The speeds of car traveling on Interstate Highway I-35 are normally distributed with a mean of 74
miles per hour and a standard deviation of 6 miles per hour.
(a) Find the percentage of the cars traveling on this highway with a speed
i. of more than 85,
ii. between 65 to 72.
(b) If a BMW is at the speed that is faster than 90 percentage of cars, what is the speed of the
BMW?

A map suggests that Atlanta is 730 miles in a direction 5.00° north of east from Dallas. The same map shows that Chicago is 560 miles in a direction 21.0° west of north from Atlanta. The figure below shows the location of these three cities. Modeling the Earth as flat, use this information to find the displacement from Dallas to Chicago.

A bus travels between two cities A and B that are 100 miles apart.Two service stations are located at mile 30 and mile 70, as well as in the cities themselves. The bus breaks down on the road. Assuming the place of breakdown is uniformly distributed between the cities, what is the probability that it is no more than 10 miles to the nearest service station? What is the expectation of the distance to the nearest service station?

The annualized cost of acquiring capacity for the new barracuda drives at Seagate is calculated as $25 per unit and the contibution margin for the product is $45 per unit.

a.) what service level(percentile of demand met) should seagate target for in building capacity? how much capacity will it build? the demand forcast for the barracuda drives is uniform between 100 and 200.

b.) how does this service level change, if Seagate outsources manufacturing to china, to arrive at an annualized cost of capacity of $10 per unit? How much capacity will it build?

c.) How does this service level change if the inbound shipment cost per unit from China to US reduces the margin per unit to $35? How much capacity will it build?

d.) Conduct the capacity calculations in all 3 scenarios above when the forcast for the barracuda drives is expected to follow the following distributions.

Let X1, X2, ..., Xn be a random sample of size from a distribution with probability density function

f(x) = λxλ−1 , 0 < x < 1, λ > 0

a) Get the method of moments estimator of λ. Calculate the estimate when x1 = 0.1, x2 = 0.2, x3 = 0.3.

b) Get the maximum likelihood estimator of λ. Calculate the estimate when x1 = 0.1, x2 = 0.2, x3 = 0.3.

An oil company wants to build a pipeline to take oil from an oil well to a refinery. Unfortunately, the well and the refinery are on either side of a straight river which is 10 miles wide, and they are 50 miles apart along the coastline (that is, if you want to go from the well to the refinery you must first cross 10 miles of river and then go 50 miles along the side of the river). The company has hired you to figure out the cheapest way to build the pipeline, and you need to clearly explain your solution so that the less mathematically-sophisticated oil people will understand.

It costs $200 per mile to lay pipe across the river but only $160 per mile to lay the pipe over land. There is also one other potential cost: It costs extra money each time that you have a bend in the pipeline. If you go straight across the river and use an L-shaped bend it costs an extra $150. If you lay the pipe diagonally across the river but hit land before you get to the refinery, you have to use a slanted L-shaped bend (like an obtuse angle). These have to be custom made and they cost $975. If you go directly diagonally across the river to the refinery without touching any land, then you do not have to pay extra for a bend (since there will not be one).

Exercise 1. Find the cheapest path to lay the pipeline by doing the following:

(a) Draw a diagram of the situation. Include any variables that you are going to use in the rest of your answer.

(b) Calculate the cost of building the pipeline for a few different situations: (i) How much would it cost to build the pipeline 10 miles straight across the river, then make a 90 degree bend and go 50 miles along the side of the river? (ii) How much would it cost to go diagonally across the river, without going on land at all? (iii) Suppose P is the point directly across the river from the oil well. How much would it cost to go diagonally across the river to a point 10 miles along the bank from P, then bend the pipeline and go the remaining 40 miles along the side of the river? (iv) How much would it cost to go diagonally across the river to a point 40 miles along the bank from P, then bend the pipeline and go the remaining distance along the side of the river?

(c) Calculate the cheapest way to build the pipeline for the situation given above by finding the minimum cost among ALL possible locations for P. (i) You will need to write a general expression for the cost of building the pipeline1 , and then use calculus to minimize the cost. (ii) Use either the first or second derivative test to prove that your result is a minimum. (d) What is the effect of the extra cost for a bend in the pipeline? If there was no extra cost for the bend, would you have a different answer for what the cheapest path would be?

Collect Customer Data - Part 2

1. Collect Mileage information:

   Prompt: "Starting Odometer Reading:\n"

   Variable: odoStart = ?

   Prompt: "Ending Odometer Reading:\n"

   Variable: odoEnd = ?

   Add code to PRINT odoStart and odoEnd variables as well as the totalMiles to check your work.

   The following data will be used as input in the test:

   1. Prompt the user to input the starting odometer reading (expect type int from the user)
   2. Prompt the user to input the ending odometer reading (expect type int from the user)
   3. Test your code.

import sys
'''
Section 1: Collect customer input
'''
#Add customer input 1 here, rentalCode = ?

#Collect Customer Data - Part 2

#4)Collect Mileage information:
#a)   Prompt the user to input the starting odometer reading and store it as the variable odoStart

#Prompt -->"Starting Odometer Reading:\n"
# odoStart = ?

#b)   Prompt the user to input the ending odometer reading and store it as the variable odoEnd

#Prompt -->"Ending Odometer Reading:"
# odoEnd = ?

