Chaper 6 AP Stats

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On averge a custmer rent 0.9797 videos more or less than the mean of 1.8 every visit

A four sided die, shaped like an asymmetrical tetrahedron, has the following roll probabilities. (Number on die, Probability): [1, 0.4] [2, 0.3] [3, 0.2] [4, 0.1] Let X = the result of a single rolla). Find P(1 < x < 4)

P( 1 < X < 4) = P(X=2 or X=3) = 0.3 + 0.2 = 0.5

P( X ≠ 3) = 1- P(X=3) = 1 - 0.2 = 0.8

A four sided die shaped like an asymmetrical tetrahedron has the following roll probabilities. Number on Die 1 2 3 4 Probability 0.4 0.3 0.2 0.1 Let X = the result of a single roll. Find P(1<X<4)

P( X=2 or X=3)=0.3+0.2=0.5

The probability of getting a question right on our Chapter 6 Quiz is 0.76 (with some studying). There are 10 questions on this quiz. (Hypothetical) What is the probability of getting a C or a B?

P(7<x<8) binomialcdf(10,76,8)-binomialcdf(10,0.76,6)=0.53

The length of human pregnancies from conception to birth varies according to a distribution that is approximately Normal with mean 266 days and standard deviation 16 days. Choose two pregnancies independently and at random. Find the probability that the difference in the lengths of the two pregnancies is greater than 25 days.

P(D<-25 or D>25)= 0.2692 normal curve with 0 as the mean and -25 and 25 marked

The probability distribution below is for the random variable X = number of mice caught in traps during a single night in small apartment building. X 0 1 2 3 4 5 P(X) 0.12 0.20 0.31 0.14 0.16 0.07 Express the event "trapping at least one mouse" in terms of X and find its probability.

P(X>1) 1-(0.12)=0.88

The Internal Revenue Service estimates that 8% of all taxpayers filling out long forms make mistakes. (b) The same IRS employee announces at lunch one day that she checked 25 forms this morning and didn't find mistakes on any of them. Is this surprising enough so that her supervisor should worry about whether she is missing errors? Explain.

P(no mistake on 25)=(1-0.08)^25= 0.1244 This probability is not low enough that if would be a rare excurrence this her supervisor should not be concerned

The probability distribution below is for the random variable X = number of mice caught in traps during a single night in small apartment building.(X, P(X)): [0, 0.12] [1, 0.20] [2, 0.31] [3, 0.14] [4, 0.16] [5, 0.07]c). Express the event "trapping at least one mouse" in terms of X and find its probability.

P(x ≥ 1)= 0.88

The probability of getting a question right on our Chapter 6 Quiz is 0.76 (with some studying). There are 10 questions on this quiz. (Hypothetical) 30. If a student got a 10% would you be surprised? Why?

P(x=1) Binomialpdf(10,0.76,0.1)=0.00002 Yes, low chance of getting only 1 problem correct

The probability of getting a question right on our Chapter 6 Quiz is 0.76 (with some studying). There are 10 questions on this quiz. (Hypothetical) If student got a 70% would you be surprised? Why?

P(x=7) Binomialpdf(10,0.76,7)=0.24 No almost 1/4 of the class will get a 70%

The probability distribution below is for the random variable X = number of mice caught in traps during a single night in a small apartment building. X 0 1 2 3 4 5 P(X) 0.12 0.20 0.31 0.14 0.16 0.07 Describe P(X>2) in words and find its value.

Probability of catching at least 2 mice 0.31+0.14+0.16+0.07=0.68 or 1-(0.12+0.20)=0.68

Joe the barber charges $32 for a shave and haircut and $20 for just a haircut. Based on experience, he determines that the probability that a randomly selected customer comes in for a shave and haircut is 0.85, the rest of his customers come in for just a haircut. Let J = what Joe charges a randomly-selected customer. (a) Give the probability distribution for J.

(J, P(J)): {20, 0.15} {32, 0.85}

The probability of getting a question right on our Chapter 6 Quiz is 0.76 (with some studying). There are 10 questions on this quiz. (Hypothetical) What is the probability of not getting a problem correct until #3?

(0.24)^2(0.76)=0.11

(0.24)^4(0.76)=0.0025

(0.24)^9(0.76)=0.000002

The length of human pregnancies from conception to birth varies according to a distribution that is approximately Normal with mean 266 days and standard deviation 16 days. Choose two pregnancies independently and at random. What is the expected difference in the lengths of the two pregnancies?

