Checkpoint: Random Variables 1 and 2

¡Supera tus tareas y exámenes ahora con Quizwiz!

A building has two elevators, an express elevator and a standard elevator. Let X be the number of passengers using the express elevator every hour, and let Y be the number of passengers using the standard elevator every hour. Assuming that μX = 55 and μY = 32, what is the mean of Z, the total number of passengers using the elevators each hour? 23 32 55 87

87 Feedback Correct. Since Z = X + Y, μZ = μX + μY = 55 + 32 = 87.

The Coff-E-Cup beverage company has opened two franchises in town. They put one location in the center of town and one at the west end of town. Let X be the number of customers that enter the center of town location in an hour, and Y be the number of customers that enter the location at the west end of town in an hour. Assuming that μX = 65 with σX = 4.3 and μY = 45 with σY = 3.2 , what is the mean of Z, the total number of customers that enter these locations in an hour? 110 65 20 45 55

Feedback Correct. Since Z = X + Y, μZ = μX + μY = 65 + 45 = 110.

The number of people entering a state park is a random variable X with a mean of 3 and a variance of 2. The admission to the park is $6 per car plus an additional $2 per person in the car. Which of the following is an expression for the total amount of money collected for a car entering the park? 2X + 6 3X + 6 3X + 2 6X + 2

Feedback Correct. The amount of money collected for a car entering the park can be represented as 2X + 6.

Salary of an employee at a large company is a random variable X with a mean of $50,000 and a variance of $2,700. The CEO of the company decides to give every employee a 2% raise and a $500 bonus. Which of the following is an expression for an employee's new salary? 500 + 0.02X 500 + 1.20X 1.02 + 500X 500 + 1.02X

Feedback Incorrect. To increase the salary by 2%, we must ADD 2% of the original salary, X, to the original salary; that is, X + 0.02X = 1.02X.

A parking garage has two entrances. Let X be the number of cars that enter the garage through door A in an hour, and Y be the number of cars that enter through door B in an hour. Assuming that μX = 15 and μY = 25, what is the mean of Z, the total number of cars that enter the garage in an hour? 10 15 25 40

10 Incorrect. Recall that since Z = X + Y, μZ = μX + μY. 10 is μY − μX.

The number of times that a hiker walks over 8 miles each day is what type of data random variable? Discrete Continuous Feedback Correct. The number of times that a hiker walks is countable and therefore is a discrete data random variable.

Discrete Correct. The number of times that a hiker walks is countable and therefore is a discrete data random variable.

Suppose Blue Cab Company charges $2.85 a ride up to 0.1 miles and $0.30 for each additional tenth of a mile. If the mean distance a passenger wants to go is 5.3 miles with a standard deviation of 1.4 miles, what is the standard deviation of the fare passengers pay? $0.42 $4.20 $9.00

$4.20 Correct. Let T be the total amount of the cab fare and the standard deviation of T is σa+bX = bσX. The key was to convert the rate to dollars per mile from dollars per tenth of a mile, so b = 3. The standard deviation of T is 3.00(1.4) = 4.20.

Safari Adventure Theme Park is a selfguided theme park in which people drive through a park filled with African wildlife. They are given a map and a written guide to the wildlife of the park. They charge $20 per car plus $2.00 per person in the car. The number of people per car can be represented by the random variable X which has a mean value μX = 3.2, and a variance σ2x = 1.4. What is the variance of the total amount of money per car that is collected entering the park? $4.00 $2.80 $26.40 $5.60

$5.60 Feedback Correct. Let T be the total amount of money that is collected from a car that enters the park. To find the variance of T, you can use the rule of variance: σ2 a + bX = b2σ2X = (2)2(1.4) = 4(1.4) = 5.60.

The random variable X, representing the number of items sold in a week, has the following probability distribution: x 0 1 2 3 4 5 6 P(X = x) 0.10 0.20 0.40 0.15 0.05 0.05 0.05 By the fourth day of a particular week, 3 items have already sold. What is the probability that there will be less than a total of 5 items sold during that week? 1.00 0.95 0.90 0.85 0.67

0.85 Recall that you are given the fact that three items sold in the first 4 days and you want to find the probability that there are less than a total of 5 items sold in the week. You should be conditioning in some way on the fact that 3 items have already sold during the week.

