uniform distribution waiting bus

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Buses run every 30 minutes without fail, hence the next bus will come any time during the next 30 minutes with evenly distributed probability (a uniform distribution). The uniform distribution is a continuous probability distribution and is concerned with events that are equally likely to occur. The probability of waiting more than seven minutes given a person has waited more than four minutes is? = 11.50 seconds and = $\sqrt{\frac{{\left(23\text{}-\text{}0\right)}^{2}}{12}}$ First, I'm asked to calculate the expected value E (X). The data follow a uniform distribution where all values between and including zero and 14 are equally likely. The LibreTexts libraries arePowered by NICE CXone Expertand are supported by the Department of Education Open Textbook Pilot Project, the UC Davis Office of the Provost, the UC Davis Library, the California State University Affordable Learning Solutions Program, and Merlot. If $X$ has a uniform distribution where $a < x < b$ or $a \leq x \leq b$, then $X$ takes on values between $a$ and $b$ (may include $a$ and $b$). 2 The probability a person waits less than 12.5 minutes is 0.8333. b. = ( That is, almost all random number generators generate random numbers on the . Let $X =$ length, in seconds, of an eight-week-old baby's smile. The longest 25% of furnace repair times take at least how long? McDougall, John A. The shuttle bus arrives at his stop every 15 minutes but the actual arrival time at the stop is random. $0.75 = k 1.5$, obtained by dividing both sides by 0.4 ba When working out problems that have a uniform distribution, be careful to note if the data is inclusive or exclusive of endpoints. The Standard deviation is 4.3 minutes. P(x12ANDx>8) P(x < k) = (base)(height) = (k 1.5)(0.4) 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. Answer Key:0.6 | .6| 0.60|.60 Feedback: Interval goes from 0 x 10 P (x < 6) = Question 11 of 20 0.0/ 1.0 Points Textbook content produced by OpenStax is licensed under a Creative Commons Attribution License . The sample mean = 7.9 and the sample standard deviation = 4.33. $P(x < 4) =$ _______. Find $a$ and $b$ and describe what they represent. Find the 90th percentile for an eight-week-old baby's smiling time. A distribution is given as X ~ U(0, 12). The graph illustrates the new sample space. It is assumed that the waiting time for a particular individual is a random variable with a continuous uniform distribution. Find the probability that a randomly chosen car in the lot was less than four years old. That is, find. k is sometimes called a critical value. It is defined by two different parameters, x and y, where x = the minimum value and y = the maximum value. A distribution is given as X ~ U (0, 20). The mean of X is $\mu =\frac{a+b}{2}$. Then $x \sim U(1.5, 4)$. Use the conditional formula, $P(x > 2 | x > 1.5) = \frac{P(x > 2 \text{AND} x > 1.5)}{P(x > 1.5)} = \frac{P(x>2)}{P(x>1.5)} = \frac{\frac{2}{3.5}}{\frac{2.5}{3.5}} = 0.8 = \frac{4}{5}$. McDougall, John A. 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 subway train on the Red Line arrives every eight minutes during rush hour. Ninety percent of the time, a person must wait at most 13.5 minutes. = 1.0/ 1.0 Points. It is generally represented by u (x,y). To find f(x): f (x) = $\frac{1}{4\text{}-\text{}1.5}$ = $\frac{1}{2.5}$ 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. = 7.5. When working out problems that have a uniform distribution, be careful to note if the data is inclusive or exclusive. P(x>1.5) 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. 1 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. 15 1 5.2 The Uniform Distribution. The age of cars in the staff parking lot of a suburban college is uniformly distributed from six months (0.5 years) to 9.5 years. 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. Get started with our course today. However, the extreme high charging power of EVs at XFC stations may severely impact distribution networks. consent of Rice University. 1 The histogram that could be constructed from the sample is an empirical distribution that closely matches the theoretical uniform distribution. P(x 2|x > 1.5) = (base)(new height) = (4 2) Find the mean and the standard deviation. The lower value of interest is 0 minutes and the upper value of interest is 8 minutes. Let $X =$ the number of minutes a person must wait for a bus. Example The data in the table below are 55 smiling times, in seconds, of an eight-week-old baby. 23 Your starting point is 1.5 minutes. 