ACT scores for the 1,171,460 members of the 2004 high school graduating class who took the test closely followed the Normal distribution with mean 20.9 and standard deviation 4.8. Choose two students independently and at random from this group. (a) What is the expected difference in their scores?

0 20.9-20.9

The probability of getting a question right on our Chapter 6 Quiz is 0.76 (with some studying). There are 10 questions on this quiz. (Hypothetical) Construct the probability distribution for this test for a student.

0 (0.76)^0 x (0.24)^10 =0 1 (0.76)^1 x (0.24)^9 = .000002(10)=.000021 2 (0.76)^2 x (0.24)^8 =.000006(45)=.00029 3 (0.76)^3 x (0.24)^7 = .00002(120)=.00242 4 (0.76)^4 x (0.24)^6 = .00006(210)=.0134 5 (0.76)^5 x (0.24)^5 = .0002(252)=.0509 6 (0.76)^6 x (0.24)^4 = .0006(210)= .134 7 (0.76)^7 x (0.24)^3 = .002(120)= .243 8 (0.76)^8 x (0.24)^2 = .006(45)=.288 9 (0.76)^9 x (0.24)^1 =.02(10)=.203 10 (0.76)^10 x (0.24)^0 =

Suppose you will when flipping these coins you will receive $1 for getting each tail you get. What is the expected value?

0(1/16)+1(1/4)+2(3/8)+3(1/4)+4(1/16) E(x)=2 We expect to get $2

A professional soccer player succeeds in scoring a goal on 84% of his penalty kicks. Assume that the success of each kick (c) What is the probability that he scores on 5 or fewer of his next 10 penalty kicks?

0.0130 binomialcdf(10,0.84,5)

The Internal Revenue Service estimates that 8% of all taxpayers filling out long forms make mistakes. (a) An IRS employee starts to randomly select forms—one at a time—to check for mistakes. What is the probability that the first form with mistakes is the 7th one she checks?

0.0485 (0.92^6)(0.08)

A professional soccer player succeeds in scoring a goal on 84% of his penalty kicks. Assume that the success of each kick (a) In a series of games, what is the probability that the first time he fails to score a goal is on his fifth penalty kick?

0.0797 (0.84^4)(0.16)

ACT scores for the 1,171,460 members of the 2004 high school graduating class who took the test closely followed the Normal distribution with mean 20.9 and standard deviation 4.8. Choose two students independently and at random from this group. (c) Find the probability that the difference in the two students' scores is greater than 6.

0.3789 P(x<6 or x>6) normal curve with 0 as the mean and -6 and 6 marked

If it cost a dollar each time is it worth playing this game? Why or why not? Now suppose I give you a weighted coin and the probability distribution is as follows... Number of tails flipped Probability 0 0.45 1 0.30 2 0.11 3 0.09 4 0.05 9. What is the mean number of tails?

0.99

A four sided die shaped like an asymmetrical tetrahedron has the following roll probabilities. Number on Die 1 2 3 4 Probability 0.4 0.3 0.2 0.1 Let X = the result of a single roll. Find P(X(not equal)3)

1-P(X=3)=1-0.2=0.8

Number of tails flipped Probability 0 0.45 1 0.30 2 0.11 3 0.09 4 0.05 10. What is the standard deviation of the number of tails?

1.17

The probability of getting a question right on our Chapter 6 Quiz is 0.76 (with some studying). There are 10 questions on this quiz. (Hypothetical) Find and interpret the mean score on the test for a student.

10(0.76)=7.6

The length of human pregnancies from conception to birth varies according to a distribution that is approximately Normal with mean 266 days and standard deviation 16 days. Choose two pregnancies independently and at random. What is the standard deviation of the difference in the lengths of the two pregnancies?

22.63

ACT scores for the 1,171,460 members of the 2004 high school graduating class who took the test closely followed the Normal distribution with mean 20.9 and standard deviation 4.8. Choose two students independently and at random from this group. (b) What is the standard deviation of the difference in their scores?

