I want to know HOW to do these problems, don't care about the answers as much. It's an online course so I need to teach myself this and I don't see any examples of this anywhere in my text or notes!
An airline sells 142 tickets for a flight that has 125 seats. The probability that a passeenger with a ticket shows up for a flight is .9.
a)Find the probability that exactly 125 passengers with tickets show up for this flight.
b)Find the probability that fewer than 125 passengers with tickets show up for this flight.
c)Find the probability that more than 125 passengers with tickets show up for this flight.
d)Find the probability that the number of passengers with tickets show up for this flight is from 120 to 125.
e) Find the mean number of passengers with tickets that show up for this flight.
f) Find the standard deviation for the number of passengers with tickets that show up for this flight.
An airline sells 142 tickets for a flight that has 125 seats. The probability that a passeenger with a ticket shows up for a flight is .9.
a)Find the probability that exactly 125 passengers with tickets show up for this flight.
b)Find the probability that fewer than 125 passengers with tickets show up for this flight.
c)Find the probability that more than 125 passengers with tickets show up for this flight.
d)Find the probability that the number of passengers with tickets show up for this flight is from 120 to 125.
e) Find the mean number of passengers with tickets that show up for this flight.
f) Find the standard deviation for the number of passengers with tickets that show up for this flight.
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a) exactly 125
binompdf(142, 0.9, 125) = 142 choose 125 * 0.9^125 * 0.1^17 = 7.68%
b) < 125
binomcdf(142, 0.9, 124) = 17.62%
c) >125
1- 7.68%-17.62% = 74.71%
d) from 120 to 125
binomcdf(142, 0.9, 125) = 25.29%
binomcdf(142, 0.9, 119) = 0.74%
binomcdf(142, 0.9, 125) - binomcdf(142, 0.9, 119)
=24.55%
e)mu = np =142*0.9 = 127.8
f) standard variation = sqrt(npq) = 3.57
binompdf(142, 0.9, 125) = 142 choose 125 * 0.9^125 * 0.1^17 = 7.68%
b) < 125
binomcdf(142, 0.9, 124) = 17.62%
c) >125
1- 7.68%-17.62% = 74.71%
d) from 120 to 125
binomcdf(142, 0.9, 125) = 25.29%
binomcdf(142, 0.9, 119) = 0.74%
binomcdf(142, 0.9, 125) - binomcdf(142, 0.9, 119)
=24.55%
e)mu = np =142*0.9 = 127.8
f) standard variation = sqrt(npq) = 3.57