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). Use the following information to answer the next eleven exercises. 2 2 f(x) = $\frac{1}{b-a}$ for a x b. 11 Solution: What does this mean? 2 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 $\frac{4}{5}$. P(A|B) = P(A and B)/P(B). a+b 15 1 Then $X \sim U(0.5, 4)$. d. What is standard deviation of waiting time? State the values of a and b. As an Amazon Associate we earn from qualifying purchases. 2 $P(x < k) = (\text{base})(\text{height}) = (k 1.5)(0.4)$ 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. Accessibility StatementFor more information contact us atinfo@libretexts.orgor check out our status page at https://status.libretexts.org. OpenStax is part of Rice University, which is a 501(c)(3) nonprofit. )=0.8333 Uniform Distribution. 23 Since the corresponding area is a rectangle, the area may be found simply by multiplying the width and the height. The possible outcomes in such a scenario can only be two. It is impossible to get a value of 1.3, 4.2, or 5.7 when rolling a fair die. = $\frac{6}{9}$ = $\frac{2}{3}$. X is now asked to be the waiting time for the bus in seconds on a randomly chosen trip. Suppose that you arrived at the stop at 10:00 and wait until 10:05 without a bus arriving. and (230) 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. Second way: Draw the original graph for X ~ U (0.5, 4). Find the probability that the value of the stock is between 19 and 22. 0.90=( So, P(x > 12|x > 8) = $\frac{\left(x>12\text{AND}x>8\right)}{P\left(x>8\right)}=\frac{P\left(x>12\right)}{P\left(x>8\right)}=\frac{\frac{11}{23}}{\frac{15}{23}}=\frac{11}{15}$. Discrete uniform distribution is also useful in Monte Carlo simulation. 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). 11 obtained by subtracting four from both sides: $k = 3.375$ What is the height of $f(x)$ for the continuous probability distribution? Find the probability that a person is born after week 40. =0.8= Use Uniform Distribution from 0 to 5 minutes. What is the average waiting time (in minutes)? 1 What has changed in the previous two problems that made the solutions different. Another simple example is the probability distribution of a coin being flipped. 1 (41.5) and (ba) We will assume that the smiling times, in seconds, follow a uniform distribution between zero and 23 seconds, inclusive. The mean of $X$ is $\mu = \frac{a+b}{2}$. 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. The Standard deviation is 4.3 minutes. If the waiting time (in minutes) at each stop has a uniform distribution with A = 0 and B = 5, then it can be shown that the total waiting time Y has the pdf $$ f(y)=\left\{\begin{array}{cc} \frac . Sketch the graph, and shade the area of interest. Post all of your math-learning resources here. 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. f ( x) = 1 12 1, 1 x 12 = 1 11, 1 x 12 = 0.0909, 1 x 12. $\mu = \frac{a+b}{2} = \frac{15+0}{2} = 7.5$. Find the third quartile of ages of cars in the lot. P(AANDB) Solution 1: The minimum amount of time youd have to wait is 0 minutes and the maximum amount is 20 minutes. 30% of repair times are 2.5 hours or less. The uniform distribution is a continuous probability distribution and is concerned with events that are equally likely to occur. 1 The waiting time at a bus stop is uniformly distributed between 1 and 12 minute. P(2 < x < 18) = 0.8; 90th percentile = 18. A student takes the campus shuttle bus to reach the classroom building. Continuous Uniform Distribution Example 2 For example, in our previous example we said the weight of dolphins is uniformly distributed between 100 pounds and 150 pounds. Find the probability. It is defined by two parameters, x and y, where x = minimum value and y = maximum value. 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(x>12) What is the theoretical standard deviation? ) The Standard deviation is 4.3 minutes. 5 $P(x > 2|x > 1.5) = (\text{base})(\text{new height}) = (4 2)(25)\left(\frac{2}{5}\right) =$ ? Find the probability that the value of the stock is more than 19. (230) 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. Your probability of having to wait any number of minutes in that interval is the same. What is the 90th percentile of square footage for homes? 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. If a person arrives at the bus stop at a random time, how long will he or she have to wait before the next bus arrives? The 90th percentile is 13.5 minutes. the 1st and 3rd buses will arrive in the same 5-minute period)? The waiting times for the train are known to follow a uniform distribution. That is . 