6.788 sqrt(4.8^2 + 4.8^2)= sqrt(46.08)=6.788

Let two disturbutions X and Y have the following probability distributions. Let D = X - Y X 2 4 6 P(X) 0.3 0.5 0.2 Y 1 3 5 P(Y) 0.4 0.2 0.4 a) Determine the probability distribution D

D -3 -1 1 3 5 P(D) 012 0.26 0.3 0.24 0.08

Number of tails flipped Probability 0 0.45 1 0.30 2 0.11 3 0.09 4 0.05 If it cost a dollar each time is it worth playing this game? Why or why not

No we will lose approx 0.01 each time we play

In order to set premiums at profitable levels, insurance companies must estimate how much they will have to pay in claims on cars of each make and model, based on the value of the car and how much damage it sustains in accidents. Let C be a random variable that represents the cost of a randomly selected car of one model to the insurance company. The probability distribution of C is given below. C $0 $1000 $4000 $10,000 PC. 0.60 0.05 0.13 0.22 The probability that the insurance company will have to pay a claim of at least $1000 for a randomly selected car is: a. 0 b. 0.05 c. 0.13 d. 0.22 e. 0.40

e. 0.40 0.05+0.13+0.22

Joe the barber, charges $32 for a shave and haircut and $20 for just a haircut. Based on experience, he determines that the probability that a randomly selected customer comes in for a shave and haircut is 0.85, the rest of his customers come in for just a haircut. Let J = what Joe charges a randomly-selected customer. (c) Find and interpret the standard deviation of J

he average distance from the mean ($30.20) for each individual customer √(.15(20-30.20)^2+ .85(32-30.20)^2) = $4.28

on averge a custmer rents 1.8 videos per visit μ 1.8

The probability of getting a question right on our Chapter 6 Quiz is 0.76 (with some studying). There are 10 questions on this quiz. (Hypothetical) What is the probability of getting an F?

p(x<5) binomialcdf(10,0.76, 5)=0.07

The probability of getting a question right on our Chapter 6 Quiz is 0.76 (with some studying). There are 10 questions on this quiz. (Hypothetical) What is the probability of getting a D?

p(x=6) binomialcdf(10,0.76, 6)=0.13

The probability of getting a question right on our Chapter 6 Quiz is 0.76 (with some studying). There are 10 questions on this quiz. (Hypothetical) What is the probability of getting a C?

p(x=7) binomialcdf(10,0.76, 7)=0.24

The probability of getting a question right on our Chapter 6 Quiz is 0.76 (with some studying). There are 10 questions on this quiz. (Hypothetical) What is the probability of getting a B?

p(x=8) binomialcdf(10,0.76, 8)=0.29

The probability of getting a question right on our Chapter 6 Quiz is 0.76 (with some studying). There are 10 questions on this quiz. (Hypothetical) What is the probability of getting at least a D?

p(x>6) 1-binomialcdf(10,0.76, 5)=0.93

The probability of getting a question right on our Chapter 6 Quiz is 0.76 (with some studying). There are 10 questions on this quiz. (Hypothetical) What is the probability of getting at least a C?

p(x>7) 1-binomialcdf(10,0.76, 6)=0.80

The probability of getting a question right on our Chapter 6 Quiz is 0.76 (with some studying). There are 10 questions on this quiz. (Hypothetical) What is the probability of getting at least a B?

p(x>8) 1-binomialcdf(10,0.76, 7)=0.56

The probability of getting a question right on our Chapter 6 Quiz is 0.76 (with some studying). There are 10 questions on this quiz. (Hypothetical) What is the probability of getting at least an A?

p(x>9) 1-binomialcdf(10,0.76, 8)=0.27 or P(9)+P(10)=0.27

The probability of getting a question right on our Chapter 6 Quiz is 0.76 (with some studying). There are 10 questions on this quiz. (Hypothetical) 20. What is the probability of getting an A?

p(x>9) 1-binomialcdf(10,0.76, 9)=0.27

A professional soccer player succeeds in scoring a goal on 84% of his penalty kicks. Assume that the success of each kick (d) Suppose that our soccer player is out of action with an injury for several weeks. When he returns, he only scores on 5 of his next 10 penalty kicks. Is this evidence that his success rate is now less than 84%? Explain.

since the probability he scores on 5 out of 10 kicks is 0.0130 that is a very low probability so he may not be back to on 84% success rate binomialcdf(10,0.84,5)

The probability of getting a question right on our Chapter 6 Quiz is 0.76 (with some studying). There are 10 questions on this quiz. (Hypothetical) Find the standard deviation for this test for a student.

