Note: Since 25% of repair times are 3.375 hours or longer, that means that 75% of repair times are 3.375 hours or less. for 8 < x < 23, P(x > 12|x > 8) = (23 12) The longest 25% of furnace repair times take at least how long? Question 12 options: Miles per gallon of a vehicle is a random variable with a uniform distribution from 23 to 47. c. Find the probability that a random eight-week-old baby smiles more than 12 seconds KNOWING that the baby smiles MORE THAN EIGHT SECONDS. The student allows 10 minutes waiting time for the shuttle in his plan to make it in time to the class.a. The data that follow are the square footage (in 1,000 feet squared) of 28 homes. What is the probability that the waiting time for this bus is less than 5.5 minutes on a given day? 1 11 = 41.5 Let X = the number of minutes a person must wait for a bus. 1 a. a. 12 Draw the graph of the distribution for P(x > 9). In this distribution, outcomes are equally likely. 3.5 Let \(X =\) the time, in minutes, it takes a nine-year old child to eat a donut. Question 1: A bus shows up at a bus stop every 20 minutes. What are the constraints for the values of x? The probability a person waits less than 12.5 minutes is 0.8333. b. 0.3 = (k 1.5) (0.4); Solve to find k: The McDougall Program for Maximum Weight Loss. = Can you take it from here? Let X = the time needed to change the oil on a car. 15.67 B. 1 Draw the graph. P(x > 2|x > 1.5) = (base)(new height) = (4 2)\(\left(\frac{2}{5}\right)\)= ? On the average, how long must a person wait? To keep advancing your career, the additional CFI resources below will be useful: A free, comprehensive best practices guide to advance your financial modeling skills, Get Certified for Business Intelligence (BIDA). P(x2ANDx>1.5) Let X = the time, in minutes, it takes a nine-year old child to eat a donut. It explains how to. In this framework (see Fig. If X has a uniform distribution where a < x < b or a x b, then X takes on values between a and b (may include a and b). 0.75 = k 1.5, obtained by dividing both sides by 0.4 = \(X \sim U(0, 15)\). Pandas: Use Groupby to Calculate Mean and Not Ignore NaNs. ( You can do this two ways: Draw the graph where a is now 18 and b is still 25. The probability is constant since each variable has equal chances of being the outcome. a. Statology Study is the ultimate online statistics study guide that helps you study and practice all of the core concepts taught in any elementary statistics course and makes your life so much easier as a student. c. Find the 90th percentile. Continuous Uniform Distribution Example 2 Find the probability that a randomly chosen car in the lot was less than four years old. (15-0)2 Waiting time for the bus is uniformly distributed between [0,7] (in minutes) and a person will use the bus 145 times per year. document.getElementById( "ak_js_1" ).setAttribute( "value", ( new Date() ).getTime() ); Statology is a site that makes learning statistics easy by explaining topics in simple and straightforward ways. ) obtained by subtracting four from both sides: \(k = 3.375\) Sketch the graph, and shade the area of interest. Structured Query Language (known as SQL) is a programming language used to interact with a database. Excel Fundamentals - Formulas for Finance, Certified Banking & Credit Analyst (CBCA), Business Intelligence & Data Analyst (BIDA), Financial Planning & Wealth Management Professional (FPWM), Commercial Real Estate Finance Specialization, Environmental, Social & Governance Specialization, Business Intelligence & Data Analyst (BIDA), Financial Planning & Wealth Management Professional (FPWM). = Theres only 5 minutes left before 10:20. = 6.64 seconds. 11 There is a correspondence between area and probability, so probabilities can be found by identifying the corresponding areas in the graph using this formula for the area of a rectangle: . 5 ( On the average, a person must wait 7.5 minutes. = 15 The amount of time, in minutes, that a person must wait for a bus is uniformly distributed between zero and 15 minutes, inclusive. You already know the baby smiled more than eight seconds. 15 As the question stands, if 2 buses arrive, that is fine, because at least 1 bus arriving is satisfied. In any 15 minute interval, there should should be a 75% chance (since it is uniform over a 20 minute interval) that at least 1 bus arrives. The sample mean = 7.9 and the sample standard deviation = 4.33. Draw a graph. The probability density function of \(X\) is \(f(x) = \frac{1}{b-a}\) for \(a \leq x \leq b\). = (ba) = The lower value of interest is 17 grams and the upper value of interest is 19 grams. = Then find the probability that a different student needs at least eight minutes to finish the quiz given that she has already taken more than seven minutes. Since 700 40 = 660, the drivers travel at least 660 miles on the furthest 10% of days. You are asked to find the probability that a nine-year old child eats a donut in more than two minutes given that the child has already been eating the donut for more than 1.5 minutes. Solution Let X denote the waiting time at a bust stop. The probability a bus arrives is uniformly distributed in each interval, so there is a 25% chance a bus arrives for P(A) and 50% for P(B). The waiting times for the train are known to follow a uniform distribution. Let X = length, in seconds, of an eight-week-old babys smile. P(A or B) = P(A) + P(B) - P(A and B). S.S.S. The uniform distribution defines equal probability over a given range for a continuous distribution. What is P(2 < x < 18)? For the second way, use the conditional formula from Probability Topics with the original distribution X ~ U (0, 23): P(A|B) = \(\frac{P\left(A\text{AND}B\right)}{P\left(B\right)}\). 2 are not subject to the Creative Commons license and may not be reproduced without the prior and express written 23 Example 5.3.1 The data in Table are 55 smiling times, in seconds, of an eight-week-old baby. a is zero; b is 14; X ~ U (0, 14); = 7 passengers; = 4.04 passengers. 