The Uniform Distribution

Transcription

1 Connexions module: m46972 The Uniform Distribution OpenStax College This work is produced by The Connexions Project and licensed under the Creative Commons Attribution License 3.0 The uniform distribution is a continuous probability distribution and is concerned with events that are equally likely to occur. When working out problems that have a uniform distribution, be careful to note if the data is inclusive or exclusive. Example The data in Table are 55 smiling times, in seconds, of an eight-week-old baby Table The sample mean =.49 and the sample standard deviation = We will assume that the smiling times, in seconds, follow a uniform distribution between zero and 23 seconds, inclusive. This means that any smiling time from zero to and including 23 seconds is equally likely. The histogram that could be constructed from the sample is an empirical distribution that closely matches the theoretical uniform distribution. Let X = length, in seconds, of an eight-week-old baby's smile. The notation for the uniform distribution is X U(a, b) where a = the lowest value of x and b = the highest value of x. The probability density function is f (x) = b a for a x b. For this example, X U(0, 23) and f (x) = 23 0 for 0 X 23. Formulas for the theoretical mean and standard deviation are (b a) 2 2 µ = a+b 2 and σ = For this problem, the theoretical mean and standard deviation are µ = (23 0) 2 =.50 seconds and σ = 2 2 = 6.64 seconds. Notice that the theoretical mean and standard deviation are close to the sample mean and standard deviation in this example. Version.4: Nov 22, 203 2:53 pm

2 Connexions module: m : Exercise (Solution on p. 20.) The data that follow are the number of passengers on 35 dierent charter shing boats. The sample mean = 7.9 and the sample standard deviation = The data follow a uniform distribution where all values between and including zero and 4 are equally likely. State the values of a and b. Write the distribution in proper notation, and calculate the theoretical mean and standard deviation Table 2 Example 2 Problem a. Refer to Example. What is the probability that a randomly chosen eight-week-old baby smiles between two and 8 seconds? Solution a. Find P(2 < x < 8). P(2 < x < 8) = (base)(height) = (8 2) ( ) ( 23 = 6 23). Figure

3 Connexions module: m Problem 2 b. Find the 90 th percentile for an eight-week-old baby's smiling time. Solution b. Ninety percent of the smiling times fall below the 90 th percentile, k, so P(x < k) = 0.90 P (x < k) = 0.90 (base) (height) = 0.90 ( ) (k 0) = k = (23) (0.90) = 20.7 Figure 2 Problem 3 c. Find the probability that a random eight-week-old baby smiles more than 2 seconds KNOW- ING that the baby smiles MORE THAN EIGHT SECONDS. Solution c. This probability question is a conditional. You are asked to nd the probability that an eight-week-old baby smiles more than 2 seconds when you already know the baby has smiled for more than eight seconds. Find P(x > 2 x > 8) There are two ways to do the problem. For the rst way, use the fact that this is a conditional and changes the sample space. The graph illustrates the new sample space. You already know the baby smiled more than eight seconds. Write a newf (x): f (x) = 23 = 8 5 for 8 < x < 23

4 Connexions module: m P(x > 2 x > 8) = (23 2) ( ) ( 5 = ) 5 Figure 3 For the second way, use the conditional formula from Probability Topics with the original distribution X U (0, 23): P (A AND B) P(A B) = P (B) For this problem, A is (x > 2) and B is (x > 8). So, P(x > 2 x > 8) = (x>2 AND x>8) P (x>8) "Introduction" < = P (x>2) P (x>8) = = 5

5 Connexions module: m Figure 4 : Exercise 5 (Solution on p. 20.) A distribution is given as X U (0, 20). What is P(2 < x < 8)? Find the 90 th percentile. Example 3 The amount of time, in minutes, that a person must wait for a bus is uniformly distributed between zero and 5 minutes, inclusive. Problem a. What is the probability that a person waits fewer than 2.5 minutes? Solution a. Let X = the number of minutes a person must wait for a bus. a = 0 and b = 5. X U(0, 5). Write the probability density function. f (x) = 5 0 = 5 for 0 x 5. Find P (x < 2.5). Draw a graph. ( ) P (x < k) = (base) (height) = (2.5 0) = The probability a person waits less than 2.5 minutes is

