I am having with the following questions and was wondering if someone might be able to show me a step by step solution to arrive at the answer. They are from Probablity and Statistics for Enginnering and the Sciences by Jay Devore 6th Edition
62) An airport limousine can accomodate up to four passengers on any one trip. The company will accept a mximum of six reservatiosn for a trip and a pasenger must have reservation. From previous records, 20% of all those making resrervations do not appear for the trip. Assume independence wherever appropriate.
a) If six reservations are made, what is the probability that at least one inidivdiual with a reservation cannot be accomodated on the trip?
b)If six reservations are made, what is the expected number of available palces whe nthe limousine departs?
c) Suppose the probabiility distribtuion of the number of reservations made is given in the accompanying table, Obtain the probabilimty mass function X if X denotes the number of passengers on a randomly selected trip.
# of reserversation 3 , 4 , 5, 6
probability .1 .2 .3 .4
56) A toll bridge charges $1.00 for passenger cars and $2.50 for other vehicles. Suppose that during daytime hours, 60% of all vehicles are passenger cars. If 25 vehicles cross the bridge during a particular daytime period, what is the resulting expected toll revenue? Hint let X = number of passengers cars then the toll revenue h(x) is a linear funciton of X.
I get an answer of 40$ for total revenue. I'm wondering if someone could double check that for me.
Finally 52 d and e
52d) The college board reports that 2% of the 2 million high schoo lstudents who take the SAT each year receive special accomodations because of documented disabilites. Consider a random sample of 25 students who have recently taken the test.
d) What is the porbaiblity that number among the 25 who reeived a special accomodation is within 2 standard deviations of the number you would expect to be accomodated? I get an answer of 1, could someone please double check.
e) Suppose that student who does not receive special accomodation is allowed 3 hours for the exam wheaeas an accomodated student receives 4.5 hours. What would you expect the average time allowed fhe 25 selected students to be?
Thanks so much if anyone can help me.
62) Construct a probability table for all possible answers. The probability that everybody shows up is .8^6. The probability that 5 show up is 6*(.8^5)*(.2^1). The probability that 4 show up is 15*(.8^4)*(.2^2). The 15 is number of combinations of 6 (people) choose 2 (that don't show up). Continue with this for all 3,2,1,0 show up. With this, the answer to a) and b) should easily fall out.
56) I too get $40.
52d) The answer is definitely not 1.
The problem has complexities.
The expected number is .02*25=.5 students. The standard deviation is sqrt(.02*.98*25) = .7 So, two standard deviations up from .5 is 1.9. So, are we asked what is the probability of getting 0 or 1 students? I get 87.6% (I hope someone could check my work)
e) .98*3 + .02*4.5 = 3.03 hrs.
16. Based on tests of the Chevrolet Cobalt, engineers have found that the miles per gallon in highway driving are normally distributed, with a mean of 32 miles per gallon and a standard deviation 3.5 miles per gallon
For 52d)
You need to find the probability that the # of students who received special accommodation is within 2*stdDev of E(x)
So: E(x) = pn = (.02)(25) = 0.5
var(x) = pn*(1-p) =(.02)(25)(1-.02) = 0.40
stdDev = sqrt(var(x)) = sqrt(.40) = 0.632
2*stdDev = 2*(0.632) = 1.264
To find probability within 2*stdDev of E(x) we want to find P( [E(x)-2*stdDev] <= X <= [E(x)+2*stdDev] )
= P(-0.764 <= X <= 1.764) --> approximate to closest interval
~ P( 0 <= X <= 2)
= b(0;25,.02)+b(1;25,.02)+b(2;25,.02)
=0.1244+.0675+.01763
=.2095
Sure! I can help you with the step-by-step solutions to these problems.
62a) To find the probability that at least one individual with a reservation cannot be accommodated on the trip, we can calculate the probability of the complement event, i.e., the probability that everyone is accommodated.
The probability that everyone shows up is (0.8)^6, since each individual has a 0.8 probability of showing up. So, the probability of everyone being accommodated is (0.8)^6.
Therefore, the probability that at least one individual cannot be accommodated is 1 - (0.8)^6.
62b) To find the expected number of available places when the limousine departs, we need to calculate the expected value of the number of passengers on a randomly selected trip.
The probability distribution of the number of reservations made is given in the accompanying table:
Number of reservations: 3, 4, 5, 6
Probability: 0.1, 0.2, 0.3, 0.4
To calculate the expected value, we multiply each possible outcome by its respective probability and sum them up.
Expected value = (3 * 0.1) + (4 * 0.2) + (5 * 0.3) + (6 * 0.4)
62c) To obtain the probability mass function X, which denotes the number of passengers on a randomly selected trip, you can simply refer to the probability distribution table provided:
Number of passengers (X): 3, 4, 5, 6
Probability: 0.1, 0.2, 0.3, 0.4
This gives you the probability of having each possible number of passengers on a randomly selected trip.
56) To calculate the expected toll revenue, we need to multiply each type of vehicle's toll price by its respective probability and sum them up.
Given that 60% of all vehicles are passenger cars and 40% are other vehicles, we can let X be the number of passenger cars crossing the bridge. The toll revenue H(X) is a linear function of X.
The expected toll revenue can be calculated as:
Expected toll revenue = (1.00 * 0.60 * X) + (2.50 * 0.40 * (25 - X))
Substituting X = 0 to 25 and summing up these values will give you the total expected toll revenue.
52d) To find the probability that the number among the 25 who received a special accommodation is within 2 standard deviations of the expected number of accommodations, calculate the expected value and standard deviation, then find the probability using the normal distribution.
The expected number of accommodations is given by 25 * 0.02 = 0.5.
The standard deviation is sqrt(25 * 0.02 * 0.98) = 0.7.
To find the probability within 2 standard deviations of the mean, calculate the cumulative probability of having fewer than or equal to 1 accommodated student (0 or 1), using the mean and standard deviation.
e) To find the expected average time allowed for the 25 selected students, multiply the probability of a student not receiving special accommodation (0.98) by the time allowed for a student without accommodation (3 hours) and add it to the probability of a student receiving special accommodation (0.02) multiplied by the time allowed for a student with accommodation (4.5 hours). This will give you the expected average time allowed for the 25 selected students.
I hope this helps! Let me know if you have any more questions.