Binomial distribution An airline always overbooks if possible. A particular has

Question

Binomial distribution
An airline always overbooks if possible. A particular has 95 seats on a flight. The airline sells 100 tickets for this flight f probability of an individual not showing up at the gate is p, where 0 pKim assuming independence, what is the probability the that the airline can accommodate all the passenfers who do not show up? Answer the question for p *0.05, 0.10, 0.15, and 0.20. Develop the probability distribution graph based on various values of p. 100 p 0.10 p(x) 0 p (0.05) 0.0576 5.8% 0.0000 1 0.0003 p (0.10) 0.5832 58.3% 0.0016 3 0.0059 p (0.15) 0.9601 96.0% 4 0.0159 0.0339 6 0.0596 0) 0.9992 100% 0.0889 8 0.1148 9 0.1304 10 0.1319 11 0.1199 12 0.0988 13 0,0743 14 0.0513 15 0.0327 16 0.0193 17 0.0106 18 0.0054 19 0.0026 20 0.0012 1.00

Explanation / Answer

observe that if number of passeger showing up is less than or equal to 95 , then everyone will get seat ,

or the number of people who don't show up is more than or equal to 5 .

for n = 100 , and various p

p=0.05 0.564018699314289 56.4 %

p= 0.10 , 0.9762889173 97.62 %

p = 0.15 0.999574486182961 99.957%

p=0.20   0.999996296826943

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