#c) Calculate total miles

#Print odoStart, odoEnd and totalMiles

# Calculate Charges 2

##   Calculate the mileage charge and store it as
# the variable mileCharge:

#a)   Code 'B' (budget) mileage charge: $0.25 for each mile driven

#b)   Code 'D' (daily) mileage charge: no charge if the average
# number of miles driven per day is 100 miles or less;
# i)   Calculate the averageDayMiles (totalMiles/rentalPeriod)

# ii)   If averageDayMiles is above the 100 mile per day
# limit:
# (1)   calculate extraMiles (averageDayMiles - 100)
# (2)   mileCharge is the charge for extraMiles,
# $0.25 for each mile

#c)   Code 'W' (weekly) mileage charge: no charge if the
# average number of miles driven per week is
# 900 miles or less;

# i)   Calculate the averageWeekMiles (totalMiles/ rentalPeriod)

# ii)   mileCharge is $100.00 per week if the average number of miles driven per week exceeds 900 miles

Im getting a Error on my Script at line 55:

LAST RUN on 9/14/2019, 8:44:31 AM

Check 1 failed

Output:

File "rental_car-customer-data-2.py", line 55 elif rentalCode == 'W' or rentalCode=='w': ^ SyntaxError: invalid syntax

Expected:

Starting Odometer Reading: Ending Odometer Reading: 1234 2222 988

My Script:

import sys
'''
Section 1: Collect customer input
'''
#Add customer input 1 here, rentalCode = ?
rentalCode = input("(B)udget, (D)aily, or (W)eekly rental?\n")
print (rentalCode)
#Collect Customer Data - Part 2

#4)Collect Mileage information:
#a)   Prompt the user to input the starting odometer reading and store it as the variable odoStart

#Prompt -->"Starting Odometer Reading:\n"
# odoStart = ?
odoStart = input('Starting Odometer Reading: ')
#b)   Prompt the user to input the ending odometer reading and store it as the variable odoEnd

#Prompt -->"Ending Odometer Reading:"
# odoEnd = ?
odoEnd = input('Ending Odometer Reading: ')
#c) Calculate total miles
totalMiles = int(odoEnd) - int(odoStart)
#Print odoStart, odoEnd and totalMiles
print (odoStart)
print (odoEnd)
print (totalMiles)

# Calculate Charges 2

##   Calculate the mileage charge and store it as
# the variable mileCharge:

#a)   Code 'B' (budget) mileage charge: $0.25 for each mile driven
if rentalCode == "B" or rentalCode=='b':
mileCharge = totalMiles * 0.25
#b)   Code 'D' (daily) mileage charge: no charge if the average
# number of miles driven per day is 100 miles or less;
# i)   Calculate the averageDayMiles (totalMiles/rentalPeriod)
elif rentalCode == "D" or rentalCode=='d':
rentalPeriod = int(input('Enter days rented: '))
averageDayMiles = totalMiles/rentalPeriod
if averageDayMiles <= 100:
mileCharge=0
# ii)   If averageDayMiles is above the 100 mile per day
# limit:
# (1)   calculate extraMiles (averageDayMiles - 100)
else:
# (2)   mileCharge is the charge for extraMiles,
# $0.25 for each mile
mileCharge = (averageDayMiles-100) *0.25

#c)   Code 'W' (weekly) mileage charge: no charge if the
# average number of miles driven per week is
# 900 miles or less;
elif rentalCode == 'W' or rentalCode=='w':
rentalWeek = int(input('Enter rental week: '))
averageWeekMiles = totalMiles / rentalWeek
if averageWeekMiles<=900:
mileCharge=0
else:
mileCharge=100*rentalWeek
print('Charges : ${}'.format(mileCharge))
# i)   Calculate the averageWeekMiles (totalMiles/ rentalPeriod)

# ii)   mileCharge is $100.00 per week if the average number of miles driven per week exceeds 900 miles

Levi-Strauss Co manufactures clothing. The quality control department measures weekly values of different suppliers for the percentage difference of waste between the layout on the computer and the actual waste when the clothing is made (called run-up). The data is in table #11.3.3, and there are some negative values because sometimes the supplier is able to layout the pattern better than the computer ("Waste run up," 2013). Do the data show that there is a difference between some of the suppliers? Test at the 1% level.

Table #11.3.3: Run-ups for Different Plants Making Levi Strauss Clothing

ONLY ANSWER PART C, D AND E

1A. Suppose The lifetime for a competing brands of tires are independent of each other and approximately normal with unknown means but known variances

σ^2= 975 miles^2 and σ^2= 965 miles^2. We gather did some testing under controlled conditions and have the following summary statistics:

n1= 75

̄x1= 3251.4

n2= 60

̄x2= 3274.7

Construct a 80% confidence interval for the difference in population means,μ1−μ2.

B.Using the same setup as part A, conduct a hypothesis test to check if the means are equal. Use α= 0.2.

C.Using the same setup as part A, but now assuming we do not know the variances but assume they are equal. Construct a 99% confidence interval for the difference in means using the same sample data as before. We now know that the sample variances are s^2= 982.2 and s^2= 967.4.

D.Using the same setup as part C, construct a 90% confidence interval for the ratio of variances σ^2/ σ^2.

E. Using the same setup as part C, conduct a hypothesis test to test whether population 1 has a smaller variance than population 2 using α= 0.1.