1 Find the third quartile of ages of cars in the lot. 14.6 - Uniform Distributions. Creative Commons Attribution License Figure 2 3.5 The data in Table 5.1 are 55 smiling times, in seconds, of an eight-week-old baby. 2.75 It means that the value of x is just as likely to be any number between 1.5 and 4.5. 238 The probability is constant since each variable has equal chances of being the outcome. What is the probability that the waiting time for this bus is less than 5.5 minutes on a given day? On the average, a person must wait 7.5 minutes. The McDougall Program for Maximum Weight Loss. Continuous Uniform Distribution - Waiting at the bus stop 1,128 views Aug 9, 2020 20 Dislike Share The A Plus Project 331 subscribers This is an example of a problem that can be solved with the. = Find the average age of the cars in the lot. The notation for the uniform distribution is. 23 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. 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P(x>1.5) A continuous uniform distribution (also referred to as rectangular distribution) is a statistical distribution with an infinite number of equally likely measurable values. 0.3 = (k 1.5) (0.4); Solve to find k: Write a new $f(x): f(x) = \frac{1}{23-8} = \frac{1}{15}$, $P(x > 12 | x > 8) = (23 12)\left(\frac{1}{15}\right) = \left(\frac{11}{15}\right)$. Answer: a. However, if you favored short people or women, they would have a higher chance of being given the $100 bill than the other passersby. = The second question has a conditional probability. 2 What is the probability that the rider waits 8 minutes or less? 238 What is the probability that the duration of games for a team for the 2011 season is between 480 and 500 hours? Therefore, each time the 6-sided die is thrown, each side has a chance of 1/6. Suppose the time it takes a nine-year old to eat a donut is between 0.5 and 4 minutes, inclusive. The 30th percentile of repair times is 2.25 hours. 2 So, $P(x > 12|x > 8) = \frac{(x > 12 \text{ AND } x > 8)}{P(x > 8)} = \frac{P(x > 12)}{P(x > 8)} = \frac{\frac{11}{23}}{\frac{15}{23}} = \frac{11}{15}$. ( Shade the area of interest. Let X = length, in seconds, of an eight-week-old babys smile. 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. The sample mean = 7.9 and the sample standard deviation = 4.33. Pandas: Use Groupby to Calculate Mean and Not Ignore NaNs. obtained by subtracting four from both sides: k = 3.375 Draw a graph. 23 The data in [link] are 55 smiling times, in seconds, of an eight-week-old baby. Find the 30th percentile for the waiting times (in minutes). b. For the first way, use the fact that this is a conditional and changes the sample space. The probability density function of $X$ is $f(x) = \frac{1}{b-a}$ for $a \leq x \leq b$. )( Find the probability that the individual lost more than ten pounds in a month. Beta distribution is a well-known and widely used distribution for modeling and analyzing lifetime data, due to its interesting characteristics. $a =$ smallest $X$; $b =$ largest $X$, The standard deviation is $\sigma = \sqrt{\frac{(b-a)^{2}}{12}}$, Probability density function: $f(x) = \frac{1}{b-a} \text{for} a \leq X \leq b$, Area to the Left of $x$: $P(X < x) = (x a)\left(\frac{1}{b-a}\right)$, Area to the Right of $x$: P($X$ > $x$) = (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 $a < x < b$. 2 So, P(x > 21|x > 18) = (25 21)$\left(\frac{1}{7}\right)$ = 4/7. . The sample mean = 11.65 and the sample standard deviation = 6.08. P(x>2) b. P(x>8) $3.375 = k$, $P(x > k) = 0.25$ (a) What is the probability that the individual waits more than 7 minutes? Uniform Distribution between 1.5 and four with shaded area between two and four representing the probability that the repair time, Uniform Distribution between 1.5 and four with shaded area between 1.5 and three representing the probability that the repair time. What percentile does this represent? 16 b. Ninety percent of the smiling times fall below the 90th percentile, k, so P(x < k) = 0.90, $\left(\text{base}\right)\left(\text{height}\right)=0.90$, $\text{(}k-0\text{)}\left(\frac{1}{23}\right)=0.90$, $k=\left(23\right)\left(0.90\right)=20.7$. Note that the length of the base of the rectangle . A distribution is given as X ~ U (0, 20). a. I was originally getting .75 for part 1 but I didn't realize that you had to subtract P(A and B). \(P\left(x

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