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ALTA - CHAPTER 5 - CONTINUOUS RANDOM VARIABLES

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Floretta is waiting for a tram. If X is the amount of time before the next tram arrives, and X is uniform with values between 1 and 7 minutes, then what is the approximate standard deviation for how long she will wait, rounded to one decimal place?

$\text{std=}1.7\text{min}$std=1.7min The standard deviation for the waiting time is σσ=(b−a)212−−−−−−−√=(b−a)12−−√ In this case we find that the standard deviation isσ=7−112−−√=612−−√=3-√≈1.7So the standard deviation for waiting is approximately 1.7 minutes.

On average, the waiting time to receive a custom ordered package from a company is 19 days. If the waiting time is exponentially distributed, what is the probability that the waiting time is between 5 and 9 days? (Find P(5<X<9).)

0.146 This probability P(5<x<9) asks us to find the area in between 5 and 9. We can also think of this area as the difference between the area to the left of 9 and the area to the left of 5. So, we can state: P(5<x<9)=P(x<9)−P(x<5) To find each of these probabilities, note that P(X<a)=1−e−λ(a), whereλ=1μ=119≈0.05263We can substitute the values of x=5, x=9, and λ=0.05263 and evaluate to find the probability.P(5<x<9)=P(x<9)−P(x<5)=(1−e−λ(9))−(1−e−λ(5))=(1−e−0.05263(9))−(1−e−0.05263(5))=(1−e−0.4737)−(1−e−0.2632)≈(1−0.6227)−(1−0.7686)≈0.3773−0.2314≈0.1459So, the probability that the waiting time is between 5 and 9 days is approximately 0.146 (or about 15%).

Let X be the waiting time to receive a custom ordered package from a company. Suppose X is exponentially distributed with average value 8 days. Find the probability that the waiting time is between 5 and 9 days.

0.211 This probability P(5<X<9) asks us to find the area in between 5 and 9. We can also think of this area as the difference between the area to the left of 9 and the area to the left of 5. So, we can state: P(5<X<9)=P(X<9)−P(X<5) To find each of these probabilities, note that P(X<a)=1−e−λ(a), whereλ=1μ=18≈0.125We can substitute the values of x=5, x=9, and λ=0.125 and evaluate to find the probability.P(5<x<9)=P(x<9)−P(x<5)=(1−e−λ(9))−(1−e−λ(5))=(1−e−0.125(9))−(1−e−0.125(5))=(1−e−1.125)−(1−e−0.625)≈(1−0.32465)−(1−0.53526)≈0.67535−0.46474≈0.211So, the probability that the wait time is between 5 and 9 days is approximately 0.211 (or about 21%).

The lifespan of a halogen light bulb is exponentially distributed with a mean of 1000 hours. Given that a halogen bulb has already worked for 400 hours, what is the probability that the bulb will continue to work for more than an additional 900 hours?

0.407 The exponential distribution has a memoryless property, stating that the probability of a time duration greater than k units from a given time is the same as the probability that the overall time duration is greater than k units.Mathematically, this can be written as P(X>k+t∣X>k)=P(X>t) for all k and t greater than or equal to zero. In this case, the probability that the light bulb will work for more than an additional 900 hours given that it has already worked for 400 hours is P(X>1300|X>400)=P(X>900)=1−(1−e−11000(900))≈0.407

If X∼U(5.5,18.5) is a continuous uniform random variable, what is P(X<9)?

7/26 We are interested in values less than 9. The probability is the area of the rectangle under this part of the curve. Area is the product of the base (length of this interval) and the height, which is the value of the probability density function(1b−a).So we findP(X<9)=(9−5.5)(1/18.5−5.5)=(3.5)(1/13)=7/26

The amount of time it takes Annie to make dinner is continuous and uniformly distributed between 19 minutes and 49 minutes. What is the probability that it takes Annie between 39 and 40 minutes given that it takes less than 44 minutes for her to make dinner?

$0.04$0.04 Let X be the time it takes to make dinner. The range of possible values is 19 to 49, so X∼U(19,49).We are given that it takes less than 44 minutes, so this means that the possible values are restricted to the range 19 to 44. This interval will help us write our probability density function (height of the rectangle). f(x)=144−19=125 We are interested in values greater than 39 and less than 40. So the range of desired outcomes is 39 to 40. This will be the length of our base, 40−39=1. The probability is the area of the shaded rectangle under the probability density function line, which is the product of the base and the height. So we findP(39<X<40∣∣X<44)=(40−39)(144−19)=(1)(125)=125=0.040

The amount of time it takes Josslyn to wait for the train is continuous and uniformly distributed between 4 minutes and 11 minutes. What is the probability that it takes Josslyn between 5 and 6 minutes given that it takes less than 8 minutes for her to wait for the train?

$0.25$0.25 Let X be the time it takes to wait for the train. The range of possible values is 4 to 11, so X∼U(4,11).We are given that it takes less than 8 minutes, so this means that the possible values are restricted to the range 4 to 8. This interval will help us write our probability density function (height of the rectangle). f(x)=18−4=14 We are interested in values greater than 5 and less than 6. So the range of desired outcomes is 5 to 6. This will be the length of our base, 6−5=1. The probability is the area of the shaded rectangle under the probability density function line, which is the product of the base and the height. So we findP(5<X<6∣∣X<8)=(6−5)(18−4)=(1)(14)=14=0.25

The waiting time to be seated at a restaurant during the evening is exponentially distributed with an average wait time of 24 minutes. Given that it has already taken 28 minutes, what is the probability that the wait time will be more than an additional 32 minutes? Round your answer to three decimal places.

$0.264$0.264 The exponential distribution has a memoryless property, stating that the probability of a time duration greater than k units from a given time is the same as the probability that the overall time duration is greater than k units. Mathematically, this can be written as P(X>k+t|X>k)=P(X>t) for all k and t greater than or equal to zero. In this case, the probability that the waiting time is more than an additional 32 minutes given that it has already taken 28 minutes is P(X>60|X>28)=P(X>32)=1−(1−e−124(32))≈0.264

On average, the time for a whale watcher to spot a whale is 14 minutes. If the time it takes to spot a whale is exponentially distributed, what is the probability that the time it takes to spot a whale is between 3 and 9 minutes? (Find P(3<X<9).) Round the final answer to three decimal places.

$0.281$0.281 This probability P(3<x<9) asks us to find the area in between 3 and 9. We can also think of this area as the difference between the area to the left of 9 and the area to the left of 3. So, we can state: P(3<x<9)=P(x<9)−P(x<3) To find each of these probabilities, note that P(X<a)=1−e−λ(a), whereλ=1μ=114≈0.07143We can substitute the values of x=3, x=9, and λ=0.07143 and evaluate to find the probability.P(3<x<9)=P(x<9)−P(x<3)=(1−e−λ(9))−(1−e−λ(3))=(1−e−0.07143(9))−(1−e−0.07143(3))=(1−e−0.6429)−(1−e−0.2143)≈(1−0.5258)−(1−0.8071)≈0.4742−0.1929≈0.2813So, the probability that the time it takes to spot a whale is between 3 and 9 minutes is approximately 0.281 (or about 28%).

Let X be the waiting time for a customer to talk to a customer service representative. Suppose X is exponentially distributed with an average value of 6 minutes. Find the probability that the wait time is between 6 and 15 minutes. Round the final answer to three decimal places.

$0.286$0.286 This probability P(6<x<15) asks us to find the area in between 6 and 15. We can also think of this area as the difference between the area to the left of 15 and the area to the left of 6. So, we can state: P(6<x<15)=P(x<15)−P(x<6) To find each of these probabilities, note that P(X<a)=1−e−λ(a), whereλ=1μ=16≈0.1667We can substitute the values of x=6, x=15, and λ=0.1667 and evaluate to find the probability.P(6<x<15)=P(x<15)−P(x<6)=(1−e−λ(15))−(1−e−λ(6))=(1−e−0.1667(15))−(1−e−0.1667(6))=(1−e−2.5)−(1−e−1)≈(1−0.08208)−(1−0.3679)≈0.9179−0.6321≈0.2858So, the probability that the wait time is between 6 and 15 minutes is approximately 0.286 (or about 29%).

The waiting time for a patient in a doctor's office is exponentially distributed with an average wait time of 18 minutes. What is the probability that the wait time is greater than 20 minutes? Round the final answer to three decimal places.

