Question

Air-USA has a policy of booking as many as 22 people on an airplane that can only seat 20 people. (Past studies have revealed that only 82% of the booked passengers actually show up for the flight.) a) Find the probability that if Air-USA books 22 people, not enough seats will be available. Round your answer to 4 decimal places. P ( X > 20 )

Answer 1

Answer:

The answer is "0.07404893".

Step-by-step explanation:

Applying the binomial distribution:

$n = 22\\\\p= 82\%=0.82\\\\q = 1-0.82 = 0.18$

Calculating the probability for not enough seats:

$=P(X>20)\\\\= P(21) + P(22)$

$= \binom{22}{21} (0.82)^{21}(0.18)^1+ \binom{22}{22} (0.82)^{22}(0.18)$

$=0 .06134598+ 0.01270295\\\\=0.07404893$