An Airliner has a capacity for 300 passengers. If the company overbook a flight with 320 passengers, What is the probability that it will not be enough seats to accommodate all passengers. Assume that the probability that a randomly selected passenger shows up to the airport is 0.96. Find the probability using the normal distribution as an approximation to the binomial distribution.

Answers

Answer 1

Answer:

The probability is   [tex]P(X  >300 ) = 0.97219 [/tex]

Step-by-step explanation:

From the question we are told that

 The capacity of  an Airliner  is  k =  300 passengers

 The sample size n =  320 passengers

  The probability the a randomly selected passenger shows up on to the airport

    [tex]p = 0.96[/tex]

Generally the mean is mathematically represented as

    [tex]\mu  =  n*  p[/tex]

  => [tex]\mu  =  320 *  0.96[/tex]

    => [tex]\mu  = 307.2[/tex]

Generally the standard deviation is  

    [tex]\sigma =  \sqrt{n *  p *  (1 -p ) }[/tex]

=>  [tex]\sigma =  \sqrt{320  *  0.96 *  (1 -0.96 ) }[/tex]

=> [tex]\sigma =3.50 [/tex]

Applying Normal approximation of binomial distribution

Generally the probability that there will not be enough seats to accommodate all passengers is mathematically represented as

  [tex]P(X  > k ) =  P( \frac{ X -\mu }{\sigma }  >  \frac{k - \mu}{\sigma } )[/tex]

Here [tex]\frac{ X -\mu }{\sigma }  =Z (The \ standardized \  value \  of  \ X )[/tex]

=>[tex]P(X  >300 ) =  P(Z >  \frac{300 - 307.2}{3.50} )[/tex]

Now applying  continuity correction we have

    [tex]P(X  >300 ) =  P(Z >  \frac{[300+0.5] - 307.2}{3.50} )[/tex]    

=>    [tex]P(X  >300 ) =  P(Z >  \frac{[300.5] - 307.2}{3.50} )[/tex]

=>    [tex]P(X  >300 ) =  P(Z >  -1.914 )[/tex]

From the z-table  

    [tex]P(Z >  -1.914 ) =  0.97219[/tex]

So

    [tex]P(X  >300 ) = 0.97219 [/tex]


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Answers

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