The random variable X, representing the number of items sold in a week, has the following probability distribution: x 0 1 2 3 4 5 6 P(X = x) 0.10 0.20 0.40 0.15 0.05 0.05 0.05 What is the probability that in a given week there will be at most 4 items sold? 0.10 1.00 0.60 0.90 0.85

0.90 Correct. P(X ≤ 4) = 0.10 + 0.20 + 0.40 + 0.15 + 0.05 = 0.85 or P(X ≤ 5) = 1 − P(X ≤ 5) = 1 − (0.05 + 0.05) = 1 − 0.10 = 0.90.

The number of people in a car that crosses a certain bridge is represented by the random variable X, which has a mean value μX = 2.7, and a variance σ2X = 1.2. The toll on the bridge is $3.00 per car plus $0.50 per person in the car. What is the mean of the total amount of money that is collected from a car that crosses the bridge? Report your answer to two decimal places. Do NOT include a dollar sign ($) when reporting your answer. What is the variance of the total amount of money that is collected from a car that crosses the bridge? Report your answer to two decimal places. Do NOT include a dollar sign ($) when reporting your answer.

Feedback Part 1 Correct. Let T be the total amount of money that is collected from a car that crosses the bridge. T is composed of two parts: the fixed toll of 3 dollars (regardless of the number of passengers) and $0.50 for every passenger, thus T = 3 + 0.50 * X. To find the mean of T, you use the rule of means: μ(a+bX) = a + b μX = 3 + 0.5(2.7) = 3 + 0.5(2.7) = 4.35. Part 2 Correct. Let T be the total amount of money that is collected from a car that crosses the bridge. T is composed of two parts: the fixed toll of $3 (regardless of the number of passengers) and $0.50 for every passenger. Thus, T = 3 + 0.50 * X. To find the variance of T, you use the rule of variances: σ2a+bX = b2 σ2X = (0.5)2(1.2) = 0.25(1.2) = 0.30.

Joe went to watch the Quakes play baseball at Epicenter stadium in Rancho Cucamonga. While he enjoyed the game, he also purchased some food. Consider whether some of the things he saw or did at the game that involve discrete or continuous variables. Which of the following is a true statement? Check all that apply. The batting average of Joe's favorite player is a continuous variable. The number of hot dogs that Joe bought is a discrete variable. The number of strikeouts in the game is a continuous variable. The Quakes standings in the league is a discrete variable.

The batting average of Joe's favorite player is a continuous variable. The number of hot dogs that Joe bought is a discrete variable. The Quakes standings in the league is a discrete variable. Feedback Correct. Batting average is a continuous variable, the number hot dogs purchased by Joe is a discrete variable, and the team's ranking in the league is a discrete variable.

According the the Pew Research Center, the probability of a randomly selected person living the United States identifying with a particular religious affiliation—Protestant Christian; Catholic Christian, Other Christian, Non-Christian (including Jewish, Muslim, Buddhist, Hindu and other religions); and Unaffiliated (the so-called "nones")—is shown in the following chart. Affiliation Protestant Christian Catholic Christian Other Christian Non-Christian Unaffiliated "none" Probability 0.466 0.208 0.033 0.06 0.233 Adapted from "America's Changing Religious Landscape," Pew Research Center, May 12, 2015, http://www.pewforum.org/religious-landscape-study/state/california/ Assuming this data is accurate and stable, what is the probability that a randomly selected Christian living in the United States would identify as Catholic? 0.294 0.208 0.659 0.707

0.208 Incorrect. You have identified the probability of a randomly selected person living in the United States identifying as Catholic, but that is from among everyone in the poll. The question is looking for the probability of people identifying as a Catholic if they already identify as being a Christian. To do this find the proportion of those identifying as Catholics and divide by the proportion of those who identify as Christian.