sqrt(np (1-p) ) = sqrt(10x0.76x0.24)=1.35

.Joe the barber charges $32 for a shave and haircut and $20 for just a haircut. Based on experience, he determines that the probability that a randomly selected customer comes in for a shave and haircut is 0.85, the rest of his customers come in for just a haircut. Let J = what Joe charges a randomly-selected customer. (b) Find and interpret the mean of J

the mean amount of money Joe can expect to make per customer in the long run = 20 0.15 32 0.85 $30.20

Suppose that the mean height of policemen is 70 inches with a standard deviation of 3 inches. And suppose that the mean height for policewomen is 65 inches with a standard deviation of 2.5 inches. If heights of policemen and policewomen are Normally distributed, find the probability that a randomly selected policewoman is taller than a randomly selected policeman.

w m Q μ: 65 μ: 70 μ: 65-70=-5 σ: 2.5 σ: 3 σ: sqrt(6.25+9)=sqrt(15.25)=3.905 σ^2: 6.25 σ^2: 9 σ^2: 6.25+9=15.25 w>m -m -m Q=w-m>0 normalcdf:(0, 1000000, -5, 3.905)=1.002

Lamar and Lawrence run a two-person lawn-care service. They have been caring for Mr. Johnson's very large lawn for several years, and they have found that the time it takes Lamar to mow the lawn itself is approximately Normally distributed with a mean of 105 minutes and a standard deviation of 10 minutes. Meanwhile, the time it takes for Lawrence to use the edger and string trimmer to attend to details is also Normally distributed with a mean of 98 minutes and a standard deviation of 15 minutes. They prefer to finish their jobs within 5 minutes of each other. What is the probability that this happens, assuming their finish times are independent?

w m z μ: 105 μ: 98 μ: 105-98=7 σ: 10 σ: 15 σ: sqrt(100+225)=sqrt(325)=18.028 σ^2: 100 σ^2: 225 σ^2: 6.25+9=100+225=325 w>m -m -m Z=w-m>0 normalcdf:(-5,5,7,18.028)=0.203

A company that rents DVDs from vending machines in grocery stores has developed the following probability distribution for the random variable X = the number of DVDs a customer rents per visit to a machine. X 1 2 3 4 P(X) 0.5 0.3 0.1 0.1 (c) Suppose the profit the company makes on each customer visit is $0.75 per DVD minus $0.05 "fixed costs." That is, if P = profit, then Use a linear transformation of your results in A. and B. to find the mean and standard deviation for P.

μ 1.3 σ=0.74 μ 0.75(1.85)-0.05 σ 0.75(1.8)

16.A game show host has developed a new game called "Grab All You Can." Here's how it works: a contestant reaches his dominant hand (i.e. right hand for right-handed people) into a jar of $10 bills and grabs as many as he can in one handful. Then he does the same thing with his non-dominant hand in a jar of $20 bills. Research with many volunteers has determined that the mean number of $10 bills drawn is 68 with a standard deviation of 9.5, and the mean number of $20 dollar bills is 58, with a standard deviation of 7.8. (c) The game's rules are that a contestant must pay $500 (from previous winnings) for one round of play (that is, one grab from each jar). If G = how much the contestant gains from one round of play, find the mean and standard deviation of G.

μ 1840-500=1340 σ: doesn't change=182.65

What is the mean number of tails?

μ 2

Suppose two brothers Mario and Luigi have a car dealership. In a typical day the following are the probability distributions for how many cars they can sell each. L P(L) M P(M) 2 0.2 0 0.1 3 0.5 1 0.25 4 0.3 2 0.35 3 0.3 c) What is the mean and standard deviation of the number of cars their company can sell in a typical day?

μ 4.95 σ: 1.1906 σ^2:1.4175

.A game show host has developed a new game called "Grab All You Can." Here is how it works: a contestant reaches his dominant hand (i.e. right hand for right-handed people) into a jar of $10 bills and grabs as many as he can in one handful. Then he does the same thing with his non-dominant hand in a jar of $20 bills. Research with many volunteers has determined that the mean number of $10 bills drawn is 68 with a standard deviation of 9.5, and the mean number of $20 dollar bills is 58, with a standard deviation of 7.8. (a) If D = the amount of money, in dollars, that a randomly-selected contestant grabs from the "$10 grab," find the mean and standard deviation of D.