6 Connexions module: m Figure 5 Problem 2 b. On the average, how long must a person wait? Find the mean, µ, and the standard deviation, σ. Solution b. µ = a + b 2 = σ = (b a) 2 2 = 2 = 7.5. On the average, a person must wait 7.5 minutes. (5 0) 2 2 = 4.3. The Standard deviation is 4.3 minutes. Problem 3 c. Ninety percent of the time, the time a person must wait falls below what value? Note: This asks for the 90 th percentile. Solution c. Find the 90 th percentile. Draw a graph. Let k = the 90 th percentile. P (x < k) = (base) (height) = (k 0) ( ) = (k) ( ) 5 k = (0.90) (5) = 3.5 k is sometimes called a critical value. The 90 th percentile is 3.5 minutes. Ninety percent of the time, a person must wait at most 3.5 minutes.

7 Connexions module: m Figure 6 : Exercise 9 (Solution on p. 20.) The total duration of baseball games in the major league in the 20 season is uniformly distributed between 447 hours and 52 hours inclusive. a.find a and b and describe what they represent. b.write the distribution. c.find the mean and the standard deviation. d.what is the probability that the duration of games for a team for the 20 season is between 480 and 500 hours? e.what is the 65 th percentile for the duration of games for a team for the 20 season? Example 4 Suppose the time it takes a nine-year old to eat a donut is between 0.5 and 4 minutes, inclusive. Let X = the time, in minutes, it takes a nine-year old child to eat a donut. Then X U (0.5, 4). a. The probability that a randomly selected nine-year old child eats a donut in at least two minutes is. Solution a

8 Connexions module: m Problem 2 b. Find the probability that a dierent 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.5 minutes. The second question has a conditional probability. You are asked to nd 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.5 minutes. Solve the problem two dierent ways (see Example ). You must reduce the sample space. First way: Since you know the child has already been eating the donut for more than.5 minutes, you are no longer starting at a = 0.5 minutes. Your starting point is.5 minutes. Write a new f (x): f (x) = 4.5 = 2 5 for.5 x 4. Find P(x > 2 x >.5). Draw a graph. Figure 7 P(x > 2 x >.5) = (base)(new height) = (4 2) ( 2 5) =? Solution b. 4 5 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.5 minutes is 4 5. Second way: Draw the original graph for X U (0.5, 4). Use the conditional formula P(x > 2 x >.5) = P (x>2 AND x>.5) P (x>.5) = P (x>2) P (x>.5) = = 0.8 = 4 5 : Exercise 2 (Solution on p. 20.) Suppose the time it takes a student to nish a quiz is uniformly distributed between six

9 Connexions module: m and 5 minutes, inclusive. Let X = the time, in minutes, it takes a student to nish a quiz. Then X U (6, 5). Find the probability that a randomly selected student needs at least eight minutes to complete the quiz. Then nd the probability that a dierent student needs at least eight minutes to nish the quiz given that she has already taken more than seven minutes. Example 5 Ace Heating and Air Conditioning Service nds that the amount of time a repairman needs to x a furnace is uniformly distributed between.5 and four hours. Let x = the time needed to x a furnace. Then x U (.5, 4). a. Find the probability that a randomly selected furnace repair requires more than two hours. b. Find the probability that a randomly selected furnace repair requires less than three hours. c. Find the 30 th percentile of furnace repair times. d. The longest 25% of furnace repair times take at least how long? (In other words: nd the minimum time for the longest 25% of repair times.) What percentile does this represent? e. Find the mean and standard deviation Solution A a. To nd f (x): f (x) = 4.5 = 2.5 so f (x) = 0.4 P(x > 2) = (base)(height) = (4 2)(0.4) = 0.8 Figure 8: Uniform Distribution between.5 and four with shaded area between two and four representing the probability that the repair time x is greater than two Solution B b. P(x < 3) = (base)(height) = (3.5)(0.4) = 0.6 The graph of the rectangle showing the entire distribution would remain the same. However the graph should be shaded between x =.5 and x = 3. Note that the shaded area starts at x =.5 rather than at x = 0; since X U (.5, 4), x can not be less than.5.

