3.375 hours is the 75th percentile of furnace repair times. You are asked to find the probability that an eight-week-old baby smiles more than 12 seconds when you already know the baby has smiled for more than eight seconds. 2 The notation for the uniform distribution is. What has changed in the previous two problems that made the solutions different. )=0.90, k=( Answer: (Round to two decimal places.) Find the 90th percentile. e. \(\mu = \frac{a+b}{2}\) and \(\sigma = \sqrt{\frac{(b-a)^{2}}{12}}\), \(\mu = \frac{1.5+4}{2} = 2.75\) hours and \(\sigma = \sqrt{\frac{(4-1.5)^{2}}{12}} = 0.7217\) hours. \(P(2 < x < 18) = 0.8\); 90th percentile \(= 18\). Write the distribution in proper notation, and calculate the theoretical mean and standard deviation. Let k = the 90th percentile. \(0.3 = (k 1.5) (0.4)\); Solve to find \(k\): Find the probability that the time is between 30 and 40 minutes. = \(\frac{15\text{}+\text{}0}{2}\) P(x>1.5) Ninety percent of the time, a person must wait at most 13.5 minutes. b. Monte Carlo simulation is often used to forecast scenarios and help in the identification of risks. Plume, 1995. 15 The longest 25% of furnace repairs take at least 3.375 hours (3.375 hours or longer). = obtained by subtracting four from both sides: k = 3.375. Example 5.3.1 The data in Table are 55 smiling times, in seconds, of an eight-week-old baby. All values \(x\) are equally likely. 0.3 = (k 1.5) (0.4); Solve to find k: 15 . = A deck of cards also has a uniform distribution. 2.75 The graph of the rectangle showing the entire distribution would remain the same. 15 This distribution is closed under scaling and exponentiation, and has reflection symmetry property . Press question mark to learn the rest of the keyboard shortcuts. 12 Uniform Distribution between 1.5 and 4 with an area of 0.25 shaded to the right representing the longest 25% of repair times. Let X = the time, in minutes, it takes a student to finish a quiz. The data follow a uniform distribution where all values between and including zero and 14 are equally likely. 11 It is defined by two parameters, x and y, where x = minimum value and y = maximum value. b. When working out problems that have a uniform distribution, be careful to note if the data is inclusive or exclusive. , it is denoted by U (x, y) where x and y are the . Sketch and label a graph of the distribution. The graph illustrates the new sample space. b. The data follow a uniform distribution where all values between and including zero and 14 are equally likely. a. Let X = the time needed to change the oil on a car. The uniform distribution is a continuous probability distribution and is concerned with events that are equally likely to occur. 23 Thus, the value is 25 2.25 = 22.75. The needed probabilities for the given case are: Probability that the individual waits more than 7 minutes = 0.3 Probability that the individual waits between 2 and 7 minutes = 0.5 How to calculate the probability of an interval in uniform distribution? This means you will have to find the value such that \(\frac{3}{4}\), or 75%, of the cars are at most (less than or equal to) that age. 5 1 The data in (Figure) are 55 smiling times, in seconds, of an eight-week-old baby. P(x 12|x > 8) = (23 12)\(\left(\frac{1}{15}\right)\) = \(\left(\frac{11}{15}\right)\). P(x>8) 2 We recommend using a . \(X =\) __________________. b is 12, and it represents the highest value of x. 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. Find the mean, \(\mu\), and the standard deviation, \(\sigma\). If you are redistributing all or part of this book in a print format, Sketch the graph of the probability distribution. then you must include on every physical page the following attribution: If you are redistributing all or part of this book in a digital format, a. The amount of time a service technician needs to change the oil in a car is uniformly distributed between 11 and 21 minutes. are not subject to the Creative Commons license and may not be reproduced without the prior and express written What percentile does this represent? 1 230 The probability that a randomly selected nine-year old child eats a donut in at least two minutes is _______. Note that the length of the base of the rectangle . Ninety percent of the time, a person must wait at most 13.5 minutes. The height is \(\frac{1}{\left(25-18\right)}\) = \(\frac{1}{7}\). )=0.90 1.5+4 a person has waited more than four minutes is? 2 What is the 90th percentile of square footage for homes? c. Find the 90th percentile. Plume, 1995. This means that any smiling time from zero to and including 23 seconds is equally likely. Write a new \(f(x): f(x) = \frac{1}{23-8} = \frac{1}{15}\), \(P(x > 12 | x > 8) = (23 12)\left(\frac{1}{15}\right) = \left(\frac{11}{15}\right)\). 1.0/ 1.0 Points. The amount of time, in minutes, that a person must wait for a bus is uniformly distributed between zero and 15 minutes, inclusive. As waiting passengers occupy more platform space than circulating passengers, evaluation of their distribution across the platform is important. 