$0.329$0.329 An exponential distribution has a decay parameter λ=1μ where μ is the mean of the random variable. In this case, the random variable has a mean of 18, so the decay parameter is λ=118. The cumulative probability density function is given by P(X<x)=1−e−λx. Since λ=118, the probability that X is greater than 20 is P(X>20)=1−P(X<20)=1−(1−e−118(20))≈0.329

Let X be the waiting time for a customer to talk to a customer service representative. Suppose X is exponentially distributed with an average value of 16 minutes. Find the probability that the wait time is between 3 and 12 minutes. Round the final answer to three decimal places.

$0.357$0.357 This probability P(3<x<12) asks us to find the area in between 3 and 12. We can also think of this area as the difference between the area to the left of 12 and the area to the left of 3. So, we can state: P(3<x<12)=P(x<12)−P(x<3) To find each of these probabilities, note that P(X<a)=1−e−λ(a), whereλ=1μ=116≈0.0625We can substitute the values of x=3, x=12, and λ=0.0625 and evaluate to find the probability.P(3<x<12)=P(x<12)−P(x<3)=(1−e−λ(12))−(1−e−λ(3))=(1−e−0.0625(12))−(1−e−0.0625(3))=(1−e−0.75)−(1−e−0.1875)≈(1−0.4724)−(1−0.829)≈0.5276−0.171≈0.3567

The waiting time for a train to pass by is exponentially distributed with an average wait time of 31 minutes. Given that it has already taken 21 minutes, what is the probability that the wait time will be more than an additional 30 minutes? Round your answer to three decimal places.

$0.380$0.380 The exponential distribution has a memoryless property, stating that the probability of a time duration greater than k units from a given time is the same as the probability that the overall time duration is greater than k units. Mathematically, this can be written as P(X>k+t∣∣X>k)=P(X>t) for all k and t greater than or equal to zero. In this case, the probability that the waiting time is more than an additional 30 minutes given that it has already taken 21 minutes isP(X>51∣∣X>21)=P(X>30)=e−131(30)≈0.380

The waiting time to receive a custom ordered package from a company is exponentially distributed with an average wait time of 39 days. Given that it has already taken 28 days, what is the probability that the wait time will be more than an additional 37 days? Round your answer to three decimal places.

$0.387$0.387 The exponential distribution has a memoryless property, stating that the probability of a time duration greater than k units from a given time is the same as the probability that the overall time duration is greater than k units. Mathematically, this can be written as P(X>k+t∣∣X>k)=P(X>t) for all k and t greater than or equal to zero. In this case, the probability that the waiting time is more than an additional 37 days given that it has already taken 28 days isP(X>65∣∣X>28)=P(X>37)=e−139(37)≈0.387

The waiting time for a fire department to get called to a house fire is exponentially distributed with an average wait time of 26 hours. Given that it has already taken 16 hours, what is the probability that the wait time will be more than an additional 22 hours? Round your answer to three decimal places.

$0.429$0.429 The exponential distribution has a memoryless property, stating that the probability of a time duration greater than k units from a given time is the same as the probability that the overall time duration is greater than k units. Mathematically, this can be written as P(X>k+t∣∣X>k)=P(X>t) for all k and t greater than or equal to zero. In this case, the probability that the waiting time is more than an additional 22 hours given that it has already taken 16 hours isP(X>38∣∣X>16)=P(X>22)=e−126(22)≈0.429

If X∼U(8,17) follows a uniform distribution, what is the mean of X?

$\text{mean=}12.5$mean=12.5 The mean of a random variable with a uniform distribution U(a,b) is μ=a+b2 (this is the midpoint of the interval [a,b]). So in this case, a=8 and b=17, so the mean isa+b2=8+17/2=25/2=12.5

An exponential distribution is formed by the time it takes for a person to choose a birthday gift. The average time it takes for a person to choose a birthday gift is 41 minutes. Given that it has already taken 24 minutes for a person to choose a birthday gift,what is the probability that it will take more than an additional 34 minutes? Round your answer to three decimal places.

$0.436$0.436 The exponential distribution has a memoryless property, stating that the probability of a time duration greater than k units from a given time is the same as the probability that the overall time duration is greater than k units. Mathematically, this can be written as P(X>k+t∣∣X>k)=P(X>t) for all k and t greater than or equal to zero. In this case, the probability that the time it takes is more than an additional 34 minutes given that it has already taken 24 minutes isP(X>58∣∣X>24)=P(X>34)=e−141(34)≈0.436

An exponential distribution is formed by the waiting times for a train to pass by. The average waiting time for this distribution is 33 minutes. Given that it has already taken 19 minutes, what is the probability that the wait time will be more than an additional 27 minutes? Round your answer to three decimal places.

$0.441$0.441 The exponential distribution has a memoryless property, stating that the probability of a time duration greater than k units from a given time is the same as the probability that the overall time duration is greater than k units. Mathematically, this can be written as P(X>k+t∣∣X>k)=P(X>t) for all k and t greater than or equal to zero. In this case, the probability that the waiting time is more than an additional 27 minutes given that it has already taken 19 minutes isP(X>46∣∣X>19)=P(X>27)=e−133(27)≈0.441

The waiting time to receive a custom ordered package from a company is exponentially distributed with an average wait time of 27 days. What is the probability that the wait time is less than 17 days? Round the final answer to three decimal places.

$0.467$0.467 An exponential distribution has a decay parameter λ=1μ where μ is the mean of the random variable. In this case, the random variable has a mean of 27, so the decay parameter is λ=127. The cumulative probability density function is given by P(X<x)=1−e−λx. Since λ=127, the probability that X is less than 17 is P(X<17)=1−e−127(17)≈0.467

Let X be the time it takes for a whale watcher to spot a whale, where X has an average value of 39 minutes. If the random variable X is known to be exponentially distributed, what is the probability that X is greater than 29 minutes? Round the final answer to three decimal places.

$0.475$0.475 An exponential distribution has a decay parameter λ=1μ where μ is the mean of the random variable. In this case, the random variable has a mean of 39, so the decay parameter is λ=139. The cumulative probability density function is given by P(X<x)=1−e−λx. Since λ=139, the probability that X is greater than 29 is P(X>29)=1−P(X<29)=1−(1−e−139(29))≈0.475

Let X be the waiting time for a patient in a doctor's office, where X has an average value of 23 minutes. If the random variable X is known to be exponentially distributed, what is the probability that the wait time is greater than 17 minutes? Round the final answer to three decimal places.

$0.478$0.478 An exponential distribution has a decay parameter λ=1μ where μ is the mean of the random variable. In this case, the random variable has a mean of 23, so the decay parameter is λ=123. The cumulative probability density function is given by P(X<x)=1−e−λx. Since λ=123, the probability that X is greater than 17 is P(X>17)=1−P(X<17)=1−(1−e−123(17))≈0.478 Go to Excel Click blank cell type the formula =EXPON.DIST(17,1/23,TRUE) =1-EXPON.DIST(17,1/23,TRUE)

The random variable X is exponentially distributed, where X represents the waiting time for a patient in the emergency room. If X has an average value of 32 minutes, what is the probability that the wait time is less than 23 minutes? Round the final answer to three decimal places.

$0.513$0.513 An exponential distribution has a decay parameter λ=1μ where μ is the mean of the random variable. In this case, the random variable has a mean of 32, so the decay parameter is λ=132. The cumulative probability density function is given by P(X<x)=1−e−λx. Since λ=132, the probability that X is less than 23 is P(X<23)=1−e−132(23)≈0.513

The amount of time it takes Josslyn to go grocery shopping is continuous and uniformly distributed between 14 minutes and 43 minutes. What is the probability that it takes Josslyn between 18 and 25 minutes given that it takes less than 27 minutes for her to go grocery shopping? Round the final answer to three decimal places.

$0.538$0.538 Let X be the time it takes to go grocery shopping. The range of possible values is 14 to 43, so X∼U(14,43).We are given that it takes less than 27 minutes, so this means that the possible values are restricted to the range 14 to 27. This interval will help us write our probability density function (height of the rectangle). f(x)=127−14=113 We are interested in values greater than 18 and less than 25. So the range of desired outcomes is 18 to 25. This will be the length of our base, 25−18=7. The probability is the area of the shaded rectangle under the probability density function line, which is the product of the base and the height. So we findP(18<X<25∣∣X<27)=(25−18)(127−14)=(7)(113)=713≈0.538

The time it takes for math students to answer a particular math question is exponentially distributed with an average value of 24 seconds. What is the probability that the time it takes for math students to answer a particular math question is less than 20 seconds? Round the final answer to three decimal places.

