mas202 applied statistics for business individual assignment fa23

Suppose that a percentage histogram for the detergent data is constructed, using "9.0 but less than 10.0 gallons" as the first class.. If a percentage histogram for the detergent data is

MAS202 – Applied Statistics for Business INDIVIDUAL ASSIGNMENT

FA23

Name: Nguyen Quoc Duy

ID: CS171997

Class: BA1705

Lecturer:

Can Tho Campus

SCENARIO 1: The table below contains the opinions of a sample of 200 people broken down by gender about the latest congressional plan to eliminate anti-trust exemptions for professional baseball

For Neutral Against Totals

1 Construct a table of row percentages

Answer:

For Neutral Against Totals

2 Construct a table of column percentages

Answer:

For Neutral Against Totals

3 Construct a table of total percentages

Answer:

For Neutral Against Totals

4 How many percent of the 200 were females who were either neutral or against the plan?

Answer:

Percent of the 200 were females who were either neutral or against the plan: [(54+12)/200] x 100% = 33%

5 How many percent of the 200 were males who were not against the plan?

Answer:

Percent of the 200 were males who were not against the plan:

[(12+36)/200] x 100% = 24%

6 How many percent of the 200 were not neutral?

Answer:

Percent of the 200 were not neutral:

[(50+60)/200] x 100% = 55%

7 How many percent of the 200 were against the plan?

Answer:

Percent of the 200 were against the plan:

(60/200) * 100 = 30%

SCENARIO 2: Given below is the stem-and-leaf display representing the amount

of detergent used in gallons (with leaves in 10ths of gallons) in a day by 25 drive-through car wash operations in Phoenix

9 147

10 02238

11 135566777

12 223489

13 02

1 Suppose that a percentage histogram for the detergent data is

constructed, using "9.0 but less than 10.0 gallons" as the first class Determine the percentage of drive-through car wash operations that use

“12.0 but less than 13.0 gallons” of detergent?

Answer:

12.0 but less than 13.0 gallons is: (6/25) x 100% = 24%

2 If a percentage histogram for the detergent data is constructed, using

"9.0 but less than 10.0 gallons" as the first class, what percentage of drive-through car wash operations use less than 12 gallons of detergent

in a day?

Answer:

less than 12 gallons is: (17/25) x 100% = 68%

3 If a relative frequency or percentage distribution for the detergent data

is, using "9.0 but less than 10.0 gallons" as the first class, what percentage of drive-through car wash operations use at least 10 gallons

of detergent in a day?

Answer:

at least 10 gallons is:(22/25) x 100% = 88%

4 Construct a relative frequency or percentage distribution for the detergent data, using "9.0 but less than 10.0" as the first class

Answer:

Purchases (gals) Percentage

5 Construct a cumulative percentage distribution for the detergent data if the corresponding frequency distribution uses "9.0 but less than 10.0"

as the first class

Answer:

Purchases (gals) Frequency Percentage

6 Construct a percentage histogram for the detergent data, using "9.0 but less than 10.0" as the first class

SCENARIO 3: Two different designs on a new line of winter jackets for the coming winter are available for your manufacturing plants Your profit (in thousands of dollars) will depend on the taste of the consumers when winter arrives The probability of the three possible different tastes of the consumers and the corresponding profits are presented in the following table

Probability Taste Design A Design B

1 What is your expected profit when Design B is chosen?

Answer:

Profit when Design B is chosen:

µ = E(X) = E(B) = (520,000 x 0.2) + (350,000 x 0.5) + (270,000 x 0.3) = $360,000

2 What is the variance of your profit when Design A is chosen?

Answer:

Profit when Design A is chosen:

µ = E(X) = E(A) = (180,000 x 0.2) + (230,000 x 0.5) + (350,000 x 0.3) = $256,000 Variance = Σ {P(x) x [x - E(x)] ^2}

= [0.2 x (180,000 – 256,000) ^2] + [0.5 x (230,000 – 256,000) ^2] + [0.3 x (350,000 – 256,000) ^2] = 4,144,000,000

3 What is the standard deviation of your profit when Design B is chosen?

Answer:

Variance = Σ {P(x) x [x - E(x)] ^2}

= [0.2 x (520,000 – 360,000) ^2] + [0.5 x (350,000 – 360,000) ^2] + [0.3 x (270,000 – 360,000) ^2]

= 7,600,000,000

Standard deviation (B) = SQRT (7,600,000,000) = 87177.98

4 What is the covariance of the profits from the two different designs?

Answer:

Covariance = ∑ [(x - μx) x (y - μy) x P (x, y)]

= [(180,000 – 256,000) x (520,000 – 360,000) x 0.2] + [(230,000 – 256,000) x (350,000-360,000) x 0.5] + [(350,000 – 256,000) x (270,000-360,000) x 0.3]

= -4,840,000,000

5 What is the expected profit, the total variance, and the total standard deviation if you increase the shift of your production lines and choose to produce both designs?