μ: 68x10=680 σ: 9.5x10=95

16.A game show host has developed a new game called "Grab All You Can." Here's how it works: a contestant reaches his dominant hand (i.e. right hand for right-handed people) into a jar of $10 bills and grabs as many as he can in one handful. Then he does the same thing with his non-dominant hand in a jar of $20 bills. Research with many volunteers has determined that the mean number of $10 bills drawn is 68 with a standard deviation of 9.5, and the mean number of $20 dollar bills is 58, with a standard deviation of 7.8. (b) If T = the total amount of money, in dollars, that a contestant grabs from both jars, find the mean and standard deviation of T.

μ:(68 x 10) + (20x58) = 1840 σ: sqrt( (10x9.5)^2 + (20x7.8)^2 ) = sqrt(9025+24336) = sqrt(33361) = $182.65

μ:2.23 σ:1.427 σ^2:2.036

Let two disturbutions X and Y have the following probability distributions. Let D = X - Y X 2 4 6 P(X) 0.3 0.5 0.2 Y 1 3 5 P(Y) 0.4 0.2 0.4 c) What is the mean and standard deviation of D.

μD=0.8 σD=2.272

μX: 1(0.4) + 2(0.3) + 3(0.2) + 4(0.1) =2σX: √(.4(1-2)^2+ .3(2-2)^2 .2(3-2)^2+ .1(4-2)^2)=1μX: is the expected mean roll if the die is rolled many times, or the expected long-run value of asingle rollσX: is the variability in the number rolled in the long run

Let two disturbutions X and Y have the following probability distributions. Let D = X - Y X 2 4 6 P(X) 0.3 0.5 0.2 Y 1 3 5 P(Y) 0.4 0.2 0.4 b)What is the mean and standard deviation of the X and Y.

μX= 3.8 μY=3 σX= 1.4 σY=1.789

What is the standard deviation number of tails?

σ=1

Make a probability distribution for the discrete random variable of number of tails when a coin is flipped 4 times.

T 0 1 2 3 4 P(T) 1/16 1/4 3/8 1/4 1/16

The event T = 4 can happen three ways: {1, 3}, {3, 1}, and {2, 2}. Probabilities for these events are, respectively, (0.4) (0.2)= 0.08; (0.2) (0.4)= 0.08; and (0.3) (0.3)= 0.09. Hence the total probability is 0.25.

A four-sided die shaped like an asymmetrical tetrahedron has the following roll probabilities. Number on Die 1 2 3 4 Probability 0.4 0.3 0.2 0.1 Let X = the result of a single roll. Describe P(x=3}x>2) in words and finds its value

The probability of rolling a 3 given that the roll is 2 or greater (0.2/0.6)=1/3

A four sided die, shaped like an asymmetrical tetrahedron, has the following roll probabilities (Number on die, Probability): [1, 0.4] [2, 0.3] [3, 0.2] [4, 0.1] Let X = the result of a single rollc). Describe P(x=3 |x ≥ 2 ) in words and find its value.

The probability of rolling a 3, given that the roll is 2 or greater; 0.2/0.6= 1/3

The probability distribution below is for the random variable X = number of mice caught in traps during a single night in small apartment building.( X, P(X)): [0, 0.12] [1, 0.20] [2, 0.31] [3, 0.14] [4, 0.16] [5, 0.07] b). Describe P(X ≥ 2) in words and find its value

The probability that two or more mice are caught during a single night; 0.68.

A four sided die shaped like an asymmetrical tetrahedron has the following roll probabilities. Number on Die 1 2 3 4 Probability 0.4 0.3 0.2 0.1 Let X = the result of a single roll. If T=the sum of two rolls, find P(T=4)

Used calculator μ 2 σ=1

Used calculator μ 2 On average you will roll a 2 σ=1 On average the number you roll will differ by 1

Suppose you will when flipping these coins you will receive $1 for getting each tail you get. If it cost a dollar each time is it worth playing this game? Why or why not?

Yes it is, if we expect to win $2, but it cost $1 we will still gain $1

The probability distribution of a continuous random variable X is given by the density curve below shaded box-shaped graph with 0, 1, 2 on x-axis and nothing on y-axis The probability that X = 1.5 is a. 0. b. very small; slightly larger than 0. c. 1/4. d. 1/3. e. 1/2.

a. 0.

