{[ promptMessage ]}

Bookmark it

{[ promptMessage ]}

Sol V - ENGM2032 Applied Probability and Statistics Winter...

Info icon This preview shows pages 1–2. Sign up to view the full content.

View Full Document Right Arrow Icon
ENGM2032, Applied Probability and Statistics Winter 2010 Assignment #5: Solutions 1) Let N n be the number of defective rods in a lot of n. Assuming that the rods are independent with a constant probability of being substandard, Bernoulli trials can be used, each rod being a trial, a substandard rod being a ‘success’, and a rod that falls within specs being a ‘failure’, p = 0 . 2 and q = 0 . 8 (note that the ‘success’ here is the exact opposite of what we would normally call a success in the situation in question). Then, N n follows a binomial distribution. a) The expected value (mean) of N 100 is then μ = E N 100 = np = 100(0 . 2) = 20 rods b) Here we are looking for the probability of exactly one ‘success’ in 12 trials, or P N 12 = 1 = 12 1 p 1 q 11 = 12(0 . 2)(0 . 8) 11 = 0 . 2062 . c) P N 10 2 = 1 P N 10 = 0 P N 10 = 1 = 1 10 0 (0 . 2) 0 (0 . 8) 10 10 1 (0 . 2) 1 (0 . 8) 9 = 0 . 6242 . d) For the next two parts, we are counting the number of trials necessary to reach the first success, T 1 . Under the assumptions made, T 1 follows a geometric distribution; hence E T 1 = 1 /p = 5 . (quite naturally, if, on average, you have 2 defective rods in 10, then you expect to test 5 rods, on average, before detecting the first defective one). e) Here we are looking for the probability that we need 5 trials to achieve the first success, P T 1 = 5 = q 4 p = (0 . 8) 4 (0 . 2) = 0 . 08192 . f) Now we are looking for the probability that the 3rd ‘success’ occurs on the 11th trial. This is given to us by the negative binomial distribution, P T 3 = 11 = 10 2 (0 . 2) 3 (0 . 8) 8 = 0 . 0604 . 2) If we assume that unlabelled apples occur independently with constant probability of occur- rence, then apples can be viewed as a sequence of Bernoulli trials with probability of ‘success’ (being unlabelled) of p = 1 / 20 = 0 . 05 and probability of ‘failure’ of q = 0 . 95.
Image of page 1

Info icon This preview has intentionally blurred sections. Sign up to view the full version.

View Full Document Right Arrow Icon
Image of page 2
This is the end of the preview. Sign up to access the rest of the document.

{[ snackBarMessage ]}

What students are saying

  • Left Quote Icon

    As a current student on this bumpy collegiate pathway, I stumbled upon Course Hero, where I can find study resources for nearly all my courses, get online help from tutors 24/7, and even share my old projects, papers, and lecture notes with other students.

    Student Picture

    Kiran Temple University Fox School of Business ‘17, Course Hero Intern

  • Left Quote Icon

    I cannot even describe how much Course Hero helped me this summer. It’s truly become something I can always rely on and help me. In the end, I was not only able to survive summer classes, but I was able to thrive thanks to Course Hero.

    Student Picture

    Dana University of Pennsylvania ‘17, Course Hero Intern

  • Left Quote Icon

    The ability to access any university’s resources through Course Hero proved invaluable in my case. I was behind on Tulane coursework and actually used UCLA’s materials to help me move forward and get everything together on time.

    Student Picture

    Jill Tulane University ‘16, Course Hero Intern