Lind Chapter 3: Exercise 60
Lind Chapter 3: Exercise 60
Owens Orchards sells apples in a large bag by weight. A sample of seven bags contained the following numbers of apples: 23,19,26,17,21,24,22.
a. Compute the mean number and median number of apples in a bag. Total number of Problems = 11
Mean = (23+19+26+17+21+24+22)/7 = 21.71 Problems correct = 8
Median = 22 CORRECT
List of # in order from low to high: 17,19,21,22,23,24,26
List of # in order from high to low: 26,24,23,22,21,19,17 b. Verify that Σ(x-xbar) = 0.
(17-21.71)+(19-21.71)+(21-21.71)+(22-21.71)+(23-21.71)+(24-21.71)+(26-21.71) ≈ 0 Lind Charpter 3: Exercise 62 …show more content…
The Citizens Banking Company is studying the number of times the ATM lcoated in a Loblaws Supermarket at the foot of Market Street is used per day. Following are the numbers of times the machine was used over each of the last 30 days. Determine the mean number of times that machine was used per day. n = 30
83 64 84 76 84 54 75 59 70 61
63 80 84 73 68 52 65 90 52 77
95 36 78 61 59 84 95 47 87 60
Mean = (83+64+84+76+84+54+75+59+70+61+63+80+84+73+68+52+65+90+52+77+95+36+78+61+59+84+95+47+87+60)/30
Mean = 2116/30 CORRECT
Mean = 70.53
The mean number of times that machine was used per day = 70.53 Lind Chapter 3: Exercise 68
The American Automobile Association checks the prices of gasoline before many holiday weekends. Listed below are the self-service prices for a sample of 15 retail outlets during the May 2003 Memorial Day weekend in the Detroit, Michigan, area.
1.44, 1.42, 1.35, 1.39, 1.49, 1.49, 1.41, 1.46,
1.41, 1.49, 1.45, 1.48, 1.39, 1.46, 1.44,
N=15
a. What is the arithmetic mean selling price? µ= (1.44+1.42+1.35+1.39+1.49+1.49+1.41+1.46+1.41+1.49+1.45+1.48+1.39+1.46+1.44)/15 µ= 21.57/15 µ= 1.44 b. What is the median selling price?
List of # in order from low to high: 1.35,1.39,1.39,1.41,1.41,1.42,1.44,1.44,1.45,1.45,1.46,1.48,1.49,1.49,1.49 CORRECT
List of # in order from high to low: 1.49,1.49,1.49,1.48,1.46,1.45,1.45,1.44,1.44,1.42,1.41,1.41,1.39,1.39,1.35
Median = 1.44 c. What is the modal selling price?
Modal = 1.49 Lind Chapter 3: Exercise 70
A recent article suggested that if you earn $25,000 a year today and the inflation rate continues at 3 percent per year, you’ll need to make $33,598 in 10 years to have the same buying power. You would need to make $44,771 if the inflation rate jumped to 6 percent.
Confirm that these statements are accurate by finding the geometric mean rate of increase.
Geometric Mean (GM) =( n√Value at end of period/value at beginning of period)-1 n = 10 years, value at beginning of period = 25000 value at end of period @ 3% inflation= 33598
GM = (10√33598/25000)-1
GM = 10√1.34-1
GM = 1.03-1
GM = 0.03 value at end of period @ 6% inflation = 44771 CORRECT
GM = (10√44771/25000)-1
GM = 10√1.79 -1
GM = 1.06-1
GM = 0.06 Lind Chapter 3: Exercise 72
The weights (in pounds) of a sample of five boxes being sent by UPS are: 12, 6, 7, 3, and 10
a. Compute the range.
Range = Largest value - Smallest value
Range = 12-3
Range = 9 b. Compute the mean deviation.