10 Connexions module: m Figure 9: Uniform Distribution between.5 and four with shaded area between.5 and three representing the probability that the repair time x is less than three Solution C c. Figure 0: Uniform Distribution between.5 and 4 with an area of 0.30 shaded to the left, representing the shortest 30% of repair times. P (x < k) = 0.30 P(x < k) = (base)(height) = (k.5)(0.4) 0.3 = (k.5) (0.4); Solve to nd k:

11 Connexions module: m = k.5, obtained by dividing both sides by 0.4 k = 2.25, obtained by adding.5 to both sides The 30 th percentile of repair times is 2.25 hours. 30% of repair times are 2.5 hours or less. Solution D d. Figure : Uniform Distribution between.5 and 4 with an area of 0.25 shaded to the right representing the longest 25% of repair times. P(x > k) = 0.25 P(x > k) = (base)(height) = (4 k)(0.4) 0.25 = (4 k)(0.4); Solve for k: = 4 k, obtained by dividing both sides by = k, obtained by subtracting four from both sides: k = The longest 25% of furnace repairs take at least hours (3.375 hours or longer). Note: Since 25% of repair times are hours or longer, that means that 75% of repair times are hours or less hours is the 75 th percentile of furnace repair times. Solution E e. µ = a+b (b a) 2 and σ = 2 2 µ = = 2.75 hours and σ = (4.5) 2 2 = hours : Exercise 4 (Solution on p. 20.) The amount of time a service technician needs to change the oil in a car is uniformly

12 Connexions module: m distributed between and 2 minutes. Let X = the time needed to change the oil on a car. a.write the random variable X in words. X =. b.write the distribution. c.graph the distribution. d.find P (x > 9). e.find the 50 th percentile. Chapter Review 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). All values x are equally likely. We write X U(a, b). The mean of X is µ = a+b 2. (b a) The standard deviation of X is σ = 2 2. The probability density function of X is f (x) = b a for a x b. The cumulative distribution function of X is P(X x) = x a b a. X is continuous. Figure 2 The probability P(c < X < d) may be found by computing the area under f (x), between c and d. Since the corresponding area is a rectangle, the area may be found simply by multiplying the width and the height. 2 Formula Review X = a real number between a and b (in some instances, X can take on the values a and b). a = smallest X; b = largest X X U (a, b) The mean is µ = a+b 2 The standard deviation is σ = (b a) 2 2

13 Connexions module: m Probability density function:f (x) = b a ( for a ) X b Area to the Left of x:p(x < x) = (x a) b a ( ) Area to the Right of x:p(x > x) = (b x) b a ( Area Between c and d:p(c < x < d) = (base)(height) = (d c) Uniform: X U(a, b) where a < x < b pdf: f (x) = b a for a x b cdf: P(X x) = x a b a mean µ = a+b 2 standard deviation σ= P(c < X < d) = (d c) (b a) 2 ( 2 b a ) b a ) 3 References McDougall, John A. The McDougall Program for Maximum Weight Loss. Plume, Use the following information to answer the next ten questions. The data that follow are the square footage (in,000 feet squared) of 28 homes Table 3 The sample mean = 2.50 and the sample standard deviation = The distribution can be written as X U(.5, 4.5). Exercise 5 What type of distribution is this? Exercise 6 (Solution on p. 2.) In this distribution, outcomes are equally likely. What does this mean? Exercise 7 What is the height of f (x) for the continuous probability distribution? Exercise 8 (Solution on p. 2.) What are the constraints for the values of x? Exercise 9 Graph P(2 < x < 3). Exercise 20 (Solution on p. 2.) What is P(2 < x < 3)? Exercise 2 What is P(x < 3.5 x < 4)?

