What is the overall efficiency for the removal with the fabric filter for a distribution of particles where 0.05% of the particles are of 0.1 microns diameter, 0.2% of particles are of 0.3 microns diameter, 2.8% of the particles are 0.7 microns and remaining particles are in the particle diameter range of 1 to 50 microns.

Suppose we have a two-armed bandit. Our estimate of the payout rate of the first arm is 0.7, and our estimate of the payout rate for the second arm is 0.6. That is, ρ1 = 0.7 and ρ2 = 0.6. Our 95% confidence intervals for θ1 and θ2 are (0.6, 0.8) and (0.3, 0.9), respectively. Suppose you used an -greedy strategy with = 0.5. How might you decide which arm to pull? Suppose you used an interval exploration strategy with 95% confidence intervals. What arm would you pull?

Miles  Freq
 0-4     3
 5-9    14
10-14   13
15-19    4

Select the most appropriate sentence corresponding to two standard deviations.

*About 68% of students drive between 5.5212 miles and 13.7730 miles to somewhere
*At least 88.9% of students drive between -2.7306 miles and 22.0248 miles to 
*About 99.7% of students drive between 1.3953 miles and 17.8989 miles to 
*About 68% of students drive less than 22.0248 miles to
*About 95% of students drive between 5.5212 miles and 13.7730 miles to
*About 99.7% of students drive between -2.7306 miles and 22.0248 miles to
*About 68% of students drive between 1.3953 miles and 17.8989 miles to
*About 99.7% of students drive between 5.5212 miles and 13.7730 miles to
*At least 75% of students drive between -2.7306 miles and 22.0248 miles to
*At least 75% of students drive less than 22.0248 miles to
*About 95% of students drive between 1.3953 miles and 17.8989 miles to
*At least 75% of students drive between 1.3953 miles and 17.8989 miles to
*About 95% of students drive less than 22.0248 miles to 
*At least 88.9% of students drive between 1.3953 miles and 17.8989 miles to
*About 99.7% of students drive less than 22.0248 miles to
*About 95% of students drive between -2.7306 miles and 22.0248 miles to
*At least 88.9% of students drive less than 22.0248 miles to
*About 68% of students drive between -2.7306 miles and 22.0248 miles to

Miles  Freq
 0-4     3
 5-9    14
10-14   13
15-19    4

Write two sentences for non-statisticians expressing Chebyshev’s Theorem. Select the most appropriate sentence corresponding to two standard deviations.

*About 68% of students drive between 5.5212 miles and 13.7730 miles to somewhere
*At least 88.9% of students drive between -2.7306 miles and 22.0248 miles to 
*About 99.7% of students drive between 1.3953 miles and 17.8989 miles to 
*About 68% of students drive less than 22.0248 miles to
*About 95% of students drive between 5.5212 miles and 13.7730 miles to
*About 99.7% of students drive between -2.7306 miles and 22.0248 miles to
*About 68% of students drive between 1.3953 miles and 17.8989 miles to
*About 99.7% of students drive between 5.5212 miles and 13.7730 miles to
*At least 75% of students drive between -2.7306 miles and 22.0248 miles to
*At least 75% of students drive less than 22.0248 miles to
*About 95% of students drive between 1.3953 miles and 17.8989 miles to
*At least 75% of students drive between 1.3953 miles and 17.8989 miles to
*About 95% of students drive less than 22.0248 miles to 
*At least 88.9% of students drive between 1.3953 miles and 17.8989 miles to
*About 99.7% of students drive less than 22.0248 miles to
*About 95% of students drive between -2.7306 miles and 22.0248 miles to
*At least 88.9% of students drive less than 22.0248 miles to
*About 68% of students drive between -2.7306 miles and 22.0248 miles to

CODE IN PYTHON:

Your task is to write a simple program that would allow a user to compute the cost of a road trip with a car. User will enter the total distance to be traveled in miles along with the miles per gallon (MPG) information of the car he drives and the per gallon cost of gas. Using these 3 pieces of information you can compute the gas cost of the trip.

User will also enter the number of days it will take to complete the trip. For each night (day-1 nights), user will need a hotel. We will ask for the number of stars for the hotel the user wants to stay. 5 star hotel costs $250 per night. 4 star hotel costs $180 per night. 3 star hotel costs $120 per night, 2 star hotel costs $100 per night and finally 1 star hotel costs $50 per night. Using the hotel costs and the number of nights users will need to stay at a hotel, you can compute the hotel cost of the trip.

Please note that 5 star and 4 star hotels give 10% discount if the user will be staying more than 2 nights -- (these are chain hotels, so even if you are staying at different locations discount apply).

We should also account for the cost of meals. This depends on the type of hotel the user is staying. In general, we take 20% of hotel cost as the meal cost (after 10% hotel discount if applicable).

Example Run #1

Distance: 350

MPG : 15

Gas Price: 3.79

Days Traveling : 3

Hotel Stars (1-5) : 3

Your total cost is $376.43333333333334

Example Run #2

Distance: 400

MPG : 20

Gas Price: 2.99

Days Traveling: 1

Hotel Stars (1-5): 5

Your total cost is $59.800000000000004.

Example Run #3

Distance: 150 MPG : 25 Gas Price: 1.99 Days Traveling : 2 Hotel Stars (1-5): 5 Your total cost is $311.94.

Under normal driving conditions, car tires can generate a fiction coefficient of approximately 0.7. Antilock braking systems can increase the effective coefficient of friction to as high as 1.0. For car traveling 65 miles an hour, determine the stopping distance for coefficients of friction of 0.7 and 1.0. How much more time do antilock brakes give the driver to respond. Also, why have antilock brakes not produced reductions in crash rates that were anticipated from this calculation?

