Question

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.

Answer 1

Answer:

The probability is   $P(X >300 ) = 0.97219$

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

    $p = 0.96$

Generally the mean is mathematically represented as

    $\mu = n* p$

  => $\mu = 320 * 0.96$

    => $\mu = 307.2$

Generally the standard deviation is  

    $\sigma = \sqrt{n * p * (1 -p ) }$

=>  $\sigma = \sqrt{320 * 0.96 * (1 -0.96 ) }$

=> $\sigma =3.50$

Applying Normal approximation of binomial distribution

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

  $P(X > k ) = P( \frac{ X -\mu }{\sigma } > \frac{k - \mu}{\sigma } )$

Here $\frac{ X -\mu }{\sigma } =Z (The \ standardized \ value \ of \ X )$

=>$P(X >300 ) = P(Z > \frac{300 - 307.2}{3.50} )$

Now applying  continuity correction we have

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

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

=>    $P(X >300 ) = P(Z > -1.914 )$

From the z-table  

    $P(Z > -1.914 ) = 0.97219$

So

    $P(X >300 ) = 0.97219$