$0.565$0.565 An exponential distribution has a decay parameter λ=1μ where μ is the mean of the random variable. In this case, the random variable has a mean of 24, so the decay parameter is λ=124. The cumulative probability density function is given by P(X<x)=1−e−λx. Since λ=124, the probability that X is less than 20 is P(X<20)=1−e−124(20)≈0.565 Go to Excel Click blank cell type the formula "=EXPON.DIST(20, 1/24,TRUE)" .565

The random variable X is exponentially distributed, where X represents the time it takes for a whale watcher to spot a whale. If X has an average value of 45 minutes, what is the probability that X is less than 45 minutes? Round the final answer to three decimal places.

$0.632$0.632 An exponential distribution has a decay parameter λ=1μ where μ is the mean of the random variable. In this case, the random variable has a mean of 45, so the decay parameter is λ=145. The cumulative probability density function is given by P(X<x)=1−e−λx. Since λ=145, the probability that X is less than 45 is P(X<45)=1−e−145(45)≈0.632

If X∼U(7,20) follows a uniform distribution, what is the mean of X?

$\text{mean=}13.5$mean=13.5 The mean of a random variable with a uniform distribution U(a,b) is μ=a+b2 (this is the midpoint of the interval [a,b]). So in this case, a=7 and b=20, so the mean isa+b2=7+20/2=27/2=13.5

Let X be the waiting time for a train to pass by, where X has an average value of 25 minutes. If the random variable X is known to be exponentially distributed, what is the probability that the wait time is less than 26 minutes? Round the final answer to three decimal places.

$0.647$0.647 An exponential distribution has a decay parameter λ=1μ where μ is the mean of the random variable. In this case, the random variable has a mean of 25, so the decay parameter is λ=125. The cumulative probability density function is given by P(X<x)=1−e−λx. Since λ=125, the probability that X is less than 26 is P(X<26)=1−e−125(26)≈0.647

Let X be the waiting time for a train to pass by, where X has an average value of 18 minutes. If the random variable X is known to be exponentially distributed, what is the probability that the wait time is less than 20 minutes? Round the final answer to three decimal places.

$0.671$0.671 An exponential distribution has a decay parameter λ=1μ where μ is the mean of the random variable. In this case, the random variable has a mean of 18, so the decay parameter is λ=118. The cumulative probability density function is given by P(X<x)=1−e−λx. Since λ=118, the probability that X is less than 20 is P(X<20)=1−e−118(20)≈0.671

Let X be the time it takes for math students to answer a particular math question, where X has an average value of 49 seconds. If the random variable X is known to be exponentially distributed, what is the probability that X is less than 55 seconds? Round the final answer to three decimal places

$0.675$0.675 An exponential distribution has a decay parameter λ=1μ where μ is the mean of the random variable. In this case, the random variable has a mean of 49, so the decay parameter is λ=149. The cumulative probability density function is given by P(X<x)=1−e−λx. Since λ=149, the probability that X is less than 55 is P(X<55)=1−e−149(55)≈0.675

The amount of time it takes Horace to do a math problem is continuous and uniformly distributed between 40 seconds and 84 seconds. What is the probability that it takes Horace less than 59 seconds given that it takes less than 62 seconds for him to do a math problem? Round the final answer to three decimal places.

$0.864$0.864 Let X be the time it takes to do a math problem. The range of possible values is 40 to 84, so X∼U(40,84).We are given that it takes less than 62 seconds, so this means that the possible values are restricted to the range 40 to 62. This interval will help us write our probability density function (height of the rectangle). f(x)=162−40=122 We are interested in values less than 59. So the range of desired outcomes is 40 to 59. This will be the length of our base, 59−40=19. The probability is the area of the shaded rectangle under the probability density function line, which is the product of the base and the height. So we findP(X<59∣∣X<62)=(59−40)(162−40)=(19)(122)=1922≈0.864

The amount of time it takes Jake to complete one homework problem from his Statistics class is continuous and uniformly distributed between 3 minutes and 13 minutes. What is the probability that it takes Jake greater than 11 minutes given that it takes greater than 9 minutes for him to complete one homework problem? Provide the final answer as a fraction.

$\frac{1}{2}$1/2 Let X be the time it takes Jake to complete one homework problem from his Statistics class. We are given that X is more than 9 minutes, so this means that the possible values are restricted to the range 9 to 13. This interval will help us write our probability density function (height of the rectangle). f(x)=113−9=14 We are interested in values greater than 11. So the range of desired outcomes is 11 to 13. This will be the length of our base, 13−11=2. The probability is the area of the shaded rectangle under the probability density function line, which is the product of the base and the height. So we findP(X>11∣∣X>9)=(13−11)(113−9)=(2)(14)=24=1/2

The amount of time it takes Michelle to make a grilled cheese sandwich is continuous and uniformly distributed between 3 minutes and 11 minutes. What is the probability that it takes Michelle between 4 and 5 minutes given that it takes less than 6 minutes for her to make a grilled cheese sandwich? Provide the final answer as a fraction.

$\frac{1}{3}$13 Let X be the time it takes to make a grilled cheese sandwich. We are given that X is less than 6 minutes, so this means that the possible values are restricted to the range 3 to 6. This interval will help us write our probability density function (height of the rectangle). f(x)=16−3=13 We are interested in values greater than 4 and less than 5. So the range of desired outcomes is 4 to 5. This will be the length of our base, 5−4=1. The probability is the area of the shaded rectangle under the probability density function line, which is the product of the base and the height. So we findP(4<X<5∣∣X<6)=(5−4)(16−3)=(1)(13)=13

The amount of time it takes Tom to make breakfast is continuous and uniformly distributed between 4 minutes and 11 minutes. What is the probability that it takes Tom between 5 and 6 minutes given that it takes less than 9 minutes for him to make breakfast? Provide the final answer as a fraction.

$\frac{1}{5}$15 Let X be the time it takes to make breakfast. We are given that X is less than 9 minutes, so this means that the possible values are restricted to the range 4 to 9. This interval will help us write our probability density function (height of the rectangle). f(x)=19−4=15 We are interested in values greater than 5 and less than 6. So the range of desired outcomes is 5 to 6. This will be the length of our base, 6−5=1. The probability is the area of the shaded rectangle under the probability density function line, which is the product of the base and the height. So we findP(5<X<6∣∣X<9)=(6−5)(19−4)=(1)(15)=15

The amount of time it takes Lexie to wait for the bus is continuous and uniformly distributed between 3 minutes and 16 minutes. What is the probability that it takes Lexie between 6 and 7 minutes given that it takes less than 11 minutes for her to wait for the bus? Provide the final answer as a fraction

$\frac{1}{8}$1/8 Let X be the time it takes to wait for the bus. The range of possible values is 3 to 16, so X∼U(3,16).We are given that it takes less than 11 minutes, so this means that the possible values are restricted to the range 3 to 11. This interval will help us write our probability density function (height of the rectangle). f(x)=111−3=18 We are interested in values greater than 6 and less than 7. So the range of desired outcomes is 6 to 7. This will be the length of our base, 7−6=1. The probability is the area of the shaded rectangle under the probability density function line, which is the product of the base and the height. So we findP(6<X<7∣∣X<11)=(7−6)(111−3)=(1)(18)=1/8

The random variable X has a uniform distribution with values between 19 and 32. What is the mean of X?

$\text{mean=}25.5$mean=25.5 The mean of a random variable with a uniform distribution U(a,b) is μ=a+b2 (this is the midpoint of the interval [a,b]). So in this case, a=19 and b=32, so the mean isa+b2=19+32/2=51/2=25.5

$\frac{1}{8}$18 Let X be the time it takes to wait for the bus. The range of possible values is 3 to 16, so X∼U(3,16).We are given that it takes less than 11 minutes, so this means that the possible values are restricted to the range 3 to 11. This interval will help us write our probability density function (height of the rectangle). f(x)=111−3=18 We are interested in values greater than 6 and less than 7. So the range of desired outcomes is 6 to 7. This will be the length of our base, 7−6=1. The probability is the area of the shaded rectangle under the probability density function line, which is the product of the base and the height. So we findP(6<X<7∣∣X<11)=(7−6)(111−3)=(1)(18)=18

The amount of time it takes Jon to wait for the bus is continuous and uniformly distributed between 3 minutes and 18 minutes. What is the probability that it takes Jon between 6 and 8 minutes given that it takes less than 16 minutes for him to wait for the bus? Provide the final answer as a fraction.