Xbar = ∑X/n
Xbar = 12+6+7+3+10/5
Xbar = 7.6
Mead Deviation (MD) = ∑|X-Xbar|/n
MD = |(12-7.6) + (6-7.6) +(7-7.6)+(3-7.6)+(10-7.6)|/5
MD = |4.4+(-1.6)+(-0.6)+(-4.6)+2.4|/5 CORRECT
MD =( 4.4+1.6+0.6+4.6+2.4)/5
MD = 2.76 c. Compute the standard deviation. s =√ ∑(X-Xbar)²/n-1 s = √[(12-7.6)² + (6-7.6)² +(7-7.6)²+(3-7.6)²+(10-7.6)²]/5-1 s= √19.36+2.56+0.36+21.16+5.76]/4 s=√12.3 s=3.5 Lind Chapter 5: Exercise 8
8. A sample of 2,000 licensed drivers revealed the following number of speeding violations.
# of Vilations # of Drivers
0 1910
1 46
2 18
3 12
4 9
5 or more 5 5
Total 2000
a. What is the experiment? number of speeding violations of 2000 licensed drivers b. List one possible event. of 1910 drivers have 0 violations c.
What is the probability that a particular driver had exactly two speeding violations?
Probability of event happening (PE)= #of favorable outcomes/total # of possible outcomes
PE = 18/2000 d. What concept of probability does this illustrate?
EMPIRICAL
Lind Chapter 5: Exercise 66
A survey of undergraduate students in the School of Business at Northern University revealed the following regarding the gender and majors of the students:
Column1 Column2 Column3 Column4 Column5
Gender Accounting Management Finance Total
Male 100 150 50 300
Female 100 50 50 200
Total 200 200 100 500 a. What is the probability of selecting a female student?
Probability of selecting a female student (PE) = # of female student / total # of student
PE = 200/500
PE = 0.4 b. What is the probability of selecting a finance or accounting major?
P(fiance major) = A = 100/500
P(accounting major) = B = 200/500
P(AorB) = P(A) + P(B)
P(AorB) = 100/500 +
200/500
P(AorB) = 300/500
P(AorB) = 3/5 or 0.6 c. What is the probability of selecting a female or an accounting major? Which rule of addition did you apply?
P(female) = A = 200/500
P(accounting major) = B = 200/500
P(accounting major and female) = 100/500
P(AorB) = P(A) +P(B) - P(AandB)
P(AorB) = 200/500 + 200/500 - 100/500
P(AorB) = 300/500
P(AorB) = 3/5 or 0.6
The General Rule of Addition was applied in this scenario. d. Are gender and major independent? Why?
Yes, gender and major are independent because gender does not determine the type of major a student chooses; or the occurrence of one event has no effect on the probability of the occurrence of another event. e. What is the probability of selecting an accounting major, given that the person selected is a male?
P(accounting major) = A = 200/500
P(male) = B = 300/500
P(accounting major and male) = (AΠB) = 100/500
P(A|B) = P(AΠB)/P(B)
P(A|B) = 100/500 /300/500
P(A|B) = 1/3 or 0.33 f. Suppose two students are selected randomly to attend a lunch with the president of the university. What is the probability that both of those selected are accounting majors?
P(accounting major) = A = 200/500
P(accounting major) = B = 200/500
P(non-accounting major) = (AUB) = 300/500
P(AΠB) = P(A) + P(B) - P(AUB)
P(AΠB) = (200/500)*(199/499)
Lind Chapter 6: Exercise 4
Which of these variables are discrete and which are continuous random variables?
a. The number of new accounts established by a salesperson in a year.
Discrete random variables b. The time between customer arrivals to a bank ATM.
Continuous random variables c. The number of customers in Big Nick’s barber shop.
DISCRETE
d. The amount of fuel in your car’s gas tank.
Continuous random variables e. The number of minorities on a jury.
Discrete random variables f. The outside temperature today.
