The distribution can be written as X ~ U(1.5, 4.5). f(X) = 1 150 = 1 15 for 0 X 15. Introduction to Statistics is our premier online video course that teaches you all of the topics covered in introductory statistics. There are several ways in which discrete uniform distribution can be valuable for businesses. 3.375 hours is the 75th percentile of furnace repair times. 1 a. P(A|B) = P(A and B)/P(B). P(x > k) = 0.25 k=(0.90)(15)=13.5 Use the following information to answer the next ten questions. a. The uniform distribution defines equal probability over a given range for a continuous distribution. \(P\left(x \(x\)) = (b x)\(\left(\frac{1}{b-a}\right)\), Area Between \(c\) and \(d\): \(P(c < x < d) = (\text{base})(\text{height}) = (d c)\left(\frac{1}{b-a}\right)\), Uniform: \(X \sim U(a, b)\) where \(a < x < b\). The probability is constant since each variable has equal chances of being the outcome. Let X = the number of minutes a person must wait for a bus. This means that any smiling time from zero to and including 23 seconds is equally likely. (In other words: find the minimum time for the longest 25% of repair times.) The total duration of baseball games in the major league in the 2011 season is uniformly distributed between 447 hours and 521 hours inclusive. 15 Let \(k =\) the 90th percentile. Then \(X \sim U(6, 15)\). It is generally represented by u (x,y). Standard deviation is (a-b)^2/12 = (0-12)^2/12 = (-12^2)/12 = 144/12 = 12 c. Prob (Wait for more than 5 min) = (12-5)/ (12-0) = 7/12 = 0.5833 d. To find \(f(x): f(x) = \frac{1}{4-1.5} = \frac{1}{2.5}\) so \(f(x) = 0.4\), \(P(x > 2) = (\text{base})(\text{height}) = (4 2)(0.4) = 0.8\), b. 1 \(\mu = \frac{a+b}{2} = \frac{15+0}{2} = 7.5\). 2.75 k 12= Ninety percent of the time, a person must wait at most 13.5 minutes. The amount of time a service technician needs to change the oil in a car is uniformly distributed between 11 and 21 minutes. A uniform distribution has the following properties: The area under the graph of a continuous probability distribution is equal to 1. The data follow a uniform distribution where all values between and including zero and 14 are equally likely. 12 The sample mean = 7.9 and the sample standard deviation = 4.33. Notice that the theoretical mean and standard deviation are close to the sample mean and standard deviation in this example. It means every possible outcome for a cause, action, or event has equal chances of occurrence. Note that the shaded area starts at x = 1.5 rather than at x = 0; since X ~ U (1.5, 4), x can not be less than 1.5. Sketch the graph of the probability distribution. The data that follow are the square footage (in 1,000 feet squared) of 28 homes. 12 15+0 The graph of the rectangle showing the entire distribution would remain the same. f ( x) = 1 12 1, 1 x 12 = 1 11, 1 x 12 = 0.0909, 1 x 12. You must reduce the sample space. 2.5 The time follows a uniform distribution. The sample mean = 7.9 and the sample standard deviation = 4.33. 0.90 How do these compare with the expected waiting time and variance for a single bus when the time is uniformly distributed on \({\rm{(0,5)}}\)? )=0.8333. 3.375 = k, Posted at 09:48h in michael deluise matt leblanc by Second way: Draw the original graph for \(X \sim U(0.5, 4)\). 0+23 The uniform distribution is a continuous probability distribution and is concerned with events that are equally likely to occur. P(0 < X < 8) = (8-0) / (20-0) = 8/20 =0.4. P(2 < x < 18) = (base)(height) = (18 2)\(\left(\frac{1}{23}\right)\) = \(\left(\frac{16}{23}\right)\). A distribution is given as X ~ U (0, 20). 15 X ~ U(a, b) where a = the lowest value of x and b = the highest value of x. Formulas for the theoretical mean and standard deviation are, \[\sigma = \sqrt{\frac{(b-a)^{2}}{12}} \nonumber\], For this problem, the theoretical mean and standard deviation are, \[\mu = \frac{0+23}{2} = 11.50 \, seconds \nonumber\], \[\sigma = \frac{(23-0)^{2}}{12} = 6.64\, seconds. State the values of a and b. = \(P(x < 4) =\) _______. c. What is the expected waiting time? Let X = the number of minutes a person must wait for a bus. b. Ninety percent of the smiling times fall below the 90th percentile, \(k\), so \(P(x < k) = 0.90\), \[(k0)\left(\frac{1}{23}\right) = 0.90\]. The Sky Train from the terminal to the rentalcar and longterm parking center is supposed to arrive every eight minutes. The concept of uniform distribution, as well as the random variables it describes, form the foundation of statistical analysis and probability theory. for 1.5 x 4. This book uses the = Refer to Example 5.2. 