14 Connexions module: m Exercise 22 (Solution on p. 2.) What is P(x =.5)? Exercise 23 What is the 90 th percentile of square footage for homes? Exercise 24 (Solution on p. 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. Use the following information to answer the next eight exercises. A distribution is given as X U(0, 2). Exercise 25 What is a? What does it represent? Exercise 26 (Solution on p. 2.) What is b? What does it represent? Exercise 27 What is the probability density function? Exercise 28 (Solution on p. 2.) What is the theoretical mean? Exercise 29 What is the theoretical standard deviation? Exercise 30 (Solution on p. 2.) Draw the graph of the distribution for P(x > 9). Exercise 3 Find P(x > 9). Exercise 32 (Solution on p. 22.) Find the 40 th percentile. Use the following information to answer the next eleven exercises. The age of cars in the sta parking lot of a suburban college is uniformly distributed from six months (0.5 years) to 9.5 years. Exercise 33 What is being measured here? Exercise 34 (Solution on p. 22.) In words, dene the random variable X. Exercise 35 Are the data discrete or continuous? Exercise 36 (Solution on p. 22.) The interval of values for x is. Exercise 37 The distribution for X is. Exercise 38 (Solution on p. 22.) Write the probability density function. Exercise 39 Graph the probability distribution.

15 Connexions module: m a. Sketch the graph of the probability distribution. Figure 3 b. Identify the following values: i. Lowest value for x: ii. Highest value for x: iii. Height of the rectangle: iv. Label for x-axis (words): v. Label for y-axis (words): Exercise 40 (Solution on p. 22.) Find the average age of the cars in the lot. Exercise 4 Find the probability that a randomly chosen car in the lot was less than four years old. a. Sketch the graph, and shade the area of interest. Figure 4

16 Connexions module: m b. Find the probability. P(x < 4) = Exercise 42 (Solution on p. 22.) Considering only the cars less than 7.5 years old, nd the probability that a randomly chosen car in the lot was less than four years old. a. Sketch the graph, shade the area of interest. Figure 5 b. Find the probability. P(x < 4 x < 7.5) = Exercise 43 What has changed in the previous two problems that made the solutions dierent? Exercise 44 (Solution on p. 22.) Find the third quartile of ages of cars in the lot. This means you will have to nd the value such that 3 4, or 75%, of the cars are at most (less than or equal to) that age. a. Sketch the graph, and shade the area of interest. Figure 6 b. Find the value k such that P(x < k) = c. The third quartile is

17 Connexions module: m Homework For each probability and percentile problem, draw the picture. Exercise 45 Births are approximately uniformly distributed between the 52 weeks of the year. They can be said to follow a uniform distribution from one to 53 (spread of 52 weeks). a. X b. Graph the probability distribution. c. f (x) = d. µ = e. σ = f. Find the probability that a person is born at the exact moment week 9 starts. That is, nd P(x = 9) = g. P(2 < x < 3) = h. Find the probability that a person is born after week 40. i. P(2 < x x < 28) = j. Find the 70 th percentile. k. Find the minimum for the upper quarter. Exercise 46 (Solution on p. 22.) A random number generator picks a number from one to nine in a uniform manner. a. X b. Graph the probability distribution. c. f (x) = d. µ = e. σ = f. P(3.5 < x < 7.25) = g. P(x > 5.67) h. P(x > 5 x > 3) = i. Find the 90 th percentile. Exercise 47 According to a study by Dr. John McDougall of his live-in weight loss program at St. Helena Hospital, the people who follow his program lose between six and 5 pounds a month until they approach trim body weight. Let's suppose that the weight loss is uniformly distributed. We are interested in the weight loss of a randomly selected individual following the program for one month. a. Dene the random variable. X = b. X c. Graph the probability distribution. d. f (x) = e. µ = f. σ = g. Find the probability that the individual lost more than ten pounds in a month. h. Suppose it is known that the individual lost more than ten pounds in a month. Find the probability that he lost less than 2 pounds in the month. i. P(7 < x < 3 x > 9) =. State this in a probability question, similarly to parts g and h, draw the picture, and nd the probability.

