Saturday

October 25, 2014

October 25, 2014

Posted by **ajones51** on Wednesday, February 7, 2007 at 10:02pm.

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 hte 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 recieves 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 dont 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.

- mathmatics probability -
**tone**, Saturday, March 10, 2012 at 10:23pm16. 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

- mathmatics probability -
**Ryan**, Wednesday, October 15, 2014 at 11:00pmFor 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

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