Continuous random variables Lind Chapter 7: Exercise 38
The accounting department at Weston Materials, Inc., a national manufacturer of unattached garages, reports that it takes two construction workers a mean of 32 hours and a standard deviation of 2 hours to erect the Red Barn model. Assume the assembly times follow the normal distribution. µ= 32, σ= 2
a. Determine the z values for 29 and 34 hours. What percent of the garages take between
32 hours and 34 hours to erect? z values at 29 and 34 hours z = (x - µ)/σ z = (29-32)/2 z = (34-32)/2 z = -1.5 z = 1.0 z values at 32 and 34 hours z = (32-32)/2 z = (34-32)/2) z = 0 z = 1.0 at 34 hours, z = 1.0, and the value is 0.3413
34.13% of the garages take between 32 and 34 hours to erect. b. What percent of the garages take between 29 hours and 34 hours to erect? at 29 hours, z = -1.5, and the value is 0.4332 at 34 hours, z = 1.0, and the value is 0.3413
0.4332+0.3413 = 0.7745 or 77.45%
77.45% of the grages take between 29 hours and 34 hours to erect. c. What percent of the garages take 28.7 hours or less to erect? z = (28.7-32)/2 z = -1.65 c. 0.0495, found by 0.5000 - 0.4505 at 28.7 hours, z = 1.65, and the value = 0.4505 or 45.05% d. 35.3, found by 32 + 1.65(2)
Maximium 45.05% of the garages take 28.7 hours or less to erect. d. Of the garages, 5 percent take how many hours or more to erect? where: z= 5, µ=32, σ=2, x =? z = (x-µ)/σ
5 = (x-32)/2
10 = x -32 x = 22
Of the garages, 5 percent takes 22 hours or more to erect. Lind Chapter 7: Exercise 44
The number of passengers on the Carnival Sensation during one-week cruises in the
Caribbean follows the normal distribution. The mean number of passengers per cruise is
1,820 and the standard deviation is 120. µ = 1820, σ= 120
a. What percent of the cruises will have between 1,820 and 1,970 passengers? z = (x-µ)/σ at 1820 passengers at 1970 passengers z = (1820-1820)/120 z = (1970-1820)/120 z = 0 z = 1.25 at 1970 passengers, z = 1.25, and the value is 0.3944
0.00% to 39.44% of the cruises will have between 1820 and 1970 passengers. b. What percent of the cruises will have 1,970 passengers or more?
39.444% or more of the cruises will have 1970 pssengers or more. c. What percent of the cruises will have 1,600 or fewer passengers?
P(X>1970) = .10565 P(X
Owens Orchards sells apples in a large bag by weight. A sample of seven bags contained the following numbers of apples: 23,19,26,17,21,24,22.
a. Compute the mean number and median number of apples in a bag. Total number of Problems = 11
Mean = (23+19+26+17+21+24+22)/7 = 21.71 Problems correct = 8
Median = 22 CORRECT
List of # in order from low to high: 17,19,21,22,23,24,26
List of # in order from high to low: 26,24,23,22,21,19,17 b. Verify that Σ(x-xbar) = 0.
(17-21.71)+(19-21.71)+(21-21.71)+(22-21.71)+(23-21.71)+(24-21.71)+(26-21.71) ≈ 0 Lind Charpter 3: Exercise 62 …show more content…
The Citizens Banking Company is studying the number of times the ATM lcoated in a Loblaws Supermarket at the foot of Market Street is used per day. Following are the numbers of times the machine was used over each of the last 30 days. Determine the mean number of times that machine was used per day. n = 30
83 64 84 76 84 54 75 59 70 61
63 80 84 73 68 52 65 90 52 77
95 36 78 61 59 84 95 47 87 60
Mean = (83+64+84+76+84+54+75+59+70+61+63+80+84+73+68+52+65+90+52+77+95+36+78+61+59+84+95+47+87+60)/30
Mean = 2116/30 CORRECT
Mean = 70.53
The mean number of times that machine was used per day = 70.53 Lind Chapter 3: Exercise 68
The American Automobile Association checks the prices of gasoline before many holiday weekends. Listed below are the self-service prices for a sample of 15 retail outlets during the May 2003 Memorial Day weekend in the Detroit, Michigan, area.
1.44, 1.42, 1.35, 1.39, 1.49, 1.49, 1.41, 1.46,
1.41, 1.49, 1.45, 1.48, 1.39, 1.46, 1.44,
N=15
a. What is the arithmetic mean selling price? µ= (1.44+1.42+1.35+1.39+1.49+1.49+1.41+1.46+1.41+1.49+1.45+1.48+1.39+1.46+1.44)/15 µ= 21.57/15 µ= 1.44 b. What is the median selling price?