Almost all U.S. light-rail systems use electric cars that run on tracks built at street level. The Federal Transit Administration claims light-rail is one of the safest modes of travel, with an accident rate of .99 accidents per million passenger miles as compared to 2.29 for buses. The following data show the miles of track and the weekday ridership in thousands of passengers for six light-rail systems.

1. Use these data to develop an estimated regression equation that could be used to predict the ridership given the miles of track.
   
   Compute b₀ and b₁ (to 2 decimals).
   b₁   
   b₀   
   
   Complete the estimated regression equation (to 2 decimals).
   =   +  x
   
2. Compute the following (to 1 decimal):

   SSE
   SST
   SSR
   MSE

   
   
3. What is the coefficient of determination (to 3 decimals)? Note: report r² between 0 and 1.
     
   
   Does the estimated regression equation provide a good fit?
   SelectYes, it even provides an excellent fitYes, it provides a good fitNo, it does not provide a good fitItem 10  
   
4. Develop a 95% confidence interval for the mean weekday ridership for all light-rail systems with 30 miles of track (to 1 decimal).
   (  ,   )
   
5. Suppose that Charlotte is considering construction of a light-rail system with 30 miles of track. Develop a 95% prediction interval for the weekday ridership for the Charlotte system (to 1 decimal).
   (  ,   )
   
   

Almost all U.S. light-rail systems use electric cars that run on tracks built at street level. The Federal Transit Administration claims light-rail is one of the safest modes of travel, with an accident rate of .99 accidents per million passenger miles as compared to 2.29 for buses. The following data show the miles of track and the weekday ridership in thousands of passengers for six light-rail systems.

1. Use these data to develop an estimated regression equation that could be used to predict the ridership given the miles of track.
   
   Compute b₀ and b₁ (to 2 decimals).
   b₁   
   b₀   
   
   Complete the estimated regression equation (to 2 decimals).
   =   +  x
   
2. Compute the following (to 1 decimal):

   SSE
   SST
   SSR
   MSE

   
   
3. What is the coefficient of determination (to 3 decimals)? Note: report r² between 0 and 1.
     
   
   Does the estimated regression equation provide a good fit?
   SelectYes, it even provides an excellent fitYes, it provides a good fitNo, it does not provide a good fitItem 10  
   
4. Develop a 95% confidence interval for the mean weekday ridership for all light-rail systems with 30 miles of track (to 1 decimal).
   (  ,   )
   
5. Suppose that Charlotte is considering construction of a light-rail system with 30 miles of track. Develop a 95% prediction interval for the weekday ridership for the Charlotte system (to 1 decimal).
   (  ,   )

Almost all U.S. light-rail systems use electric cars that run on tracks built at street level. The Federal Transit Administration claims light-rail is one of the safest modes of travel, with an accident rate of .99 accidents per million passenger miles as compared to 2.29 for buses. The following data show the miles of track and the weekday ridership in thousands of passengers for six light-rail systems.

1. Use these data to develop an estimated regression equation that could be used to predict the ridership given the miles of track.
   
   Compute b₀ and b₁ (to 2 decimals).
   b₁
   b₀
   
   Complete the estimated regression equation (to 2 decimals).
   =  +  x
   
2. Compute the following (to 1 decimal):

   SSE
   SST
   SSR
   MSE

   
   
3. What is the coefficient of determination (to 3 decimals)? Note: report r² between 0 and 1.
   
   
   Does the estimated regression equation provide a good fit?
   SelectYes, it even provides an excellent fitYes, it provides a good fitNo, it does not provide a good fitItem 10
   
4. Develop a 95% confidence interval for the mean weekday ridership for all light-rail systems with 30 miles of track (to 1 decimal).
   (  ,  )
   
5. Suppose that Charlotte is considering construction of a light-rail system with 30 miles of track. Develop a 95% prediction interval for the weekday ridership for the Charlotte system (to 1 decimal).
   (  ,  )

A young engineering student tries to decide what she will do on Sunday evening. She has three options; going to cinema (C), going to bowling saloon (B), going to a restaurant (R). She may choose C option with probability 0.4, B option with probability 0.2 and R option with probability 0.4. If she chooses cinema option, she may go to three movies; a thriller (CT), a cartoon (CC) and romantic comedy (CR). Conditional probabilities associated with these three movie options are 0.6, 0.35 and 0.05. If she chooses to go to bowling saloon, she has two options; Rocker House (BR) with probability 0.7 and Empire Bowling (BE) with probability 0.3. Restaurant options are Chinese (RC) with probability 0.1, Burger House (RB) with probability 0.2 and Urfa Kebap (RU) with probability 0.7. a) (16 Points) Draw the decision trees diagram for these options.A young industrial engineering student tries to decide what she will do on Sunday evening. She has three options; going to cinema (C), going to bowling saloon (B), going to a restaurant (R). She may choose C option with probability 0.4, B option with probability 0.2 and R option with probability 0.4. If she chooses cinema option, she may go to three movies; a thriller (CT), a cartoon (CC) and romantic comedy (CR). Conditional probabilities associated with these three movie options are 0.6, 0.35 and 0.05. If she chooses to go to bowling saloon, she has two options; Rocker House (BR) with probability 0.7 and Empire Bowling (BE) with probability 0.3. Restaurant options are Chinese (RC) with probability 0.1, Burger House (RB) with probability 0.2 and Kebap (RK) with probability 0.7

. a) Draw the decision trees diagram for these options.

b) Compute the probabilities for whole options