$\frac{2}{13}$2/13 Let X be the time it takes to wait for the bus. We are given that X is less than 16 minutes, so this means that the possible values are restricted to the range 3 to 16. This interval will help us write our probability density function (height of the rectangle). f(x)=116−3=113 We are interested in values greater than 6 and less than 8. So the range of desired outcomes is 6 to 8. This will be the length of our base, 8−6=2. The probability is the area of the shaded rectangle under the probability density function line, which is the product of the base and the height. So we findP(6<X<8∣∣X<16)=(8−6)(116−3)=(2)(113)=2/13

The amount of time it takes Judy to complete one homework problem from her Algebra class is continuous and uniformly distributed between 1 minutes and 9 minutes. What is the probability that it takes Judy greater than 7 minutes given that it takes greater than 6 minutes for her to complete one homework problem? Provide the final answer as a fraction.

$\frac{2}{3}$2/3 Let X be the time it takes to complete one homework problem from Algebra class. We are given that X is more than 6 minutes, so this means that the possible values are restricted to the range 6 to 9. This interval will help us write our probability density function (height of the rectangle). f(x)=1/9−6=1/3 We are interested in values greater than 7. So the range of desired outcomes is 7 to 9. This will be the length of our base, 9−7=2. The probability is the area of the shaded rectangle under the probability density function line, which is the product of the base and the height. So we findP(X>7∣∣X>6)=(9−7)(1/9−6)=(2)(1/3)=2/3

The amount of time it takes Jessica to do a math problem is continuous and uniformly distributed between 45 seconds and 76 seconds. What is the probability that it takes Jessica less than 48 seconds given that it takes less than 74 seconds for her to do a math problem? Provide the final answer as a fraction.

$\frac{3}{29}$329 Let X be the time it takes to do a math problem. The range of possible values is 45 to 76, so X∼U(45,76).We are given that it takes less than 74 seconds, so this means that the possible values are restricted to the range 45 to 74. This interval will help us write our probability density function (height of the rectangle). f(x)=174−45=129 We are interested in values less than 48. So the range of desired outcomes is 45 to 48. This will be the length of our base, 48−45=3. The probability is the area of the shaded rectangle under the probability density function line, which is the product of the base and the height. So we findP(X<48∣∣X<74)=(48−45)(174−45)=(3)(129)=329

The amount of time it takes George to wait for a taxi is continuous and uniformly distributed between 5 minutes and 12 minutes. What is the probability that it takes George less than 9 minutes given that it takes less than 10 minutes for him to wait for a taxi? Provide the final answer as a fraction.

$\frac{4}{5}$4/5 Let X be the time it takes to wait for a taxi. We are given that X is less than 10 minutes, so this means that the possible values are restricted to the range 5 to 10. This interval will help us write our probability density function (height of the rectangle). f(x)=110−5=15 We are interested in values less than 9. So the range of desired outcomes is 5 to 9. This will be the length of our base, 9−5=4. The probability is the area of the shaded rectangle under the probability density function line, which is the product of the base and the height. So we findP(X<9∣∣X<10)=(9−5)(110−5)=(4)(15)=4/5

The amount of time it takes Victoria to solve crossword puzzles can be modeled by a continuous and uniformly distributed random time between 4 minutes and 12 minutes. What is the probability that it will take Victoria greater than 8 minutes (total) for a puzzle that she has already been working on for 7 minutes? Provide the final answer as a fraction

$\frac{4}{5}$45 Let X be the time it takes to solve a crossword puzzle. We are told that the probability density function for X is uniform, so we can conclude that the conditional probability density function for X given that X>7 is also a uniform distribution. The possible values are restricted to the range 7 to 12. This interval will help us write the conditional probability density function (height of the rectangle). f(x)=112−7=15 We are interested in values greater than 8. So the range of desired outcomes is 8 to 12. This will be the length of the base, 12−8=4. The probability is the area of the shaded rectangle under the probability density function line, which is the product of the base and the height. So we findP(X>8∣∣X>7)=(12−8)(112−7)=(4)(15)=45

The amount of time it takes Jessica to wait for the train is continuous and uniformly distributed between 3 minutes and 11 minutes. What is the probability that it takes Jessica more than 6 minutes given that it takes more than 4 minutes for her to wait for the train? Provide the final answer as a fraction.

$\frac{5}{7}$5/7 Let X be the time it takes to wait for the train. The range of possible values is 3 to 11, so X∼U(3,11).We are given that it takes more than 4 minutes, so this means that the possible values are restricted to the range 4 to 11. This interval will help us write our probability density function (height of the rectangle). f(x)=111−4=17 We are interested in values greater than 6. So the range of desired outcomes is 6 to 11. This will be the length of our base, 11−6=5. The probability is the area of the shaded rectangle under the probability density function line, which is the product of the base and the height. So we findP(X>6∣∣X>4)=(11−6)(111−4)=(5)(17)=5/7

If X follows a uniform distribution between 19 and 33, what is the mean of X?

$\text{mean=}26$mean=26 The mean of a random variable with a uniform distribution U(a,b) is μ=a+b2 (this is the midpoint of the interval [a,b]). So in this case, a=19 and b=33, so the mean isa+b2=19+332=522=26

The amount of time it takes Felicia to eat an apple is continuous and uniformly distributed between 3 minutes and 18 minutes. What is the probability that it takes Felicia less than 9 minutes given that it takes less than 10 minutes to eat an apple? Provide the final answer as a fraction.

$\frac{6}{7}$6/7 Let X be the time it takes to eat an apple. We are given that X is less than 10 minutes, so this means that the possible values are restricted to the range 3 to 10. This interval will help us write our probability density function (height of the rectangle). f(x)=110−3=17 We are interested in values less than 9. So the range of desired outcomes is 3 to 9. This will be the length of our base, 9−3=6. The probability is the area of the shaded rectangle under the probability density function line, which is the product of the base and the height. So we findP(X<9∣∣X<10)=(9−3)(110−3)=(6)(17)=6/7

The amount of time it takes Evelyn to go grocery shopping is continuous and uniformly distributed between 20 minutes and 44 minutes. What is the probability that it takes Evelyn less than 38 minutes given that it takes less than 41 minutes for her to go grocery shopping? Provide the final answer as a fraction.

$\frac{6}{7}$67 Let X be the time it takes to go grocery shopping. The range of possible values is 20 to 44, so X∼U(20,44).We are given that it takes less than 41 minutes, so this means that the possible values are restricted to the range 20 to 41. This interval will help us write our probability density function (height of the rectangle). f(x)=141−20=121 We are interested in values less than 38. So the range of desired outcomes is 20 to 38. This will be the length of our base, 38−20=18. The probability is the area of the shaded rectangle under the probability density function line, which is the product of the base and the height. So we findP(X<38∣∣X<41)=(38−20)(141−20)=(18)(121)=1821=67

Let X be the waiting time for a fisherman to feel a bite on the fishing line, where X has an average value of 11 minutes. If the random variable X is known to be exponentially distributed, what are the parameters of this exponential distribution? Enter your answer for λ as a fraction.

$\lambda=1/11,\ \ \mu=11,\ \ \sigma=11$λ=1/11, μ=11, σ=11 An exponential distribution has a decay parameter λ=1μ where μ is the mean of the random variable. In this case, the random variable has a mean of 11 based on the information given, so the decay parameter is λ=111. The standard deviation, σ, of an exponential distribution is equal to the mean, so σ=11.

Let X be the waiting time for a fisherman to feel a bite on the fishing line, where X has an average value of 15 minutes. If the random variable X is known to be exponentially distributed, what are the parameters of this exponential distribution? Enter your answer for λ as a fraction.