18 Connexions module: m Exercise 48 (Solution on p. 23.) A subway train on the Red Line arrives every eight minutes during rush hour. We are interested in the length of time a commuter must wait for a train to arrive. The time follows a uniform distribution. a. Dene the random variable. X = b. X c. Graph the probability distribution. d. f (x) = e. µ = f. σ = g. Find the probability that the commuter waits less than one minute. h. Find the probability that the commuter waits between three and four minutes. i. Sixty percent of commuters wait more than how long for the train? State this in a probability question, similarly to parts g and h, draw the picture, and nd the probability. Exercise 49 The age of a rst grader on September at Garden Elementary School is uniformly distributed from 5.8 to 6.8 years. We randomly select one rst grader from the class. a. Dene the random variable. X = b. X c. Graph the probability distribution. d. f (x) = e. µ = f. σ = g. Find the probability that she is over 6.5 years old. h. Find the probability that she is between four and six years old. i. Find the 70 th percentile for the age of rst graders on September at Garden Elementary School. Use the following information to answer the next three exercises. The Sky Train from the terminal to the rentalcar and longterm parking center is supposed to arrive every eight minutes. The waiting times for the train are known to follow a uniform distribution. Exercise 50 (Solution on p. 23.) What is the average waiting time (in minutes)? a. zero b. two c. three d. four Exercise 5 Find the 30 th percentile for the waiting times (in minutes). a. two b. 2.4 c d. three Exercise 52 (Solution on p. 23.) The probability of waiting more than seven minutes given a person has waited more than four minutes is?

19 Connexions module: m a b c. 0.5 d Exercise 53 The time (in minutes) until the next bus departs a major bus depot follows a distribution with f (x) = 20 where x goes from 25 to 45 minutes. a. Dene the random variable. X = b. X c. Graph the probability distribution. d. The distribution is (name of distribution). It is (discrete or continuous). e. µ = f. σ = g. Find the probability that the time is at most 30 minutes. Sketch and label a graph of the distribution. Shade the area of interest. Write the answer in a probability statement. h. Find the probability that the time is between 30 and 40 minutes. Sketch and label a graph of the distribution. Shade the area of interest. Write the answer in a probability statement. i. P(25 < x < 55) =. State this in a probability statement, similarly to parts g and h, draw the picture, and nd the probability. j. Find the 90 th percentile. This means that 90% of the time, the time is less than minutes. k. Find the 75 th percentile. In a complete sentence, state what this means. (See part j.) l. Find the probability that the time is more than 40 minutes given (or knowing that) it is at least 30 minutes. Exercise 54 (Solution on p. 23.) Suppose that the value of a stock varies each day from $6 to $25 with a uniform distribution. a. Find the probability that the value of the stock is more than $9. b. Find the probability that the value of the stock is between $9 and $22. c. Find the upper quartile - 25% of all days the stock is above what value? Draw the graph. d. Given that the stock is greater than $8, nd the probability that the stock is more than $2. Exercise 55 A reworks show is designed so that the time between reworks is between one and ve seconds, and follows a uniform distribution. a. Find the average time between reworks. b. Find probability that the time between reworks is greater than four seconds. Exercise 56 (Solution on p. 24.) The number of miles driven by a truck driver falls between 300 and 700, and follows a uniform distribution. a. Find the probability that the truck driver goes more than 650 miles in a day. b. Find the probability that the truck drivers goes between 400 and 650 miles in a day. c. At least how many miles does the truck driver travel on the furthest 0% of days?

