Air America Research Paper
MAT540 Quantitative Methods of Science
Assignment #1
Quantitative Methods
July 28, 2012
#1
Overbooking Flights Air America has a policy of booking as many as 160 persons on an airplane that can seat only 145 (past studies have revealed that only 85% of the booked passengers actually arrive for the flight). Find the probability that if Air America books 160 persons, not enough seats will be available. Is this probability low enough so that overbooking is not real concern for passengers?
For calculations, we are going to use binominal distribution formula and computer.
N=160
P= 0.85
P(X) = n!/x!(n-x)! * p! (1-p)n-x
P(x≤145) = P (x=0) + P (x=1) + P (x=2) + … + P (x=145) = 0.989
P (x>145) = 0.013 = 1-0.989=0.011
[pic]
Answer: Probability that if Air America books 160 persons, not enough seats will be available is 1%, which is very low. Yes, this probability is low enough so that overbooking is not real concern for passengers
#2 …show more content…
The State of Virginia has implemented as Standard of Learning (SOL) test that all public school students must pass before they can graduate from high school.
A passing grade is 75. Montgomery County High School administrators want to gauge how well their students might do on te SOL test, but they don’t want to take the time to test the whole student population. Instead, they selected 20 students at random and gave them the test. (Results presented on paper
handout) µ= (83+48+72+66+58+79+92+71+83+95+56+37+92+81+67+93+45+71+80+78) / 20 = 72.35 γ (Standard Deviation)=16.67
P (x>75)p (x=0) + p (x=1) + … + p (x=75)
P(x) = ( x- µ ) / γ = = (75 – 72.35) / 16.26 = = 0.16298 = = .4364 = Probability that a student at the high school will pass the test = 43.64%
P (z> (75-72.35)/16.67)= P (z>0.16)[pic]
http://25yearsofprogramming.com/javascript/probability.htm
Assignment #1
Quantitative Methods
July 28, 2012
#1
Overbooking Flights Air America has a policy of booking as many as 160 persons on an airplane that can seat only 145 (past studies have revealed that only 85% of the booked passengers actually arrive for the flight). Find the probability that if Air America books 160 persons, not enough seats will be available. Is this probability low enough so that overbooking is not real concern for passengers?
For calculations, we are going to use binominal distribution formula and computer.
N=160
P= 0.85
P(X) = n!/x!(n-x)! * p! (1-p)n-x
P(x≤145) = P (x=0) + P (x=1) + P (x=2) + … + P (x=145) = 0.989
P (x>145) = 0.013 = 1-0.989=0.011
[pic]
Answer: Probability that if Air America books 160 persons, not enough seats will be available is 1%, which is very low. Yes, this probability is low enough so that overbooking is not real concern for passengers
#2 …show more content…
The State of Virginia has implemented as Standard of Learning (SOL) test that all public school students must pass before they can graduate from high school.
A passing grade is 75. Montgomery County High School administrators want to gauge how well their students might do on te SOL test, but they don’t want to take the time to test the whole student population. Instead, they selected 20 students at random and gave them the test. (Results presented on paper
handout) µ= (83+48+72+66+58+79+92+71+83+95+56+37+92+81+67+93+45+71+80+78) / 20 = 72.35 γ (Standard Deviation)=16.67
P (x>75)p (x=0) + p (x=1) + … + p (x=75)
P(x) = ( x- µ ) / γ = = (75 – 72.35) / 16.26 = = 0.16298 = = .4364 = Probability that a student at the high school will pass the test = 43.64%
P (z> (75-72.35)/16.67)= P (z>0.16)[pic]
http://25yearsofprogramming.com/javascript/probability.htm