List of # in order from low to high: 1.35,1.39,1.39,1.41,1.41,1.42,1.44,1.44,1.45,1.45,1.46,1.48,1.49,1.49,1.49 CORRECT
List of # in order from high to low: 1.49,1.49,1.49,1.48,1.46,1.45,1.45,1.44,1.44,1.42,1.41,1.41,1.39,1.39,1.35
Median = 1.44 c. What is the modal selling price?
Modal = 1.49 Lind Chapter 3: Exercise 70
A recent article suggested that if you earn $25,000 a year today and the inflation rate continues at 3 percent per year, you’ll need to make $33,598 in 10 years to have the same buying power. You would need to make $44,771 if the inflation rate jumped to 6 percent.
Confirm that these statements are accurate by finding the geometric mean rate of increase.
Geometric Mean (GM) =( n√Value at end of period/value at beginning of period)-1 n = 10 years, value at beginning of period = 25000 value at end of period @ 3% inflation= 33598
GM = (10√33598/25000)-1
GM = 10√1.34-1
GM = 1.03-1
GM = 0.03 value at end of period @ 6% inflation = 44771 CORRECT
GM = (10√44771/25000)-1
GM = 10√1.79 -1
GM = 1.06-1
GM = 0.06 Lind Chapter 3: Exercise 72
The weights (in pounds) of a sample of five boxes being sent by UPS are: 12, 6, 7, 3, and 10
a. Compute the range.
Range = Largest value - Smallest value
Range = 12-3
Range = 9 b. Compute the mean deviation.
Xbar = ∑X/n
Xbar = 12+6+7+3+10/5
Xbar = 7.6
Mead Deviation (MD) = ∑|X-Xbar|/n
MD = |(12-7.6) + (6-7.6) +(7-7.6)+(3-7.6)+(10-7.6)|/5
MD = |4.4+(-1.6)+(-0.6)+(-4.6)+2.4|/5 CORRECT
MD =( 4.4+1.6+0.6+4.6+2.4)/5
MD = 2.76 c. Compute the standard deviation. s =√ ∑(X-Xbar)²/n-1 s = √[(12-7.6)² + (6-7.6)² +(7-7.6)²+(3-7.6)²+(10-7.6)²]/5-1 s= √19.36+2.56+0.36+21.16+5.76]/4 s=√12.3 s=3.5 Lind Chapter 5: Exercise 8
8. A sample of 2,000 licensed drivers revealed the following number of speeding violations.
# of Vilations # of Drivers
0 1910
1 46
2 18
3 12
4 9
5 or more 5 5
Total 2000
a. What is the experiment? number of speeding violations of 2000 licensed drivers b. List one possible event. of 1910 drivers have 0 violations c.
What is the probability that a particular driver had exactly two speeding violations?
Probability of event happening (PE)= #of favorable outcomes/total # of possible outcomes
PE = 18/2000 d. What concept of probability does this illustrate?
EMPIRICAL
Lind Chapter 5: Exercise 66
A survey of undergraduate students in the School of Business at Northern University revealed the following regarding the gender and majors of the students:
Column1 Column2 Column3 Column4 Column5
Gender Accounting Management Finance Total
Male 100 150 50 300
Female 100 50 50 200
Total 200 200 100 500 a. What is the probability of selecting a female student?
Probability of selecting a female student (PE) = # of female student / total # of student
PE = 200/500
PE = 0.4 b. What is the probability of selecting a finance or accounting major?
P(fiance major) = A = 100/500
P(accounting major) = B = 200/500
P(AorB) = P(A) + P(B)
P(AorB) = 100/500 +
200/500
P(AorB) = 300/500
P(AorB) = 3/5 or 0.6 c. What is the probability of selecting a female or an accounting major? Which rule of addition did you apply?