According the the Pew Research Center, the probability of a randomly selected person living in the United States identifying with a particular religious affiliation—Protestant Christian; Catholic Christian, Other Christian, Non-Christian (including Jewish, Muslim, Buddhist, Hindu and other religions); and Unaffiliated (the so-called "nones")—is shown in the following chart. Affiliation Protestant Christian Catholic Christian Other Christian Non-Christian Unaffiliated "none" Probability 0.466 0.208 0.033 0.06 0.233 Adapted from "America's Changing Religious Landscape," Pew Research Center, May 12, 2015, http://www.pewforum.org/religious-landscape-study/state/california/ Assuming this data is accurate and stable, what is the probability that a randomly selected person living in the United States would not identify as Christian? 0.293 0.707 0.06

0.293 Correct. The probability of finding someone who does not identify as Christian can be found two ways: either as the sum of the probabilities of being in the Non-Christian and Unaffiliated groups or as 1 minus the sum of probabilities of being in one of three groups that identify as being Christian.

The probability distribution for the number of defects during an eight hour shift on the assembly line at Wanda's Wooden Widgets is as shown in the chart below. x 0 1 2 3 4 5 P(X = x) 0.50 0.25 0.15 0.06 0.03 0.01 On average, how many defects are found during an 8-hour shift? 5.3 2.5 0.9 0.50 0.1667

0.9 Correct. We need to find the mean of X, μx. μx = (0 * 0.50) + (1 * 0.25) + (2 * 0.15) + (3 * 0.06) + (4 * 0.03) + (5 * 0.01) = 0.9.

The probability distribution for the number of defects during an 8-hour shift on the assembly line at Wanda's Wooden Widgets is as shown in the chart below. x 0 1 2 3 4 5 P(X = x) 0.50 0.25 0.15 0.06 0.03 0.01 What is the probability that in a given shift there will be at most 2 defects? 0.90 0.75 0.40 0.10 1.00

0.90 Correct. The probability of at most 2 defects is the sum of the probabilities for the number of defects equal to or less than 2: P(X ≤ 2) = 0.50 + 0.25 + 0.15 = 0.90 or P(X ≤ 2) = 1 − P(X ≥ 3) = 1 − (0.06 + 0.03+ 0.01) = 1 − 0.10 = 0.90.

The random variable X, representing the number of accidents in a certain intersection in a week, has the following probability distribution: x 0 1 2 3 4 5 P(X = x) 0.20 0.30 0.20 0.15 0.10 0.05 On average, how many accidents are there in the intersection in a week? 5.3 2.5 1.8 0.30 0.1667 Feedback Correct. We need to find the mean of X, μx. μx = (0 * 0.20) + (1 * 0.30) + (2 * 0.20) + (3 * 0.15) + (4 * 0.10) + (5 * 0.05) = 1.8.

1.8 Correct. We need to find the mean of X, μx. μx = (0 * 0.20) + (1 * 0.30) + (2 * 0.20) + (3 * 0.15) + (4 * 0.10) + (5 * 0.05) = 1.8.

The random variable X, representing the number of items sold in a week, has the following probability distribution: x 0 1 2 3 4 5 6 P(X = x) 0.10 0.20 0.40 0.15 0.05 0.05 0.05 On average, how many items are sold in a week? 6.3 3.0 2.2 2.0 0.1428 Feedback Correct. We need to find the mean of X, μx. μx = (0 * 0.10) + (1 * 0.20) + (2 * 0.40) + (3 * 0.15) + (4 * 0.05) + (5 * 0.05) + (6 * 0.05) = 2.2.

2.2 Correct. We need to find the mean of X, μx. μx = (0 * 0.10) + (1 * 0.20) + (2 * 0.40) + (3 * 0.15) + (4 * 0.05) + (5 * 0.05) + (6 * 0.05) = 2.2.

