq 60) = P(N(I) \leq 3) , where I = (0, 60] = sum_{k=0}^{3} P(N(I) = k) , where N(I) has Poisson dist = e^(-lambda t)[1+(lambda t)+(lambda t)^2/2 +(lambda t)^3/6] where lambda = 5% and t = L(I) = 60. __________________________________________________