Answer:

Expected profit = E(A) + E(B) = 256,000 + 360,000 = $616,000

Variance (A) = Σ {P(x) x [x - E(x)]^2}

= [0.2 x (180,000– 256,000)^2] + [0.5 x (230,000 – 256,000)^2] + [0.3 x (350,000 – 256,000)^2]= 4,144,000,000

Variance (B) = Σ {P(x) x [x - E(x)]^2}

= [0.2 x (520,000 – 360,000)^2] + [0.5 x (350,000 – 360,000)^2] + [0.3 x (270,000 – 360,000)^2]

= 7,600,000,000

Total variance = Variance (A) + Variance (B) = 4,144,000,000 + 7,600,000,000 = 11,744,000,000,000

Total satandal deviation = sqrt(11,744,000,000,000) =108369.74

SCENARIO 4: Given below are the rating and performance scores of 15 laptop computers

Performance Score 115 191 153 194 236 184 184 216

Performance Score 185 183 189 202 192 141 187 189

1 What is the sample covariance between the performance scores and the rating?

Answer:

= (115 + 191 + 153 + 194 + 236 + 184 + 184 + 216 + 185 + 183 + 189 + X

202 + 192 + 141 + 187 + 189)/ 16 ≈ 183.81

= (74 + 78 + 79 + 80 + 84 + 76 + 77 + 95 + 83 + 78 + 77 + 78 + 79 + 75 + Ȳ

77 + 79)/ 16 ≈ 79.31

Cov (X, Y) = [(Xi - ) (Yi - )]/(n-1) = [(115 –183.81) (74 -79.31) + (191 –X Ȳ 183.81) (78 – 79.31) + …+(187 – 183.81) (77 – 79.31) + (189 – 183.81) (79 – 79.31)]/15 ≈ 84.26

2 What is the sample correlation coefficient between the performance scores and the rating?

Answer:

Σ(Xi - )^2 = (115 – 183.81)^2 + (191 – 183.81)^2 + (153 – 183.81)^2 + (194 – X

183.81)^2 + (236 – 183.81)^2 + …+ (189 – 183.81)^2 = 11896.44

sX = sqrt {[Σ(Xi - )^2] / (n - 1)}X

= sqrt (11896.44)/(16 – 1) ≈ 28.16

Σ(Yi - )^2 = (74 -79.31)^2 + (78 – 79.31)^2 +…+ (77 – 79.31)^2 + (79 – Ȳ 79.31)^2 ≈ 361.44

sY = sqrt {[Σ(Yi - )^2] / (n - 1)}Ȳ

= sqrt (361.44) / (16-1) ≈ 4.91

r = cov(X,Y) / (sX x sY) = 84.26 / (28.16 * 4.91) ≈ 0.61

3 How will you classify the linear relationship between the performance scores and the rating?

Answer:

The correlation coefficient here is 0.61, which is greater than 0 and too close to 1

So we can confirm that if the value of one variable increases, the value of the other variable also increases this correlation is strong

SCENARIO 5: Mothers Against Drunk Driving is a very visible group whose focus is to educate the public about the harm caused by drunk drivers A study was recently done that emphasized the problem we all face with drinking and driving Four hundred accidents that occurred on a Saturday night were analyzed Two items noted were the number of vehicles involved and whether alcohol played a role in the accident The numbers are shown below:

Did alcohol

play a role?

Number of Vehicles Involved Totals

1 What proportion of accidents involved more than one vehicle?

Answer:

Accidents involved more than one vehicle: [(275+100)/450] x 100% = 83.33%

2 What proportion of accidents involved alcohol and a single vehicle?

Answer:

Accidents involved alcohol and a single vehicle: (50/450) x 100% = 11.11%

3 What proportion of accidents involved alcohol or a single vehicle?

Answer:

Accidents involved alcohol or a single vehicle:

[(50+25+100+40)/450] x 100% = 47.78%

4 Given alcohol was involved, what proportion of accidents involved a single vehicle?

Answer:

Given alcohol was involved, accidents involved a single vehicle:

(50/190) x 100% = 26.32%

5 Given that multiple vehicles were involved, what proportion of accidents involved alcohol?

Answer:

Given that multiple vehicles were involved, accidents involved alcohol:

[(100+40)/(275+100)] x 100% = 37.33%

bags per 1,000 customers In a recent year, the hotel chain had 5.6 mishandled bags per 1,000 customers Assume that the number of mishandled bags has a Poisson distribution

1 What is the probability that in the next 1,000 customers, the hotel chain will have no mishandled bags?

Answer:

P(x = 0) = [e - (5.6) x (5.6) 0

]/ 0! = 0.0063

2 What is the probability that in the next 1,000 customers, the hotel chain will have at least two mishandled bags?

P(x≥2) = 1 – P(x < 2) =1-[P(x = 0) + P(x = 1)]

= 1 – {0.0063 +[ e −(5.6 )

x (5.6) 1 ]/1!} = 0.96

3 What is the probability that in the next 1,000 customers, the hotel chain will have between two and four inclusive mishandled bags?

P(2 ≤ x ≤ 4 ¿ = P(x < 5) – P(x < 2)

= P(x = 2) + P(x = 3) + P(x = 4)

= 0.0580 + 0.1082 + 0.1515

= 0.39

4 What is the probability that in the next 1,000 customers, the hotel chain will have less than two or more than eight mishandled bags?

P(x < 2 U x > 8) = P(x = 0) + P(x = 1) + P(x > 8)

= P(x = 0) + P(x = 1) + 1 – [P(x = 0) + P(x = 1) + P(x = 0) + P(x = 2) + + P(x

= 8)]

= 1- [P(x = 2) + P(x = 3) + P(x = 4) + P(x = 5) + P(x = 6) + P(x = 7) + P(x = 8)] = 0.11

5 What is the probability that in the next 1,000 customers, the hotel chain will have less than two and more than eight mishandled bags.

P(x < 2 ∩ x > 8) = 0