Let the random variable X represent the profit made on a randomly selected day by a certain store. Assume that X is Normal with mean $360 and standard deviation $50. What is P(X > $400)? a. 0.2119 b. 0.2881 c. 0.5319 d. 0.7881 e. 0.8450

a. 0.2119 normalcdf(400,1000000,360,50)

Which of the following is the most important reason for randomly assigning subjects to treatment groups in an experiment? a. To create groups that are as similar as possible. b. To eliminate bias. c. To create a double-blind design d. To reduce random variation. e. To counteract the placebo effect.

a. To create groups that are as similar as possible.

A professional soccer player succeeds in scoring a goal on 84% of his penalty kicks. Assume that the success of each kick (b) How many kicks should he make before he eventually misses one?

about 6.25 kicks

A rock concert producer has scheduled an outdoor concert. If it is warm that day, she expects to make a $20,000 profit. If it is cool that day, she expects to make a $5000 profit. If it is very cold that day, she expects to suffer a $12,000 loss. Based upon historical records, the weather office has estimated the chances of a warm day to be 0.60; the chances of a cool day to be 0.25. What is the producer's expected profit? a. $5,000 b. $11,450 c. $13,000 d. $13,250 e. $15,050

b. $11,450 0.6(20,000)+0.25(5,000)+0.15(-12,000)

In the Florida scratch-card lottery, the numbers and values of prizes awarded for every 100,000 cards sold are given in the table below. Number of cards = 10, Payoff = $1,000 Number of cards = 1000, Payoff = $50 Number of cards = 5000, Payoff = $5 Use Scenario 6-4. The probability that a random scratch-card will pay off is a. .0250. b. .0601. c. .2500. d. .6010. e. .8500.

b. .0601.

"Insert tab A into slot B" is something you might read in the assembly instructions for pre-fabricated bookshelves. Suppose that tab A varies in size according to a Normal distribution with a mean of 30 mm. and a standard deviation of 0.5 mm., and the size of slot B is also Normally distributed, with a mean of 32 mm. and a standard deviation of 0.8 mm. The two parts are randomly and independently selected for packaging. What is the probability that tab A won't fit into slot B? a. 0.0007 b. 0.0170 c. 0.0618 d. 0.9382 e. 0.9830

b. 0.0170

random variable X has a probability distribution as follows: X 0 1 2 3 P(X) 2k 3k 13k 2k Where k is a positive constant. The probability P(X < 2.0) is equal to a. 0.90 b. 0.25 c. 0.65 d. 0.15 e. 1.00

b. 0.25 2k+3k+13k+2k=1 20k=1 1/20 P(X=0)+P(X=1) 2(1/20)+3(1/20)=1/4

The probability distribution of a continuous random variable X is given by the density curve below shaded box-shaped graph with 0, 1, 2 on x-axis and nothing on y-axis The probability that X is at least 1.5 is a. 0. b. 1/4. c. 1/3. d. 1/2. e. 3/4.

b. 1/4.

Jen's commute to work requires that she take the Blue subway line, then transfer to the Red line. The length of the trip on the Blue line has a mean of 18 minutes with a standard deviation of 2 minutes. The Red line trip takes 12 minutes with a standard deviation of 1 minute. The waiting time between when she gets off the Blue line and her Red line train arrives has mean of 10 minutes and a standard deviation of 5 minutes. Assume (perhaps unrealistically) that these times are independent random variables. What are the mean and standard deviation of her entire commute? a. Mean = 40 minutes; Standard deviation = 8 minutes b. Mean = 40 minutes; Standard deviation = 5.48 minutes c. Mean = 40 minutes; Standard deviation = 2.83 minutes d. Mean = 30 minutes; Standard deviation = 5.48 minutes e. Mean = 30 minutes; Standard deviation = 8 minutes

b. Mean = 40 minutes; Standard deviation = 5.48 minutes mean: 18+12+10=40 SD: sqrt( (2^2)+(1^2)+(5^2) )=sqrt(30)=5.477

To set premiums at profitable levels, insurance companies must estimate how much they will have to pay based on the value of the car and how much damage it sustains in accidents. Let C be a random variable that represents the cost of a randomly selected car. C $0 $1000 $4000 $10,000 PC. 0.60 0.05 0.13 0.22 Which is the best interpretation of expected value? (In the choices below, "ExpC." represents the expected value). a. If the company insures ten cars of this model, they know they will incur 10ExpC. in costs. b. The maximum cost to the company for insuring this car model is ExpC. Per car. c. The company must insure at least ExpC. Of these cars to make a profit. d. If the company insures a large number of these cars, they can expect the cost per car to average approximately ExpC. e. If the company insures a large number of these cars, they can expect the variabilterm-48ity in cost per car to average approximately E

b. The maximum cost to the company for insuring this car model is ExpC. per car.