$\lambda=1/15,\ \ \mu=15,\ \ \sigma=15$λ=1/15, μ=15, σ=15 An exponential distribution has a decay parameter λ=1μ where μ is the mean of the random variable. In this case, the random variable has a mean of 15 based on the information given, so the decay parameter is λ=115. The standard deviation, σ, of an exponential distribution is equal to the mean, so σ=15.

The amount of time it takes a caller to wait for customer service has an exponential distribution with an average time of 24 minutes. Identify the parameters of this exponential distribution. Enter your answer for λ as a fraction.

$\lambda=1/24,\ \ \mu=24,\ \ \sigma=24$λ=1/24, μ=24, σ=24 An exponential distribution has a decay parameter λ=1μ where μ is the mean of the random variable. In this case, the random variable has a mean of 24 based on the information given, so the decay parameter is λ=124. The standard deviation, σ, of an exponential distribution is equal to the mean, so σ=24 also.

Let X be the waiting time for a fire department to get called to a house fire, where X has an average value of 32 hours. If the random variable X is known to be exponentially distributed, what are the parameters of this exponential distribution? Enter your answer for λ as a fraction.

$\lambda=1/32,\ \ \mu=32,\ \ \sigma=32$λ=1/32, μ=32, σ=32 An exponential distribution has a decay parameter λ=1μ where μ is the mean of the random variable. In this case, the random variable has a mean of 32 based on the information given, so the decay parameter is λ=132. The standard deviation, σ, of an exponential distribution is equal to the mean, so σ=32.

The time it takes for a fisherman to catch a fish is exponentially distributed with an average wait time of 33 minutes. Identify the parameters of the exponential distribution. Enter your answer for λ as a fraction.

$\lambda=1/33,\ \ \mu=33,\ \ \sigma=33$λ=1/33, μ=33, σ=33 An exponential distribution has a decay parameter λ=1μ where μ is the mean of the random variable. In this case, the random variable has a mean of 33 based on the information given, so the decay parameter is λ=133. The standard deviation, σ, of an exponential distribution is equal to the mean, so σ=33.

The random variable X is exponentially distributed, where X represents the waiting time for a customer in line at the county recorder's office. If X has an average value of 40 seconds, what are the parameters of the exponential distribution? Enter your answer for λ as a fraction.

$\lambda=1/40,\ \ \mu=40,\ \ \sigma=40$λ=1/40, μ=40, σ=40 An exponential distribution has a decay parameter λ=1μ where μ is the mean of the random variable. In this case, the random variable has a mean of 40 based on the information given, so the decay parameter is λ=140. The standard deviation, σ, of an exponential distribution is equal to the mean, so σ=40.

Let X be the time it takes for math students to answer a particular math question, where X has an average value of 47 seconds. If the random variable X is known to be exponentially distributed, what are the parameters of this exponential distribution? Enter your answer for λ as a fraction.

$\lambda=1/47,\ \ \mu=47,\ \ \sigma=47$λ=1/47, μ=47, σ=47 An exponential distribution has a decay parameter λ=1μ where μ is the mean of the random variable. In this case, the random variable has a mean of 47 based on the information given, so the decay parameter is λ=147. The standard deviation, σ, of an exponential distribution is equal to the mean, so σ=47.

Alice is waiting for a shuttle. If X is the amount of time before the next shuttle arrives, and X is uniform with values between 6 and 17 minutes, then what is the average (mean) time for how long she will wait?

$\text{mean=}11.5\text{min}$mean=11.5min The mean of a random variable with a uniform distribution U(a,b) is μ=a+b2 (this is the midpoint of the interval [a,b]). So in this case, a=6 and b=17, so the mean isa+b2=6+172=232=11.5So the average waiting time is 11.5 minutes.

Floretta is waiting for a tram. If X is the amount of time before the next tram arrives, and X is uniform with values between 1 and 7 minutes, then what is the average (mean) time for how long she will wait?

$\text{mean=}4\text{min}$mean=4min The mean of a random variable with a uniform distribution U(a,b) is μ=a+b2 (this is the midpoint of the interval [a,b]). So in this case, a=1 and b=7, so the mean isa+b2=1+72=82=4So the average waiting time is 4 minutes.

Lexie is waiting for a train. If X is the amount of time before the next train arrives, and X is uniform with values between 4 and 11 minutes, then what is the average (mean) time for how long she will wait?

$\text{mean=}7.5\text{min}$mean=7.5min The mean of a random variable with a uniform distribution U(a,b) is μ=a+b2 (this is the midpoint of the interval [a,b]). So in this case, a=4 and b=11, so the mean isa+b2=4+112=152=7.5So the average waiting time is 7.5 minutes.

Lexie is waiting for a train. If X is the amount of time before the next train arrives, and X is uniform with values between 4 and 11 minutes, then what is the approximate standard deviation for how long she will wait, rounded to one decimal place?

$\text{std=}2.0\text{min}$std=2.0min The standard deviation for the waiting time is σσ=(b−a)212−−−−−−−√=(b−a)12−−√ In this case we find that the standard deviation isσ=11−412−−√=712−−√=73-√6≈2.0So the standard deviation for waiting is approximately 2.0 minutes.

If X∼U(8,17) follows a uniform distribution, what is the standard deviation of X? Round to 3 decimal places. Provide your answer below:

$\text{std=}2.598$std=2.598 The standard deviation is σσ=(b−a)212−−−−−−−√=(b−a)12−−√ In this case we find that the standard deviation isσ=17−8/12−−√=9/12−−√=33-√2≈2.598

Alice is waiting for a shuttle. If X is the amount of time before the next shuttle arrives, and X is uniform with values between 6 and 17 minutes, then what is the approximate standard deviation for how long she will wait, rounded to one decimal place

$\text{std=}3.2\text{min}$std=3.2min The standard deviation for the waiting time is σσ=(b−a)212−−−−−−−√=(b−a)12−−√ In this case we find that the standard deviation isσ=17−612−−√=1112−−√=113-√6≈3.2So the standard deviation for waiting is approximately 3.2 minutes.

If X∼U(7,20) follows a uniform distribution, what is the standard deviation of X, rounded to three decimal places?

$\text{std=}3.753$std=3.753 The standard deviation is σσ=(b−a)212−−−−−−−√=(b−a)12−−√ In this case we find that the standard deviation isσ=20−712−−√=1312−−√=133-√6≈3.753

The random variable X has a uniform distribution with values between 19 and 32. What is the standard deviation of X, rounded to three decimal places?

$\text{std=}3.753$std=3.753 The standard deviation is σσ=(b−a)212−−−−−−−√=(b−a)12−−√ In this case we find that the standard deviation isσ=32−19/12−−√=1312−−√=133-√6≈3.753

If X follows a uniform distribution between 19 and 33, what is the standard deviation of X, rounded to three decimal places?

$\text{std=}4.041$std=4.041 The standard deviation is σσ=(b−a)212−−−−−−−√=(b−a)12−−√ In this case we find that the standard deviation isσ=33−19/12−−√=1412−−√=73-√3≈4.041

The amount of time it takes Ariana to do a math problem is continuous and uniformly distributed between 38 seconds and 79 seconds. What is the probability that it takes Ariana between 58 and 70 seconds to do a math problem?

0.293 Let X be the time it takes to do a math problem. The range of possible values is 38 to 79, so X∼U(38,79).We are interested in values greater than 58 and less than 70. The probability is the area of the rectangle under this part of the curve. Area is the product of the base (length of this interval) and the height, which is the value of the probability density function(1b−a).So we findP(58<X<70)=(70−58)(179−38)=(12)(141)=0.293

The amount of time it takes Isabella to solve a Rubik's cube is continuous and uniformly distributed between 5 minutes and 11 minutes. What is the probability that it takes Isabella more than 9 minutes to solve a Rubik's cube?

0.333 Let X be the time it takes to solve a Rubik's cube. The range of possible values is 5 to 11, so X∼U(5,11). We are interested in values greater than 9. The probability is the area of the rectangle under this part of the curve. Area is the product of the base (length of this interval) and the height, which is the value of the probability density function(1b−a). So we find P(X>9)=(11−9)(111−5)=(2)(16)=0.333

An exponential distribution is formed by the waiting times for a train to pass by. The average waiting time for this distribution is 26 minutes. Given that it has already taken 19 minutes, what is the probability that the wait time will be more than an additional 23 minutes?

