CHAPTER 8. TRIP DISTRIBUTION NPTEL May 3‚ 2007 Chapter 8 Trip distribution 8.1 Overview The decision to travel for a given purpose is called trip generation. These generated trips from each zone is then distributed to all other zones based on the choice of destination. This is called trip distribution which forms the second stage of travel demand modeling. There are a number of methods to distribute trips among destinations; and two such methods are growth factor model and gravity
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goal of the distribution of income is to achieve economic equality‚ which is to give every citizen the opportunity of earning a decent living. However‚ our current system’s inability to better allocate the resources we have at our disposal has widened the gap between the wealthy and the poor especially during the past 20 years. The primary benefit of the distribution of income is to transfer wealth‚ with the help of the government‚ to those who are less fortunate. The current distribution of income
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Normal Distribution:- A continuous random variable X is a normal distribution with the parameters mean and variance then the probability function can be written as f(x) = - < x < ‚ - < μ < ‚ σ > 0. When σ2 = 1‚ μ = 0 is called as standard normal. Normal distribution problems and solutions – Formulas: X < μ = 0.5 – Z X > μ = 0.5 + Z X = μ = 0.5 where‚ μ = mean σ = standard deviation X = normal random variable Normal Distribution Problems and Solutions – Example
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25‚ 28 26‚ 28‚ 26‚ 28‚ 31‚ 30‚ 26‚ 26 the information is to be organized into a frequency distribution. A. How many classes would you recommend? b. What class interval would you suggest? C .what lower limit would you recommend for the first class? d. organize the information into a frequency distribution and determine the relative frequency distribution. e. comment on the shape of the distribution. 15. Molly’s Candle Shop has several retail stores in the coastal areas of North and South
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solve k = 20.275 d) P ( 17 < X < 21) P ( (17 -18)/2.5 < Z < ( 21-18)/2.5) P ( -0.4 < Z < 1.2) = 0.8849 – 0.3446 = 0.5403 ( 4 decimal places) 4. In a sample of 25 observations from a Normal Distribution with mean 98.6 and standard deviation 17.2‚ find: Ans: a) n = 25‚ [pic] = ( = 98.6‚ [pic] = /n = 17.2/(25 = 3.44 [pic]( N
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R) (b) Find P(V < 200) (c) Find P(150≦V≦300) (d) Find the mean of V (e) Find the median of V (f) Find the standard deviation of V Question 3: [20 points] The number of traffic accidents at a certain intersection is thought to be well modeled by a Poisson process with a mean of 3 accidents per year. (a) Find the mean waiting time between accidents. (b) Find the standard deviation of the waiting times between accidents. (c) Find the probability that more than one year elapses between accidents. (d)
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Mathematics): Survival distributions Age-at-death random variable T0 – age-at-death (lifetime for newborn) random variable To completely determine the distribution of T0 ‚ we may use (for t ≥ 0)‚ (1) (cumulative) distribution function: F0 (t) = Pr(T0 ≤ t) (2) survival function: s0 (t) = 1 − F0 (t) = Pr(T0 > t) (3) probability density function: f0 (t) = F0 (t) = (4) force of mortality: µ0 (t) = d F0 (t) dt f0 (t) −s0 (t) = 1 − F0 (t) s0 (t) Requirements: (1) For distribution function‚
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A population of measurements is approximately normally distributed with mean of 25 and a variance of 9. Find the probability that a measurement selected at random will be between 19 and 31. Solution: The values 19 and 31 must be transformed into the corresponding z values and then the area between the two z values found. Using the transformation formula from X to z (where µ = 25 and σ √9 = 3)‚ we have z19 = (19 – 25) / 3 = -2 and z31 = (31 - 25) / 3 = +2 From the area between z =±2 is 2(0
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Chapter 13: Chi-Square Applications SHORT ANSWER 1. When samples of size n are drawn from a normal population‚ the chi-square distribution is the sampling distribution of = ____________________‚ where s2 and are the sample and population variances‚ respectively. ANS: PTS: 1 OBJ: Section 13.2 2. Find the chi-square value for each of the right-tail areas below‚ given that the degrees of freedom are 7: A) 0.95 ____________________ B) 0.01 ____________________ C) 0.025 ____________________
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Probability distribution Definition with example: The total set of all the probabilities of a random variable to attain all the possible values. Let me give an example. We toss a coin 3 times and try to find what the probability of obtaining head is? Here the event of getting head is known as the random variable. Now what are the possible values of the random variable‚ i.e. what is the possible number of times that head might occur? It is 0 (head never occurs)‚ 1 (head occurs once out of 2 tosses)
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