X and Y are independent random variables, and a and b are constants. Which one of the following statements is true? a. σ X + Y = σ X + σ Y b. Var(X − Y) = Var(X) + Var(Y) c. Var(a + bX) = bVar(X) d. σX−Y = σ X − σ Y e. Var(X + Y) = sqrt( Var(X^2) + Var(Y^2) )

b. Var(X − Y) = Var(X) + Var(Y)

In the Florida scratch-card lottery, the numbers and values of prizes awarded for every 100,000 cards sold are given in the table below. Number of cards = 10, Payoff = $1,000 Number of cards = 1000, Payoff = $50 Number of cards = 5000, Payoff = $5 Use Scenario 6-4. The expected payoff per card is a. $1.00. b. $.90. c. $.85. d. $.50. e. $.25.

c. $.85.

Consider the following game. You pay me an entry fee of x dollars; then I roll a fair die. If the die shows a number less than 3, I pay you nothing; if the die shows a 3 or 4, I give you back your entry fee of x dollars; if the die shows a 5, I will pay you $1; and if the die shows a 6, I pay you $3. What value of x makes the game fair (in terms of expected value) for both of us? a. $2 b. $4 c. $1 d. $0.75 e. $0.5

c. $1

In order to set premiums at profitable levels, insurance companies must estimate how much they will have to pay in claims on cars of each make and model, based on the value of the car and how much damage it sustains in accidents. Let C be a random variable that represents the cost of a randomly selected carterm-47 of one model to the insurance company. The probability distribution of C is given below. C $0 $1000 $4000 $10,000 PC. 0.60 0.05 0.13 0.22 The expected value of C is a. $0 b. $554 c. $2770 d. $3750 e. $4057

c. $2770 0(0.6)+1000(0.05)+4000(0.13)+10000(0.22)

The probability distribution of a continuous random variable X is given by the density curve below shaded box-shaped graph with 0, 1, 2 on x-axis and nothing on y-axis The probability that X is between 0.5 and 1.5 is a. 1/4. b. 1/3. c. 1/2. d. 3/4. e. 1.

c. 1/2.

A business evaluates a proposed venture as follows. It stands to make a profit of $10,000 with probability 3/20, to make a profit of $5000 with probability 9/20, to break even with probability 5/20, and to lose $5000 with probability 3/20. The expected profit in dollars is a. 1500 b. 0 c. 3000 d. 3250 e. -1500

c. 3000 10,000(3/20)+5,000(9/20)+0(5/20)+-5,000(3/20)

Which of the following random variables is geometric? a. The number of phone calls received in a one-hour period b. The number of times I have to roll a six-sided die to get two 5s. c. The number of digits I will read beginning at a randomly selected starting point in a table of random digits until I find a 7. d. The number of 7s in a row of 40 random digits. e. All four of the above are geometric random variables.

c. The number of digits I will read beginning at a randomly selected starting point in a table of random digits until I find a 7.

If A = result of a single roll of a six-sided die and B = result of a single roll of an 8-sided die, then μ A = 3.5;σ A= 1.71;μ B= 4.5;σ C= 2.29. If D = the difference B - A, then which of the following is true? a. μ D= 1;σ D= 1.71 + 2.29 b. μ D= 1;σ D= 1.71 − 2.29 c. μ D = 1;σ D = sqrt(1.712^2 + 2.292^2) d. μ D = 1;σ D = sqrt(1.712^2 - 2.292^2) e. μ D= 1;σ D=sqrt(1.712^2 + 2.292^2)

c. μ D = 1;σ D = sqrt(1.712^2 + 2.292^2) μ: 3.5-4.5 σ: sqrt(1.712^2 + 2.292^2)