0.413 The exponential distribution has a memoryless property, stating that the probability of a time duration greater than k units from a given time is the same as the probability that the overall time duration is greater than k units. Mathematically, this can be written as P(X>k+t∣∣X>k)=P(X>t) for all k and t greater than or equal to zero. In this case, the probability that the waiting time is more than an additional 23 minutes given that it has already taken 19 minutes isP(X>42∣∣X>19)=P(X>23)=e−126(23)≈0.413

In 2011, New York's Emergency Medical Services estimated that they receive 180 calls in an hour on average. Let X be the time between 911 calls, where X has an average value of 20 seconds. If the random variable X is known to be exponentially distributed, what is the probability that the time between two calls is less than 12 seconds?

0.451 An exponential distribution has a decay parameter λ=1μ where μ is the mean of the random variable. In this case, the random variable has a mean of 20, so the decay parameter is λ=120=0.05. The cumulative probability density function is given byP(X<x)=1−e−λx. Since λ=0.05, the probability that X is less than 12 is P(X<12)=1−e−0.05(12)≈0.451.

The amount of time it takes Marc to make dinner is continuous and uniformly distributed between 24.5 minutes and 53.5 minutes. What is the probability that it takes Marc more than 46 minutes to make dinner?

15/58 Let X be the time it takes to make dinner. The range of possible values is 24.5 to 53.5, so X∼U(24.5,53.5).We are interested in values greater than 46. The probability is the area of the rectangle under this part of the curve. Area is the product of the base (length of this interval) and the height, which is the value of the probability density function(1b−a).So we findP(X>46)=(53.5−46)(153.5−24.5)=(7.5)(129)=1558

The time it takes for a person to choose a birthday gift is exponentially distributed with an average value of 30 minutes. Given that it has already taken 12 minutes for a person to choose a birthday gift, what is the probability that it will take more than an additional 22 minutes?

0.480 The exponential distribution has a memoryless property, stating that the probability of a time duration greater than k units from a given time is the same as the probability that the overall time duration is greater than k units. Mathematically, this can be written as P(X>k+t∣∣X>k)=P(X>t) for all k and t greater than or equal to zero. In this case, the probability that the time it takes is more than an additional 22 minutes given that it has already taken 12 minutes isP(X>34∣∣X>12)=P(X>22)=e−130(22)≈0.480

The amount of time it takes Evelyn to solve a Rubik's cube is continuous and uniformly distributed between 3 minutes and 12 minutes. What is the probability that it takes Evelyn more than 8 minutes given that it takes more than 4 minutes for her to solve a Rubik's cube?

0.500 Let X be the time it takes to solve a Rubik's cube. The range of possible values is 3 to 12, so X∼U(3,12). We are given that it takes more than 4 minutes, so this means that the possible values are restricted to the range 4 to 12. This interval will help us write our probability density function (height of the rectangle). f(x)=1/12−4=1/8 We are interested in values greater than 8. So the range of desired outcomes is 8 to 12. This will be the length of our base, 12−8=4. The probability is the area of the shaded rectangle under the probability density function line, which is the product of the base and the height. So we find P(X>8∣∣X>4)=(12−8)(112−4)=(4)(18)=48=0.500

0.500 Let X be the time it takes to solve a Rubik's cube. The range of possible values is 3 to 12, so X∼U(3,12). We are given that it takes more than 4 minutes, so this means that the possible values are restricted to the range 4 to 12. This interval will help us write our probability density function (height of the rectangle). f(x)=112−4=18 We are interested in values greater than 8. So the range of desired outcomes is 8 to 12. This will be the length of our base, 12−8=4. The probability is the area of the shaded rectangle under the probability density function line, which is the product of the base and the height. So we find P(X>8∣∣X>4)=(12−8)(112−4)=(4)(18)=48=0.500

The amount of time it takes Jamie to wait in line at the bank is continuous and uniformly distributed between 8 minutes and 19 minutes. What is the probability that it takes Jamie less than 12 minutes, given that it will take her less than 15 minutes for her to wait in line?

0.571 Let X be the time it takes to wait in line at the bank. The range of possible values is 8 to 19, so X∼U(8,19). We are given that it takes Jamie less than 15 minutes in line, so this means that the possible values are restricted to the range 8 to 15. This interval will help us write our probability density function (height of the rectangle). f(x)=115−8=17 . We are interested in values less than 12. So the range of desired outcomes is 8 to 12. This will be the length of our base, 12−8=4. The probability is the area of the shaded rectangle under the probability density function line, which is the product of the base and the height. So we find P(X<12∣∣X<15)=(12−8)(115−8)=(4)(17)=47≈0.571

Let X be the waiting time for a patient in a doctor's office, where X has an average value of 25 minutes. If the random variable X is known to be exponentially distributed, what is the probability that the wait time is less than 35 minutes?

0.753 An exponential distribution has a decay parameter λ=1μ where μ is the mean of the random variable. In this case, the random variable has a mean of 25, so the decay parameter is λ=125. The cumulative probability density function is given by P(X<x)=1−e−λx. Since λ=125, the probability that X is less than 35 is P(X<35)=1−e−125(35)≈0.753

The figure above shows the probability density function for the random variable x. The shaded area represents P(1≤x≤5). What is P(1≤x≤5)?

0.92 The probability that x is between 1 and 5 is equal to the area of the shaded region. Notice that the shaded region can be separated into a right triangle from the upper portion and a rectangle from the lower portion. The rectangle has dimensions 4 by 0.11, so its area is 4⋅0.11=0.44. The triangle has a width of 4 and a height of 0.24, so its area is 12(4)(0.24)=0.48. The probability that x is between 1 and 5 is the sum of these two areas: 0.44+0.48=0.92.

If X∼U(2.5,8.5) is a continuous uniform random variable, what is P(X<3)?

1/12 We are interested in values less than 3. The probability is the area of the rectangle under this part of the curve. Area is the product of the base (length of this interval) and the height, which is the value of the probability density function(1b−a).So we findP(X<3)=(3−2.5)(18.5−2.5)=(0.5)(1/6)=1/12

If X is a continuous uniform random variable with values between 1.5 and 12.5, what is P(X<2)?

1/22 We are interested in values less than 2. The probability is the area of the rectangle under this part of the curve. Area is the product of the base (length of this interval) and the height, which is the value of the probability density function(1b−a).So we find P(X<2)=(2−1.5)(1/12.5−1.5)=(0.5)(1/11)=1/22

The probability density function for the random variable X is shown below. Find P(X>5)

1/4 Remember that the probability of a range of values is given by the area under the graph over that range of values. So we want the area under the curve for the values of x greater than 5. Note that this region is a triangle, with base 10−5=5 and height 110, so the area (and hence the probability) is P(X>5)=1/2(5)(1/10)=1/4

The probability density function for the random variable X is shown below. Find P(X>11).

1/9 Remember that the probability of a range of values is given by the area under the graph over that range of values. So we want the area under the curve for the values of x greater than 11. A coordinate system has a horizontal x-axis labeled from 0 to 15 in increments of 1 and a vertical y-axis labeled at StartFraction 1 Over 18 EndFraction and StartFraction 1 Over 12 EndFraction. A random probability function f left-parenthesis x right-parenthesis is depicted by two solid line segments. The first solid line segment connects the points left-parenthesis 0 comma StartFraction 1 Over 12 EndFraction right-parenthesis and left-parenthesis 9 comma StartFraction 1 Over 12 EndFraction right-parenthesis. The second solid line segment decreases from left to right connecting the points left-parenthesis 9 comma StartFraction 1 Over 12 EndFraction right-parenthesis to the point left-parenthesis 15 comma 0 right-parenthesis. A dashed horizontal line segment goes from the point left-parenthesis 0 comma StartFraction 1 Over 18 EndFraction right-parenthesis to the point left-parenthesis 11 comma StartFraction 1 Over 18 EndFraction right-parenthesis. A vertical dashed line segment connects the points left-parenthesis 9 comma 0 right-parenthesis and left-parenthesis 9 comma StartFraction 1 Over 12 EndFraction right-parenthesis. Another vertical dashed line segment connects the points left-parenthesis 11 comma 0 right-parenthesis and left-parenthesis 11 comma StartFraction 1 Over 18 EndFraction right-parenthesis. The triangular region enclosed by the points left-parenthesis 11 comma StartFraction 1 Over 18 EndFraction right-parenthesis, left-parenthesis 11 comma 0 right-parenthesis, and left-parenthesis 15 comma 0 right-parenthesis is shaded. Note that this region is a triangle, with base 15−11=4 and height 118, so the area (and hence the probability) is P(X>11)=12(4)(118)=1/9

The amount of time it takes Lexie to go grocery shopping is continuous and uniformly distributed between 14.5 minutes and 43.5 minutes. What is the probability that it takes Lexie between 19 and 41 minutes to go grocery shopping?

