CEE 202 HW 5 SOLUTION

# 00095 x 00545 x 45 0 a mall is planning for snow

This preview shows page 1. Sign up to view the full content.

This is the end of the preview. Sign up to access the rest of the document.

Unformatted text preview: rm is 1 = (Σ x2 P(x))  ­ 2µ2 + µ2 = (Σ x2 P(x))  ­ µ2 QED 3) A European train arrives at a station every 30 minutes. American tourists have not checked the schedule and each person’s waiting time is a random variable with a continuous uniform distribution. Let’s get the distribution first 4/7 CEE 202 – Engineering Risk &amp; Uncertainty Spring 2011 (Bond) Homework 5 Expected Values, More Distributions a) What is the probability that a tourist waits more than 20 minutes? The tourist won’t wait more than 30 minutes (because the pdf is 0 after that point), so we want P(20≤X≤30). Both of these values are within the range 0 ­30. d − c 30 − 20 =0.333 P (c ≤ X ≤ d ) = = B − A 30 − 0 b) What is the probability that two tourists not traveling together each wait between 4 and 10 minutes? 10 − 4 P ( 4 ≤ X ≤ 10) = = 0.2 30 =0.333 If two tourists are not traveling together, it means that they are statistically independent, the probability that two tourists each wait between 4 and 10 minutes is € € 4) In snowier Chicago, the probability density function of annual snowfall, in inches, is # x &lt; 10 0, % 10 &quot; x &lt; 45 f ( x ) = \$ &quot;0.00095 x + 0.0545, % x # 45 0, &amp; A mall is planning for snow removal. They expect to spend \$400 per inch plus a flat fee of \$1000 per year. a) What are the expected value and standard deviation of annual !...
View Full Document

Ask a homework question - tutors are online