Let the random variable X represent the profit made on a randomly selected day by a certain store. Assume X is Normal with a mean of $360 and standard deviation $50. The probability is approximately 0.6 that on a randomly selected day the store will make less than which of the following amounts? a. $330.00 b. $347.40 c. $361.30 d. $372.60 e. $390.00

d. $372.60

Roll one 10-sided die 12 times. The probability of getting exactly 4 eights in those 12 rolls is given by

d. (12 4) x (1/10)^4 x (9/10)^3

9.The variance of sum of two random variables X and Y is a. σ x + σ Y b. (σ x)^2 + (σ Y)^2 c. σ x + σ Y, but only if X and Y are independent d. (σ x)^2 + (σ Y)^2, but only if X and Y are independent e. None of these

d. (σ x)^2 + (σ Y)^2, but only if X and Y are independent

Let the random variable X represent the profit made on a randomly selected day by a certain store. Assume X is Normal with a mean of $360 and standard deviation $50. The value of P(X > $400) is a. 0.2881. b. 0.8450. c. 0.7881. d. 0.2119. e. 0.1600.

d. 0.2119.

marketing survey compiled data on the total number of televisions in households. If X = the number of televisions in a randomly-selected household, and we omit the rare cases of more than 5 televisions, then X has the following distribution: X 0 1 2 3 4 5 P(X) 0.24 0.37 0.20 0.11 0.05 0.03 What is the probability that a randomly chosen household has at least two televisions? a. 0.19 b. 0.20 c. 0.29 d. 0.39 e. 0.61

d. 0.39 0.20+0.11+0.05+0.03

If P (A) = 0.6;P(B) = 0.2 and P(AandB)= 0.10 which of the following must be true? a. A and B are independent and mutually exclusive. b. A and B are not independent, but they are mutually exclusive. c. A and B are independent, but they are not mutually exclusive. d. A and B are neither independent nor mutually exclusive. e. A and B are mutually exclusive, but there is not enough information to determine if they are independent.

d. A and B are neither independent nor mutually exclusive.

Which of the following is true about every random variable I. It takes on numerical or categorical values. II. It describes the results of a random phenomenon. III. Its behavior can be described by a probability distribution. a. I only b. II only c. III only d. II and III e. All three statements are true

d. II and III

Songs on Leila's smartphone have a mean length of 4.3 minutes and a standard deviation of 1.2 minutes. The phone inserts a period of silence between songs that has a mean length of 0.2 minutes and a standard deviation of 0.1 minutes. What are the mean and standard deviation of time it takes for the phone to play a randomly selected two-song sequence a. Mean = 4.5 minutes; Standard Deviation - 2.5 b. Mean = 8.8 minutes; Standard Deviation - 2.5 c. Mean = 4.5 minutes; Standard Deviation - 1.7 d. Mean = 8.8 minutes; Standard Deviation - 1.7 e. Mean = 4.5 minutes; Standard Deviation - 1.58

d. Mean = 8.8 minutes; Standard Deviation - 1.7 mean: 4.3+0.2+4.3=8.8 SD:sqrt( (1.2^2+0.1^2+1.2^2) )=sqrt(2.89)=1.7

Suppose two brothers Mario and Luigi have a car dealership. In a typical day the following are the probability distributions for how many cars they can sell each. L P(L) M P(M) 2 0.2 0 0.1 3 0.5 1 0.25 4 0.3 2 0.35 3 0.3 a) Determine the probability distribution for the company for the number of cars sold per day.

B P(B) 2 0.2x0.1 3 0.2x0.25 + 0.1x0.5 4 0.1x0.3 + 0.25x0.5 + 0.35x0.2 5 0.25x0.3 + 0.35x0.5 + 0.3x0.2 6 0.35x0.3 + 0.3x0.5 7 0.3x0.3

Number of tails flipped Probability 0 0.45 1 0.30 2 0.11 3 0.09 4 0.05 13. Suppose you will when flipping these coins you will receive $1 for getting each tail you get. What is the expected value?

E(x)=0(0.45)+1(0.3)+2(0.11)+3(0.09)+4(0.05) E(x)=0.99

The probability of getting a question right on our Chapter 6 Quiz is 0.76 (with some studying). There are 10 questions on this quiz. (Hypothetical) What are the four conditions for a binomial probability experiment?

Independent pass/fail some prob. of success set # of trails

Suppose two brothers Mario and Luigi have a car dealership. In a typical day the following are the probability distributions for how many cars they can sell each. L P(L) M P(M) 2 0.2 0 0.1 3 0.5 1 0.25 4 0.3 2 0.35 3 0.3 b)What is the mean and standard deviation of the number of cars each sell individually in a typical day.

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