22/29 Let X be the time it takes to go grocery shopping. The range of possible values is 14.5 to 43.5, so X∼U(14.5,43.5). We are interested in values greater than 19 and less than 41. The probability is the area of the rectangle under this part of the curve. Area is the product of the base (length of this interval) and the height, which is the value of the probability density function(1b−a). So we find P(19<X<41)=(41−19)(143.5−14.5)=(22)(129)=22/29

If X∼U(4.5,18.5) is a continuous uniform random variable, what is P(X>6)?

25/28 We are interested in values greater than 6. The probability is the area of the rectangle under this part of the curve. Area is the product of the base (length of this interval) and the height, which is the value of the probability density function(1b−a).So we find P(X>6)=(18.5−6)(118.5−4.5)=(12.5)(114)=25/28

The probability density function for the random variable X is shown below. Find P(X>4).

25/81 Remember that the probability of a range of values is given by the area under the graph over that range of values. So we want the area under the curve for the values of x greater than 4. A coordinate system has a horizontal x-axis labeled from 0 to 9 in increments of 1 and a vertical y-axis labeled at 0, StartFraction 10 Over 81 EndFraction, and StartFraction 2 Over 9 EndFraction. A solid line segment labeled f left-parenthesis x right-parenthesis falls from the point left-parenthesis 0 comma StartFraction 2 Over 9 EndFraction right-parenthesis to the point left parenthesis 9 comma 0 right-parenthesis. A vertical dashed line segment extends from the point left-parenthesis 4 comma 0 right-parenthesis to the point left-parenthesis 4 comma StartFraction 10 Over 81 EndFraction right-parenthesis, and a horizontal dashed line segment connects the points left-parenthesis 0 comma StartFraction 10 Over 81 EndFraction right-parenthesis to the point left-parenthesis 4 comma StartFraction 10 Over 81 EndFraction right-parenthesis. The region between the points left-parenthesis 4 comma 0 right-parenthesis and left-parenthesis 9 comma 0 right-parenthesis, below the solid line segment, and above the x-axis is shaded. Note that this region is a triangle, with base 9−4=5 and height 1081, so the area (and hence the probability) isP(X>4)=12(5)(1081)=25/81

The probability density function for the random variable X is shown below. Find P(X<2).

29 Remember that the probability for a range of values is equal to the area under the curve. So we want the area under the curve for the values of x less than 2. A coordinate system has a horizontal x-axis labeled from 0 to 11 in increments of 1 and a vertical y-axis labeled from 0 to StartFraction 1 Over 9 EndFraction. A random probability function f left-parenthesis x right-parenthesis is depicted by two solid line segments. The first solid line segment goes from the point left-parenthesis 0 comma StartFraction 1 Over 9 EndFraction right-parenthesis to the point left-parenthesis 8 comma StartFraction 1 Over 9 EndFraction right-parenthesis. The second solid line segment falls from left to right from the point left-parenthesis 8 comma StartFraction 1 Over 9 EndFraction right-parenthesis to the point left-parenthesis 10 comma 0 right-parenthesis. A vertical dashed line extends from the point left-parenthesis 8 comma 0 right-parenthesis to the point left-parenthesis 8 comma StartFraction 1 Over 9 EndFraction right-parenthesis. The region between the points left-parenthesis 0 comma 0 right-parenthesis and left-parenthesis 2 comma 0 right-parenthesis, below the solid horizontal line segment, and above the x-axis is shaded. Note that this region is a rectangle, with base 2 and height 19, so the area (and hence the probability) isP(X<2)=(base)(height)=(2)(19)=29So we find that P(X<2)=29.

The amount of time it takes William to wait for the bus is continuous and uniformly distributed between 4.5 minutes and 14.5 minutes. What is the probability that it takes William more than 7 minutes to wait for the bus?

3/4 Let X be the time it takes to wait for the bus. The range of possible values is 4.5 to 14.5, so X∼U(4.5,14.5).We are interested in values greater than 7. The probability is the area of the rectangle under this part of the curve. Area is the product of the base (length of this interval) and the height, which is the value of the probability density function(1b−a).So we findP(X>7)=(14.5−7)(114.5−4.5)=(7.5)(110)=3/4

The probability density function for the random variable X is shown below. Find P(X<6).

6/7 Remember that the probability for a range of values is equal to the area under the curve. So we want the area under the curve for the values of x less than 6. A coordinate system has a horizontal axis labeled from 0 to 9 in increments of 1 and a vertical axis labeled from 0 to StartFraction 1 Over 7 EndFraction. A solid line segment labeled f left-parenthesis x right-parenthesis starts at the point left parenthesis 0 comma StartFraction 1 Over 7 EndFraction right-parenthesis and extends horizontally to the right until it reaches the point left-parenthesis 6 comma StartFraction 1 Over 7 EndFraction right-parenthesis. The line segment then decreases from left to right connecting the points left-parenthesis 6 comma StartFraction 1 Over 7 EndFraction right-parenthesis and left-parenthesis 8 comma StartFraction 1 Over 7 EndFraction right-parenthesis. A dashed vertical line segment connects the points left-parenthesis 6 comma 0 right-parenthesis and left-parenthesis 6 comma StartFraction 1 Over 7 EndFraction right-parenthesis. A rectangular region to the left of the vertical line segment and underneath the solid line segment is shaded. Note that this region is a rectangle, with base 6 and height 17, so the area (and hence the probability) isP(X<6)=(base)(height)=(6)(17)=67So we find that P(X<6)=67.

The amount of time it takes Isabella to do a math problem is continuous and uniformly distributed between 38.5 seconds and 79.5 seconds. What is the probability that it takes Isabella between 51 and 60 seconds to do a math problem?

9/41 Let X be the time it takes to do a math problem. The range of possible values is 38.5 to 79.5, so X∼U(38.5,79.5).We are interested in values greater than 51 and less than 60. The probability is the area of the rectangle under this part of the curve. Area is the product of the base (length of this interval) and the height, which is the value of the probability density function(1b−a).So we findP(51<X<60)=(60−51)(179.5−38.5)=(9)(141)=9/41

What is the probability that a computer part lasts exactly 7 years? (Find P(X=7).)

We might be tempted in a situation like this to plug in 7 to the probability density function f(x)=0.1e−0.1x. However, because the exponential distribution is continuous, the probability of any particular value is 0. So P(X=7)=0. Another way to think about this is to remember that with continuous distributions, the probability of a range of values is represented by the area under the probability density function curve. At a single value, the width of the area under the curve is 0 (because a point has no width), so the area is 0.

The probability density function for the random variable X is shown below. Find P(X>5).

9/64 Remember that the probability of a range of values is given by the area under the graph over that range of values. So we want the area under the curve for the values of x greater than 5. A coordinate system has a horizontal x-axis labeled from 0 to 9 in increments of 1 and a vertical y-axis labeled at 0, StartFraction 3 Over 32 EndFraction, and StartFraction 1 Over 4 EndFraction. A solid line segment labeled f left-parenthesis x right-parenthesis falls from the point left-parenthesis 0 comma StartFraction 1 Over 4 EndFraction right-parenthesis to the point left-parenthesis 8 comma 0 right-parenthesis. A horizontal dashed line segment extends from the point left-parenthesis 0 comma StartFraction 3 Over 32 EndFraction right-parenthesis to the point left-parenthesis 5 comma StartFraction 3 Over 32 EndFraction right-parenthesis, and a vertical dashed line segment connects the points left-parenthesis 5 comma 0 right-parenthesis and left-parenthesis 5 comma StartFraction 3 Over 32 EndFraction right-parenthesis.. The area to the right of the point left-parenthesis 5 comma 0 right-parenthesis, below the solid line segment, and above the horizontal axis is shaded. Note that this region is a triangle, with base 8−5=3 and height 332, so the area (and hence the probability) isP(X>5)=12(3)(332)=9/64

Continuous distribution function (abbreviated as cdf): a probability distribution in which the variable of interest can be any value Exponential distribution: a continuous distribution that has an exponential shape, often concerning the amount of time until an event occursExponential distribution is also known negative exponential distribution Decay parameter: a value that relates the rate at which probability tends toward zero, denoted by the Greek letter λ Memoryless property: knowing what happened in the past has no effect on future probabilitiesMemoryless property is also known as forgetfulness property

Exponential distribution: a continuous distribution that has an exponential shape, often concerning the amount of time until an event occursIn many sources, the Exponential distribution is called the Negative exponential distribution Decay parameter: a value that relates the rate at which probability tends toward zero, denoted by the Greek letter λ

Suppose it takes a child between 0.5 and 4 minutes to eat a donut. What is the probability that a different child eats a donut in greater than 2 minutes, given that the child has already been eating the donut for more than 1.5 minutes?