P(female) = A = 200/500
P(accounting major) = B = 200/500
P(accounting major and female) = 100/500
P(AorB) = P(A) +P(B) - P(AandB)
P(AorB) = 200/500 + 200/500 - 100/500
P(AorB) = 300/500
P(AorB) = 3/5 or 0.6
The General Rule of Addition was applied in this scenario. d. Are gender and major independent? Why?
Yes, gender and major are independent because gender does not determine the type of major a student chooses; or the occurrence of one event has no effect on the probability of the occurrence of another event. e. What is the probability of selecting an accounting major, given that the person selected is a male?
P(accounting major) = A = 200/500
P(male) = B = 300/500
P(accounting major and male) = (AΠB) = 100/500
P(A|B) = P(AΠB)/P(B)
P(A|B) = 100/500 /300/500
P(A|B) = 1/3 or 0.33 f. Suppose two students are selected randomly to attend a lunch with the president of the university. What is the probability that both of those selected are accounting majors?
P(accounting major) = A = 200/500
P(accounting major) = B = 200/500
P(non-accounting major) = (AUB) = 300/500
P(AΠB) = P(A) + P(B) - P(AUB)
P(AΠB) = (200/500)*(199/499)
Lind Chapter 6: Exercise 4
Which of these variables are discrete and which are continuous random variables?
a. The number of new accounts established by a salesperson in a year.
Discrete random variables b. The time between customer arrivals to a bank ATM.
Continuous random variables c. The number of customers in Big Nick’s barber shop.
DISCRETE
d. The amount of fuel in your car’s gas tank.
Continuous random variables e. The number of minorities on a jury.
Discrete random variables f. The outside temperature today.
Continuous random variables Lind Chapter 7: Exercise 38
The accounting department at Weston Materials, Inc., a national manufacturer of unattached garages, reports that it takes two construction workers a mean of 32 hours and a standard deviation of 2 hours to erect the Red Barn model. Assume the assembly times follow the normal distribution. µ= 32, σ= 2
a. Determine the z values for 29 and 34 hours. What percent of the garages take between
32 hours and 34 hours to erect? z values at 29 and 34 hours z = (x - µ)/σ z = (29-32)/2 z = (34-32)/2 z = -1.5 z = 1.0 z values at 32 and 34 hours z = (32-32)/2 z = (34-32)/2) z = 0 z = 1.0 at 34 hours, z = 1.0, and the value is 0.3413
34.13% of the garages take between 32 and 34 hours to erect. b. What percent of the garages take between 29 hours and 34 hours to erect? at 29 hours, z = -1.5, and the value is 0.4332 at 34 hours, z = 1.0, and the value is 0.3413
0.4332+0.3413 = 0.7745 or 77.45%
77.45% of the grages take between 29 hours and 34 hours to erect. c. What percent of the garages take 28.7 hours or less to erect? z = (28.7-32)/2 z = -1.65 c. 0.0495, found by 0.5000 - 0.4505 at 28.7 hours, z = 1.65, and the value = 0.4505 or 45.05% d. 35.3, found by 32 + 1.65(2)
Maximium 45.05% of the garages take 28.7 hours or less to erect. d. Of the garages, 5 percent take how many hours or more to erect? where: z= 5, µ=32, σ=2, x =? z = (x-µ)/σ
5 = (x-32)/2
10 = x -32 x = 22
Of the garages, 5 percent takes 22 hours or more to erect. Lind Chapter 7: Exercise 44
The number of passengers on the Carnival Sensation during one-week cruises in the
Caribbean follows the normal distribution. The mean number of passengers per cruise is
1,820 and the standard deviation is 120. µ = 1820, σ= 120
a. What percent of the cruises will have between 1,820 and 1,970 passengers? z = (x-µ)/σ at 1820 passengers at 1970 passengers z = (1820-1820)/120 z = (1970-1820)/120 z = 0 z = 1.25 at 1970 passengers, z = 1.25, and the value is 0.3944
0.00% to 39.44% of the cruises will have between 1820 and 1970 passengers. b. What percent of the cruises will have 1,970 passengers or more?
39.444% or more of the cruises will have 1970 pssengers or more. c. What percent of the cruises will have 1,600 or fewer passengers?
P(X>1970) = .10565 P(X