Question

An airline finds that 5% of the persons making reservations on a certain flight will not show up for the flight. If the airline sells 105 tickets for a flight that has only 100 seats, what is the probability that a seat will be available for every person holding a reservation and planning to fly? (Round your answer to four decimal place

Answer 1

Answer:

The answer to the question is;

The probability that a seat will be available for every person holding a reservation and planning to fly is 0.63307.

Step-by-step explanation:

Let the sample size =n = 100

The success probability = 5 % = 0.05

Number of tickets sold = 105 tickets

In the case where there the airline has found that 5 % will not show up, then every passenger should have  a seat, we have  

A Binomial distribution is appropriate where there is a chance for a certain number of successful outcomes from a number of independent trails

However n·p and n·q must be ≥ 5 for there to be a normal approximation of a Binomial distribution thus

n·p = 105×0.05 =  5.25 ≥ 5

and n·q = n(1 - p) = 105 (1 - 0.05) = 99.75 ≥ 5

As the requirements are met, we can proceed with the approximation of the Binomial distribution by the normal distribution

 $z = \frac{x-np}{\sqrt{np(1-p)} } = \frac{4.5 - 105*0.05}{\sqrt{105*0.05(1-0.05)} }$ =  - 0.3358

We therefore have P(x ≥ 5) = P( x > 4.5) = P(z > -0.34) = 1 - P(z < -0.34) = 1 -0.36693 = 0.63307

Another way to solve the question is as follows

p = 0.95 q = 0.05

μ = np = 0.95*105 = 99.75, σ = $\sqrt{npq}$ = 2.233

P (x≤100) = P$(z$ = P(z<0.34) = 0.63307.