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. The probability density function is 1), travelers have different characteristics: trip length l L, desired arrival time, t a T a, and scheduling preferences c, c, and c associated to their socioeconomic class c C.The capital and curly letter . 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. Question 2: The length of an NBA game is uniformly distributed between 120 and 170 minutes. For example, we want to predict the following: The amount of timeuntilthe customer finishes browsing and actually purchases something in your store (success). Suppose that the value of a stock varies each day from 16 to 25 with a uniform distribution. What does this mean? 233K views 3 years ago This statistics video provides a basic introduction into continuous probability distribution with a focus on solving uniform distribution problems. 12 On the average, a person must wait 7.5 minutes. Lets suppose that the weight loss is uniformly distributed. Note that the length of the base of the rectangle . Example The data in the table below are 55 smiling times, in seconds, of an eight-week-old baby. \[P(x < k) = (\text{base})(\text{height}) = (12.50)\left(\frac{1}{15}\right) = 0.8333\]. P(x>12) e. \(\mu =\frac{a+b}{2}\) and \(\sigma =\sqrt{\frac{{\left(b-a\right)}^{2}}{12}}\), \(\mu =\frac{1.5+4}{2}=2.75\) a is zero; b is 14; X ~ U (0, 14); = 7 passengers; = 4.04 passengers. Beta distribution is a well-known and widely used distribution for modeling and analyzing lifetime data, due to its interesting characteristics. When working out problems that have a uniform distribution, be careful to note if the data is inclusive or exclusive. Find the probability that the commuter waits less than one minute. Create an account to follow your favorite communities and start taking part in conversations. 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. Heres how to visualize that distribution: And the probability that a randomly selected dolphin weighs between 120 and 130 pounds can be visualized as follows: The uniform distribution has the following properties: We could calculate the following properties for this distribution: Use the following practice problems to test your knowledge of the uniform distribution. Your starting point is 1.5 minutes. ba Answer: a. 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). The uniform distribution is a probability distribution in which every value between an interval from a to b is equally likely to occur. 2 obtained by subtracting four from both sides: \(k = 3.375\) \(P(x > 2|x > 1.5) = (\text{base})(\text{new height}) = (4 2)(25)\left(\frac{2}{5}\right) =\) ? What is the height of f(x) for the continuous probability distribution? Press J to jump to the feed. 1.5+4 This is a conditional probability question. a= 0 and b= 15. The uniform distribution is a continuous probability distribution and is concerned with events that are equally likely to occur. . A random number generator picks a number from one to nine in a uniform manner. The uniform distribution is a continuous probability distribution and is concerned with events that are equally likely to occur. \(0.3 = (k 1.5) (0.4)\); Solve to find \(k\): Let k = the 90th percentile. This paper addresses the estimation of the charging power demand of XFC stations and the design of multiple XFC stations with renewable energy resources in current . a. 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. Find the 90th percentile for an eight-week-old baby's smiling time. The probability a person waits less than 12.5 minutes is 0.8333. b. ) = The second question has a conditional probability. Textbook content produced by OpenStax is licensed under a Creative Commons Attribution License . Find the probability that a different 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. 0+23 Find the probability that she is over 6.5 years old. 23 Legal. A good example of a discrete uniform distribution would be the possible outcomes of rolling a 6-sided die. The amount of time a service technician needs to change the oil in a car is uniformly distributed between 11 and 21 minutes. What is P(2 < x < 18)? Write the random variable \(X\) in words. 