12= Find the probability that a randomly selected furnace repair requires less than three hours. The Continuous Uniform Distribution in R. You may use this project freely under the Creative Commons Attribution-ShareAlike 4.0 International License. \(b\) is \(12\), and it represents the highest value of \(x\). a = 0 and b = 15. )( (In other words: find the minimum time for the longest 25% of repair times.) 2 Use the following information to answer the next three exercises. and The probability is constant since each variable has equal chances of being the outcome. This means you will have to find the value such that \(\frac{3}{4}\), or 75%, of the cars are at most (less than or equal to) that age. There are two types of uniform distributions: discrete and continuous. The answer for 1) is 5/8 and 2) is 1/3. for 0 x 15. What is the theoretical standard deviation? The possible outcomes in such a scenario can only be two. What is the probability that the duration of games for a team for the 2011 season is between 480 and 500 hours? 1999-2023, Rice University. = P(155 < X < 170) = (170-155) / (170-120) = 15/50 = 0.3. If you are waiting for a train, you have anywhere from zero minutes to ten minutes to wait. 2.5 (In other words: find the minimum time for the longest 25% of repair times.) Let X = the number of minutes a person must wait for a bus. = a+b P(B) = P(x>8) P(x>8) The McDougall Program for Maximum Weight Loss. f(x) = \(\frac{1}{b-a}\) for a x b. 41.5 When working out problems that have a uniform distribution, be careful to note if the data is inclusive or exclusive of endpoints. A distribution is given as X ~ U(0, 12). What is the probability that the duration of games for a team for the 2011 season is between 480 and 500 hours? For this problem, \(\text{A}\) is (\(x > 12\)) and \(\text{B}\) is (\(x > 8\)). A continuous uniform distribution is a statistical distribution with an infinite number of equally likely measurable values. 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. A distribution is given as \(X \sim U(0, 20)\). P(A|B) = P(A and B)/P(B). Find the 90thpercentile. Lets suppose that the weight loss is uniformly distributed. The second question has a conditional probability. Find the probability that a randomly selected home has more than 3,000 square feet given that you already know the house has more than 2,000 square feet. Draw the graph. Would it be P(A) +P(B) + P(C) - P(A and B) - P(A and C) - P(B and C) - P(A and B and C)? a+b 1 \(X \sim U(a, b)\) where \(a =\) the lowest value of \(x\) and \(b =\) the highest value of \(x\). 1 For example, it can arise in inventory management in the study of the frequency of inventory sales. However the graph should be shaded between \(x = 1.5\) and \(x = 3\). 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. The shaded rectangle depicts the probability that a randomly. Find P(x > 12|x > 8) There are two ways to do the problem. 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. Find the probability that a randomly chosen car in the lot was less than four years old. This page titled 5.3: The Uniform Distribution is shared under a CC BY 4.0 license and was authored, remixed, and/or curated by OpenStax via source content that was edited to the style and standards of the LibreTexts platform; a detailed edit history is available upon request. Uniform Distribution between 1.5 and 4 with an area of 0.30 shaded to the left, representing the shortest 30% of repair times. The amount of time a service technician needs to change the oil in a car is uniformly distributed between 11 and 21 minutes. = The probability a person waits less than 12.5 minutes is 0.8333. b. = Definitions of Statistics, Probability, and Key Terms, Data, Sampling, and Variation in Data and Sampling, Frequency, Frequency Tables, and Levels of Measurement, Stem-and-Leaf Graphs (Stemplots), Line Graphs, and Bar Graphs, Histograms, Frequency Polygons, and Time Series Graphs, Independent and Mutually Exclusive Events, Probability Distribution Function (PDF) for a Discrete Random Variable, Mean or Expected Value and Standard Deviation, Discrete Distribution (Playing Card Experiment), Discrete Distribution (Lucky Dice Experiment), The Central Limit Theorem for Sample Means (Averages), A Single Population Mean using the Normal Distribution, A Single Population Mean using the Student t Distribution, Outcomes and the Type I and Type II Errors, Distribution Needed for Hypothesis Testing, Rare Events, the Sample, Decision and Conclusion, Additional Information and Full Hypothesis Test Examples, Hypothesis Testing of a Single Mean and Single Proportion, Two Population Means with Unknown Standard Deviations, Two Population Means with Known Standard Deviations, Comparing Two Independent Population Proportions, Hypothesis Testing for Two Means and Two Proportions, Testing the Significance of the Correlation Coefficient. 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