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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standard deviation = square root of variance = sqrt(846) = 29.086 4. If we have the following data 34‚ 38‚ 22‚ 21‚ 29‚ 37‚ 40‚ 41‚ 22‚ 20‚ 49‚ 47‚ 20‚ 31‚ 34‚ 66 Draw a stem and leaf. Discuss the shape of the distribution. Solution: 2 3 4 5 6 | | | | | 219200 48714 0197 6 This distribution is right skewed (positively skewed) because the “tail” extends to the right. 5. What type of relationship is shown by this scatter plot? 45 40 35 30 25 20 15 10 5 0 0 5 10 15 20 Solution: Weak positive linear
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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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The Poisson distribution is a discrete distribution. It is often used as a model for the number of events (such as the number of telephone calls at a business‚ number of customers in waiting lines‚ number of defects in a given surface area‚ airplane arrivals‚ or the number of accidents at an intersection) in a specific time period. It is also useful in ecological studies‚ e.g.‚ to model the number of prairie dogs found in a square mile of prairie. The major difference between Poisson and Binomial
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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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expected average outcome over many observations.The common symbol for the mean (also known as the expected value of X) is ‚ formally defined by Variance - The variance of a discrete random variable X measures the spread‚ or variability‚ of the distribution‚ and is defined by The standard deviation is the square root of the variance. Expectation - The expected value (or mean) of X‚ where X is a discrete random variable‚ is a weighted average of the possible values that X can take‚ each value
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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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