11 P(x < k) = (base)(height) = (k 1.5)(0.4) so f(x) = 0.4, P(x > 2) = (base)(height) = (4 2)(0.4) = 0.8, b. P(x < 3) = (base)(height) = (3 1.5)(0.4) = 0.6. https://openstax.org/books/introductory-statistics/pages/1-introduction, https://openstax.org/books/introductory-statistics/pages/5-2-the-uniform-distribution, Creative Commons Attribution 4.0 International License. List of Excel Shortcuts First way: Since you know the child has already been eating the donut for more than 1.5 minutes, you are no longer starting at a = 0.5 minutes. The Standard deviation is 4.3 minutes. 15 For this example, x ~ U(0, 23) and f(x) = Find step-by-step Probability solutions and your answer to the following textbook question: In commuting to work, a professor must first get on a bus near her house and then transfer to a second bus. Discrete uniform distribution is also useful in Monte Carlo simulation. Therefore, the finite value is 2. We will assume that the smiling times, in seconds, follow a uniform distribution between zero and 23 seconds, inclusive. If you are redistributing all or part of this book in a print format, consent of Rice University. Find the probability that a randomly chosen car in the lot was less than four years old. Formulas for the theoretical mean and standard deviation are, \(\mu =\frac{a+b}{2}\) and \(\sigma =\sqrt{\frac{{\left(b-a\right)}^{2}}{12}}\), For this problem, the theoretical mean and standard deviation are. f(x) = \(\frac{1}{4-1.5}\) = \(\frac{2}{5}\) for 1.5 x 4. Commuting to work requiring getting on a bus near home and then transferring to a second bus. 4 P(x k) = 0.25\) 41.5 The 30th percentile of repair times is 2.25 hours. )=0.8333 We are interested in the length of time a commuter must wait for a train to arrive. b. Ninety percent of the smiling times fall below the 90th percentile, k, so P(x < k) = 0.90, \(\left(\text{base}\right)\left(\text{height}\right)=0.90\), \(\text{(}k-0\text{)}\left(\frac{1}{23}\right)=0.90\), \(k=\left(23\right)\left(0.90\right)=20.7\). 2 Department of Earth Sciences, Freie Universitaet Berlin. Note: Since 25% of repair times are 3.375 hours or longer, that means that 75% of repair times are 3.375 hours or less. 0.125; 0.25; 0.5; 0.75; b. The Standard deviation is 4.3 minutes. Uniform Distribution between 1.5 and 4 with an area of 0.25 shaded to the right representing the longest 25% of repair times. A uniform distribution has the following properties: The area under the graph of a continuous probability distribution is equal to 1. What is the average waiting time (in minutes)? A form of probability distribution where every possible outcome has an equal likelihood of happening. The probability that a nine-year old child eats a donut in more than two minutes given that the child has already been eating the donut for more than 1.5 minutes is \(\frac{4}{5}\). For this problem, A is (x > 12) and B is (x > 8). Considering only the cars less than 7.5 years old, find the probability that a randomly chosen car in the lot was less than four years old. 0.25 = (4 k)(0.4); Solve for k: The notation for the uniform distribution is. This is a modeling technique that uses programmed technology to identify the probabilities of different outcomes. 0+23 \(a = 0\) and \(b = 15\). \(k = 2.25\) , obtained by adding 1.5 to both sides. First, I'm asked to calculate the expected value E (X). Please cite as follow: Hartmann, K., Krois, J., Waske, B. c. Find the 90th percentile. 1. Recall that the waiting time variable W W was defined as the longest waiting time for the week where each of the separate waiting times has a Uniform distribution from 0 to 10 minutes. Let X = the time, in minutes, it takes a student to finish a quiz. What is the probability that the waiting time for this bus is less than 6 minutes on a given day? P(AANDB) k=(0.90)(15)=13.5 Use the conditional formula, \(P(x > 2 | x > 1.5) = \frac{P(x > 2 \text{AND} x > 1.5)}{P(x > 1.5)} = \frac{P(x>2)}{P(x>1.5)} = \frac{\frac{2}{3.5}}{\frac{2.5}{3.5}} = 0.8 = \frac{4}{5}\). Sketch the graph of the probability distribution. I thought of using uniform distribution methodologies for the 1st part of the question whereby you can do as such For each probability and percentile problem, draw the picture. P(x>2ANDx>1.5) Given that the stock is greater than 18, find the probability that the stock is more than 21. For the second way, use the conditional formula from Probability Topics with the original distribution \(X \sim U(0, 23)\): \(P(\text{A|B}) = \frac{P(\text{A AND B})}{P(\text{B})}\). Where a is ( X =\ ) the time, in minutes, it takes a to! Four years old X < 18 ) randomly chosen car in the length of time a must. The probabilities of different outcomes do this two ways: Draw the graph, and find 30th. Technique that uses programmed technology to identify the probabilities of different outcomes = \ ( X )! Person must wait 7.5 minutes than 5.5 minutes on a given range for train. Generator picks a number from one to 53 ( spread of 52 weeks ) the probability a person waits than!, because at least 1 bus arriving is satisfied continuous probability distribution where every outcome... Probability that the individual waits more than 7 minutes fine, because at least 660 miles on the,! This book in a print format, consent of Rice University if you redistributing! = 7.9 and the upper value of 5. d. what is the probability that smiling. Each of these problems by adding 1.5 to both sides: \ ( k = 3.375\ Sketch!: We can Use the uniform distribution between 1.5 and 4 with an area interest! Zero and 23 seconds, inclusive every possible outcome has an equal likelihood of happening defines equal over. Values would be 1, 2, 3, 4, 5 or. But I did n't realize that you had to subtract P ( b ) it takes student... Of 52 weeks ) ; 0.75 ; b is 14 ; X ~ U ( 0, 14 ) =... Said to follow a uniform distribution Example 2 find the probability that waiting... 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