--- title: "Trip time normal distribution probabilities" subject: "Transportation Engineering II" chapter: 1 key: "semester_7/transportation_engineering_ii:trip-time-normal-distribution-probability" source: https://www.bctengineer.com/q/semester_7/transportation_engineering_ii/trip-time-normal-distribution-probability site: BCT Engineering --- # Trip time normal distribution probabilities Transportation Engineering II · Chapter 1: Traffic Engineering - Asked in: 2071 Chaitra, 8 marks - Video: [#Normaldistribution, #statistics, Normal Distributions and its applications in Traffic and Highways](https://www.youtube.com/watch?v=JG3-Ke2jZto) — Transportation Engg Talks | Satish IITR - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/transportation_engineering_ii/trip-time-normal-distribution-probability - Whole chapter: https://www.bctengineer.com/md/semester_7/transportation_engineering_ii/chapter-1 ## Question Average trip time for office is 30 minutes with standard deviation of 5 min. Assuming normal distribution of trip time, calculate the followings: a) Probability of trip time being at least 35 minutes b) If the working hour starts at 10:00 AM and trip starts at 9:40 AM what is the probability of being late? ## Short answer Convert to standard normal z-scores. (a) For at least 35 min, z = 1.00, probability = 0.1587 (15.9%). (b) With only 20 min available, z = –2.00, probability of being late = 0.9772 (97.7%). ## Answer Trip time $T$ is normally distributed with mean $\mu = 30$ min and standard deviation $\sigma = 5$ min. Use the standard normal variable $z = (T - \mu)/\sigma$. #### (a) Probability that the trip time is at least 35 minutes $$ z = \frac{35 - 30}{5} = 1.00,\quad P(T \ge 35) = 1 - \Phi(1.00) = 1 - 0.8413 = 0.1587 $$ **Answer (a): 0.1587 (about 15.9 %).** #### (b) Probability of being late Work starts at 10:00 AM and the trip starts at 9:40 AM, so only 20 minutes are available. The person is late if the trip takes more than 20 minutes: $$ z = \frac{20 - 30}{5} = -2.00,\quad P(T > 20) = 1 - \Phi(-2.00) = \Phi(2.00) = 0.9772 $$ **Answer (b): 0.9772 (about 97.7 %).** Leaving at 9:40 makes the person late almost every day. To be on time with 95 percent confidence, the trip must allow $30 + 1.645 \times 5 = 38.2$ min, so it must start by about 9:21 AM (10:00 minus 38.2 min). ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For mean 30 min and standard deviation 5 min, what is z for a trip time of 35 min? A. 1.00 B. 7.00 C. 0.20 D. -1.00 Answer: A. z = (T - mu)/sigma = (35 - 30)/5 = 1.00. (Key point: Standardising the variable) 2. What is P(T >= 35 min)? A. 0.8413 B. 0.1587 C. 0.5000 D. 0.0228 Answer: B. P = 1 - Phi(1.00) = 1 - 0.8413 = 0.1587, about 15.9 %. (Key point: Upper-tail probability) 3. The trip starts at 9:40 AM and work starts at 10:00 AM. When is the person late? A. When the trip takes more than 30 min B. When the trip takes more than 20 min C. When the trip takes less than 20 min D. When the trip takes more than 40 min Answer: B. Only 20 minutes are available, so a trip longer than 20 min makes the person late. (Key point: Time available defines lateness) 4. Why is the probability of being late as high as 0.9772? A. P(T>20) = 1 - Phi(2) = 0.0228 B. P(T>20) = Phi(-2) = 0.0228 C. z = -2, so P(T>20) = Phi(2) = 0.9772, as 20 min is far below the mean D. 20 min is below the mean, so the probability is exactly 0.5 Answer: C. z = (20 - 30)/5 = -2, and the area to the right of z = -2 equals Phi(2) = 0.9772. 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