20 Connexions module: m Solutions to Exercises in this Module Solution to Exercise (p. 2) a is zero; b is 4; X U (0, 4); µ = 7 passengers; σ = 4.04 passengers Solution to Exercise (p. 5) P(2 < x < 8) = 0.8; 90 th percentile = 8 Solution to Exercise (p. 7) a. a is 447, and b is 52. a is the minimum duration of games for a team for the 20 season, and b is the maximum duration of games for a team for the 20 season. b. X U (447, 52). c. µ = 484, and σ = 2.36 Figure 7 d. P(480 < x < 500) = e. 65 th percentile is 495. hours. Solution to Exercise (p. 8) P (x > 8) = P (x > 8 x > 7) = Solution to Exercise (p. ) a. Let X = the time needed to change the oil in a car. b. X U (, 2).

21 Connexions module: m c. Figure 8 d. P (x > 9) = 0.2 e. the 50 th percentile is 6 minutes. Solution to Exercise (p. 3) It means that the value of x is just as likely to be any number between.5 and 4.5. Solution to Exercise (p. 3).5 x 4.5 Solution to Exercise (p. 3) Solution to Exercise (p. 4) zero Solution to Exercise (p. 4) 0.6 Solution to Exercise (p. 4) b is 2, and it represents the highest value of x. Solution to Exercise (p. 4) six Solution to Exercise (p. 4)

22 Connexions module: m Figure 9 Solution to Exercise (p. 4) 4.8 Solution to Exercise (p. 4) X = The age (in years) of cars in the sta parking lot Solution to Exercise (p. 4) 0.5 to 9.5 Solution to Exercise (p. 4) f (x) = 9 where x is between 0.5 and 9.5, inclusive. Solution to Exercise (p. 5) µ = 5 Solution to Exercise (p. 6) a. Check student's solution. b Solution to Exercise (p. 6) a. Check student's solution. b. k = 7.25 c Solution to Exercise (p. 7) a. X U(, 9) b. Check student's solution. c. f (x) = 8 where x 9 d. ve e. 2.3 f g h. 2 3 i. 8.2

23 Connexions module: m Solution to Exercise (p. 8) a. X represents the length of time a commuter must wait for a train to arrive on the Red Line. b. X U(0, 8) c. f (x) = 8 where x 8 d. four e. 2.3 f. 8 g. 8 h. 3.2 Solution to Exercise (p. 8) d Solution to Exercise (p. 8) b Solution to Exercise (p. 9) a. The probability density function of X is 25 6 = 9. P(X > 9) = (25 9) ( ) 9 = 6 9 = 2 3. Figure 20 b. P(9 < X < 22) = (22 9) ( ) 9 = 3 9 = 3.

24 Connexions module: m Figure 2 c. The area must be 0.25, and 0.25 = (width) ( 9), so width = (0.25)(9) = Thus, the value is = d. This is a conditional probability question. P(x > 2 x > 8). You can do this two ways: Draw the graph where a is now 8 and b is still 25. The height is (25 8) = 7 So, P(x > 2 x > 8) = (25 2) ( 7) = 4/7. Use the formula: P(x > 2 x > 8) = = P (x>2) P (x>8) = (25 2) (25 8) = 4 7. Solution to Exercise (p. 9) P (x>2 AND x>8) P (x>8) a. P(X > 650) = = = 8 = b. P(400 < X < 650) = = = c. 0.0 = width, so width = 400(0.0) = 40. Since = 660, the drivers travel at least 660 miles on the furthest 0% of days. Glossary Denition : Conditional Probability the likelihood that an event will occur given that another event has already occurred.