Let X = the time, in minutes, it takes a child. Then X∼U(0.5,4). We know this is conditional probability because there is a given condition: "given that the child has already been eating the donut for more than 1.5 minutes." Since we know the child has already been eating the donut for more than 1.5 minutes, we are no longer starting at a=0.5 minutes. The starting point is 1.5 minutes. This condition affects: the probability density function (height) the interval (base) So, the probability density function is calculated using the new interval. 1.5 to 4. f(x)=14−1.5=25 This is theheightof the desired area. The base is the distance of the interval that we are looking for, from 2 to 4. A coordinate system has a horizontal x-axis labeled at 0, 1.5, 2, and 4 and a vertical y-axis labeled from 0 to StartFraction 2 Over 5 EndFraction. Vertical line segments at x-axis values 1.5 and 4 extend up from the horizontal axis to y-axis value StartFraction 2 Over 5 EndFraction. A horizontal line segment at y-axis value StartFraction 2 Over 5 EndFraction connects the two vertical line segments. The area under the horizontal line segment, between x-axis values 2 and 4, and above the horizontal axis is shaded. The probability P(x>2∣x>1.5) is the product of the base and the height (because we are looking for the area of a rectangle). P(x>2∣x>1.5)=(4−2)(14−1.5)=(2)(25)=(45) So, the probability that a different child eats a donut in greater than 2 minutes, given that the child has already been eating the donut for more than 1.5 minutes is 45=0.8 or 80%.

The amount of time it takes Hugo to solve a Rubik's cube is continuous and uniformly distributed between 4 minutes and 11 minutes. What is the probability that it takes Hugo less than 5 minutes given that it takes less than 6 minutes for him to solve a Rubik's cube? Provide the final answer as a fraction.

Let X be the time it takes to solve a Rubik's cube. The range of possible values is 4 to 11, so X∼U(4,11).We are given that it takes less than 6 minutes, so this means that the possible values are restricted to the range 4 to 6. This interval will help us write our probability density function (height of the rectangle). f(x)=16−4=12 We are interested in values less than 5. So the range of desired outcomes is 4 to 5. This will be the length of our base, 5−4=1. The probability is the area of the shaded rectangle under the probability density function line, which is the product of the base and the height. So we find P(X<5∣∣X<6)=(5−4)(16−4)=(1)(12)=1/2

Let X be the time it takes to solve a crossword puzzle. We are told that the probability density function for X is uniform, so we can conclude that the conditional probability density function for X given that X>7 is also a uniform distribution. The possible values are restricted to the range 7 to 12. This interval will help us write the conditional probability density function (height of the rectangle). f(x)=112−7=15 We are interested in values greater than 8. So the range of desired outcomes is 8 to 12. This will be the length of the base, 12−8=4. The probability is the area of the shaded rectangle under the probability density function line, which is the product of the base and the height. So we find P(X>8∣∣X>7)=(12−8)(1/12−7)=(4)(1/5)=4/5

The amount of time it takes John to wait in line at the bank is continuous and uniformly distributed between 7 minutes and 17 minutes. What is the probability that it takes John between 9 and 12 minutes to wait in line at the bank?

Let X be the time it takes to wait in line at the bank. The range of possible values is 7 to 17, so X∼U(7,17).We are interested in values greater than 9 and less than 12. The probability is the area of the rectangle under this part of the curve. Area is the product of the base (length of this interval) and the height, which is the value of the probability density function(1b−a).So we find P(9<X<12)=(12−9)(117−7)=(3)(110)=0.300

The height of a particular type of potted bamboo plant can grow to around 4 feet in height, but can vary. Let H be a random variable that represents the height of a potted bamboo plant. The probability density function of H is given below. Move the left and right sliders so that the area of the blue shaded region represents the probability that a potted bamboo plant is at least 4 feet tall, but no more than 6 feet tall. Using the information you obtained from the graph of the continuous probability density function above, what is the probability that a potted bamboo plant is at least 4 feet tall, but no more than 6 feet tall? Round your answer to the nearest tenth of a percent.

Since H is the random variable that represents the height of a potted bamboo plant, we want H to be at least 4, but no more than 6. That is, we want P(4≤H≤6). To compute this, we need to find the area under the probability density function of H between 4 feet tall and 6 feet tall. So, the left slider should be at 4, and the right slider should be at 6. $31.1\text{ %}$31.1 % The probability that a potted bamboo plant is at least 4 feet tall, but no more than 6 feet tall is equal to the area beneath its density curve between 4 and 6. From the graph above, we see that the area is 0.31113753. Writing as a percent and rounding to the nearest tenth, we have P(4≤H≤6)=31.1%.

What is the probability that a computer part lasts between 9 and 11 years? (Find P(9<X<11).)

This probability P(9<X<11) asks us to find the area in between 9 and 11. The graph below shows the curve of the probability density function and the shaded area that represents the desired probability. We can also think of this area as the difference between the area to the left of 11 and the area to the left of 9. So, we can state: P(9<X<11)=P(X<11)−P(X<9) We can substitute the values of x=11 and x=9 and λ=0.10 to find the probability. P(9<X<11)=P(X<11)−P(X<9)=(1−e−λ(11))−(1−e−λ(9))=(1−e−0.10(11))−(1−e−0.10(9))=(1−e−1.1)−(1−e−0.9)≈(1−0.334)−(1−0.407)≈0.666−0.593≈0.073 So, the probability that a computer part lasts between 9 and 11 years is approximately 0.073 (or about 7%).

What is the probability that a computer part lasts at most 7 years? (Find P(X<7).)

This probability P(X<7) asks us to find the area to the left of 7. The graph below shows the curve of the probability density function and the shaded area that represents the desired probability. We can find the area to the left of the curve by using the cumulative distribution function: P(X<x)=1−e−λx. We can substitute 7 in for x to find the probability. P(X<x)=1−e−λx=1−e−0.10(7)=1−e−0.70≈1−0.497≈0.503 So, the probability that a computer part lasts at most 7 years is approximately 0.503 (or about 50%).

What is the probability that a computer part lasts at least 7 years? (Find P(X>7).)

This probability P(X>7) asks us to find the area to the right of 7. The graph below shows the curve of the probability density function and the shaded area that represents the desired probability. In order to find P(X>7), we need to use the complement rule, since we are looking for the area to the right of the value. So we are subtracting the area to the left of 7 from 1 (the greatest possible probability). We can find the area to the right of the curve by using the complement rule: P(X>x)=1−P(X<x)=1−(1−e−λx) We can substitute the values of x=7 and λ=0.10 to find the probability. P(X<x)=1−(1−e−λx)=1−(1−e−0.10(7))=1−(1−e−0.70)≈1−(1−0.497)&≈1−(0.503)&≈0.497 So, the probability that a computer part lasts at least 7 years is approximately 0.497 (or about 49.7%).

Uniform distribution: happens when each of the values within an interval are equally likely to occur, so each value has the exact same probability as the others over the entire interval givenA Uniform distribution may also be referred to as a Rectangular distribution

Let X be the time it takes for a person to choose a birthday gift, where X has an average value of 27 minutes. If the random variable X is known to be exponentially distributed, what are the parameters of this exponential distribution?

λ=127, μ=27, σ=27 An exponential distribution has a decay parameter λ=1μ where μ is the mean of the random variable. In this case, the random variable has a mean of 27 based on the information given, so the decay parameter is λ=127. The standard deviation, σ, of an exponential distribution is equal to the mean, so σ=27.

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