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

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.

ForNeutralAgainstTotals

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

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 14710 0223811 13556677712 22348913 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?

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?

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?

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

Purchases (gals)Percentage

5 Construct a cumulative percentage distribution for the detergent data ifthe corresponding frequency distribution uses "9.0 but less than 10.0" as the first class

Purchases (gals)FrequencyPercentage

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.

ProbabilityTasteDesign ADesign B

= [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?

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]

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 toproduce both designs?

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

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

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 Ȳ183.81) (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 – X183.81)^2 + (236 – 183.81)^2 + …+ (189 – 183.81)^2 = 11896.44sX = 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 – Ȳ79.31)^2 ≈ 361.44

sY = sqrt {[Σ(Yi - )^2] / (n - 1)}Ȳ = sqrt (361.44) / (16-1) ≈ 4.91

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 alcoholplay a role?

Number of Vehicles InvolvedTotals

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

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

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

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

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

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 bagsper 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?

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

P(x < 2 ∩ x > 8) = 0