The number of computer monitors manufactured per day by CompScreens Inc. is a random variable X, with a mean μx = 120 and a variance of σ2x = 36. The cost of manufacturing the monitors is $500 base cost plus $15 per monitor. What is the variance of the total cost per day for CompScreens Inc. to manufacture computer monitors? 540 1040 8100 8600

8100

The probability distribution for the number of defects during an 8-hour shift on the assembly line at Wanda's Wooden Widgets is as shown in the chart below. x 0 1 2 3 4 5 P(X = x) 0.50 0.25 0.15 0.06 0.03 0.01 After 2 hours of a particular shift, 2 defects have already been detected. What is the probability that there will be less than a total of 4 defects detected during the entire shift? 1.00 0.96 0.84 0.25 Feedback Incorrect. It seems that you have found either the probability that in a given shift there are at most 3 defects or the probability that given that 2 defects have already occurred, the probability that less than 5 defects will occur in that shift. Recall that you want to find: given that 2 defects have already occurred, the probability that less than 4 defects will occur in the entire shift.

0.96 Incorrect. It seems that you have found either the probability that in a given shift there are at most 3 defects or the probability that given that 2 defects have already occurred, the probability that less than 5 defects will occur in that shift. Recall that you want to find: given that 2 defects have already occurred, the probability that less than 4 defects will occur in the entire shift.

Daily rainfall is an example of what type of data random variable? Discrete Continuous

Continuous Correct. Rainfall is measured and therefore is a continuous data random variable.

The Coff-E-Cup beverage company has opened two franchises in town. They put one location in the center of town and one at the west end of town. Let X be the number of customers that enter the center of town location in an hour, and Y be the number of customers that enter the location at the west end of town in an hour. Assuming that μX = 65 with σX = 4.3 and μY = 45 with σY = 3.2 , what is the standard deviation of Z, the total number of customers that enter these locations in an hour? 28.73 7.50 5.36 3.75

Correct. Since Z = X + Y, σ2X+Y = σ2X + σ2Y and σX+Y = √(σ2X+Y). This equals √(4.32 + 3.22) = √(28.73.) = 5.36.

The number of people entering a state park is a random variable X with a mean of 3 and a variance of 2. The admission to the park is $6 per car plus an additional $1.50 per person in the car. What is the mean of the total amount of money collected for a car entering the park? Report your answer to two decimal places. Do NOT include a dollar sign ($) when reporting your answer.

Feedback Correct. Let T be the total amount of money that is collected from a car that enters the park. T is composed of two parts: the fixed fee of 6 dollars (regardless of the number of passengers) and $1.50 for every passenger. The total amount of money can be written as T = 6 + 1.5X. To find the mean of T you use the rule of means: μ(a+bX) = a + b μX = 6 + 1.5(3) = 10.5.

A national park has 2 entrance gates, one at the north end and a second gate at the south end of the park. Let X be the number of cars entering the North gate per hour and let Y be the number of cars entering the South gate per hour. Assume that X is random variable with mean μx = 23 and standard deviation σx = 3. Assume Y is a random variable with mean μy = 18 and standard deviation σy = 4. Assume that the two gates operate independently. What is the standard deviation of Z, the total number of cars entering the park each hour? √7 4 5 7

Feedback Correct. When X and Y are independent, the standard deviation of their total, Z = X + Y is σz = σz = √(σ2x + σ2y) = √(32 + 42) = √(25) = 5.

Which of the following is a characteristic of a discrete random variable? It is something you count. It is something you measure. It is something you both measure and count.

It is something you count. Correct. Discrete data is countable. A discrete random variable is something you count.

The following three histograms represent the probability distributions of the three random variables X, Y, and Z. Which of the three random variables has the largest standard deviation? X Y Z All three random variables have the same standard deviation.

X Feedback Although the random variables take on the same values, they do not have equal standard deviations. For example, random variable X will have a larger standard deviation than random variable Y because it takes on values 2 and 4 more often.


Conjuntos de estudio relacionados

Chapter 13 - Viruses Viroids and Prions - MIC 205A

View Set

Earth and Environmental Practice Quiz

View Set

Basic Medical Histology (GMS6630) Unit 1 - Multiple Choice Questions

View Set

EXAM 2 Textbook Questions Drugs and Behavior

View Set