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. P(A or B) = P(A) + P(B) - P(A and B). Answer: (Round to two decimal places.) As the question stands, if 2 buses arrive, that is fine, because at least 1 bus arriving is satisfied. The distribution is ______________ (name of distribution). a. Suppose that you arrived at the stop at 10:00 and wait until 10:05 without a bus arriving. P(x > 2|x > 1.5) = (base)(new height) = (4 2)\(\left(\frac{2}{5}\right)\)= ? What is the height of \(f(x)\) for the continuous probability distribution? Draw a graph. 2 Let X = the time, in minutes, it takes a nine-year old child to eat a donut. You must reduce the sample space. The 30th percentile of repair times is 2.25 hours. Accessibility StatementFor more information contact us atinfo@libretexts.orgor check out our status page at https://status.libretexts.org. (In other words: find the minimum time for the longest 25% of repair times.) ( 5 It is defined by two parameters, x and y, where x = minimum value and y = maximum value. b. (41.5) Write a newf(x): f(x) = \(\frac{1}{23\text{}-\text{8}}\) = \(\frac{1}{15}\), P(x > 12|x > 8) = (23 12)\(\left(\frac{1}{15}\right)\) = \(\left(\frac{11}{15}\right)\). b. The sample mean = 2.50 and the sample standard deviation = 0.8302. . A student takes the campus shuttle bus to reach the classroom building. X = The age (in years) of cars in the staff parking lot. All values \(x\) are equally likely. ( The data follow a uniform distribution where all values between and including zero and 14 are equally likely. P(120 < X < 130) = (130 120) / (150 100), The probability that the chosen dolphin will weigh between 120 and 130 pounds is, Mean weight: (a + b) / 2 = (150 + 100) / 2 =, Median weight: (a + b) / 2 = (150 + 100) / 2 =, P(155 < X < 170) = (170-155) / (170-120) = 15/50 =, P(17 < X < 19) = (19-17) / (25-15) = 2/10 =, How to Plot an Exponential Distribution in R. Your email address will not be published. (In other words: find the minimum time for the longest 25% of repair times.) What percentage of 20 minutes is 5 minutes?). I was originally getting .75 for part 1 but I didn't realize that you had to subtract P(A and B). However the graph should be shaded between x = 1.5 and x = 3. (a) The probability density function of X is. The longest 25% of furnace repairs take at least 3.375 hours (3.375 hours or longer). 15 The longest 25% of furnace repairs take at least 3.375 hours (3.375 hours or longer). 23 What is the average waiting time (in minutes)? a+b the 1st and 3rd buses will arrive in the same 5-minute period)? Write the probability density function. 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. 12 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: . a = 0 and b = 15. 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. ( \(k = 2.25\) , obtained by adding 1.5 to both sides. 2 That is . Then X ~ U (6, 15). 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Concept of uniform distribution has the following properties: the uniform distribution waiting bus time the! The campus shuttle bus to reach the classroom building textbook content produced by OpenStax is licensed a! And B ) by U ( 6, 15 ) \ ) miles driven by a truck driver between. His stop every 20 minutes is useful in Monte Carlo simulation if the data follow a uniform distribution a... With a focus on solving uniform distribution ( x ) = \ ( k = 2.25\ ), obtained adding!, obtained by adding 1.5 to both sides 0.8 ; 90th percentile the. The shuttle bus arrives at his stop every 15 minutes but the actual arrival time at bus. On solving uniform distribution can be written as x ~ U ( 1.5, 4.5 ) values between and zero. < 18 ) and 12 minute probability a person must wait at most 13.5 minutes a service technician to! Times. data, due to its interesting characteristics between an interval from to! Distribution with a focus on solving uniform distribution, as well as the question stands, if buses. And 12 minute needs at least 3.375 hours is the height of \ ( X\ ) are equally to! Commuter waits less than 12.5 minutes is 5 minutes? ) time between... Eat a donut value of interest is 170 minutes a uniform distribution would remain the same 5-minute )... Is supposed to arrive every eight uniform distribution waiting bus is 5 minutes? ) x U!
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