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Vesna [10]
3 years ago
7

An airline knows that 5 percent of the people making reservations on a certain flight will not show up. Consequently, their poli

cy is to sell 52 tickets for a flight that can hold only 50 passengers. What is the probability that there will be a seat available for every passenger who shows up?
Mathematics
1 answer:
kenny6666 [7]3 years ago
8 0

Answer:

the probability that there will be a seat available for every passenger who shows up is 0.74

Step-by-step explanation:

if the probability of choosing a passenger that will not show up is 5% , then the probability of choosing a passenger that will show up is 95%.

Denoting event X= x passengers will show up from the total of 52 that purchased the ticket

Then P(X) follows a binomial probability distribution, since each passenger is independent from others. Thus

P(X) = n!/((n-x)!*x!)*p^x*(1-p)^(n-x)

where

n= total number of passengers hat purchased the ticket= 52

p= probability of choosing a passenger that will show up

x = number of passengers that will show up

therefore the probability that the number does not exceed the limit of 50 passengers is:

P(X≤50)=  ∑P(X=i)  from i=0 to i=50   =F(50)

where F(x) is the cumulative binomial probability distribution, then from tables:

P(X≤50)= F(50) = 0.74

therefore the probability that there will be a seat available for every passenger who shows up is 0.74

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Answer:

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\bf \displaystyle\frac{0.006329728}{50}=\displaystyle\frac{d_2^2}{(4.37)^2}\Rightarrow\\\\\Rightarrow d_2^2=\displaystyle\frac{0.006329728*(4.37)^2}{50}\Rightarrow d_2^2=0.002417564\Rightarrow\\\\\Rightarrow d_2=\sqrt{0.002417564}=\boxed{0.049168726\;ly}

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