Question

An airline company is considering a new policy of booking as many as 253 persons on an airplane that can seat only 220. (Past studies have revealed that only 80% of the booked passengers actually arrive for the flight.) Estimate the probability that if the company books 253 persons. not enough seats will be available.

Answer 1

Answer:

0.23% probability that not enough seats will be available.

Step-by-step explanation:

I am going to use the normal approximation to the binomial to solve this question.

Binomial probability distribution

Probability of exactly x sucesses on n repeated trials, with p probability.

Can be approximated to a normal distribution, using the expected value and the standard deviation.

The expected value of the binomial distribution is:

$E(X) = np$

The standard deviation of the binomial distribution is:

$\sqrt{V(X)} = \sqrt{np(1-p)}$

Normal probability distribution

Problems of normally distributed samples can be solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

When we are approximating a binomial distribution to a normal one, we have that $\mu = E(X)$, $\sigma = \sqrt{V(X)}$.

In this problem, we have that:

We are estimating the number of passengers that show up for the flight.

$n = 253, p = 0.8$

So

$\mu = E(X) = np = 253*0.8 = 202.4$

$\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{253*0.8*0.2} = 6.3624$

Estimate the probability that if the company books 253 persons. not enough seats will be available.

This is the probability that more than 220 people arrive for the flight.

Using continuity correction, this is P(X > 220 + 0.5) = P(X > 220.5), which is the pvalue of Z when X = 220.5. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{220.5 - 202.4}{6.3624}$

$Z = 2.84$

$Z = 2.84$ has a pvalue of 0.9977

1 - 0.9977 = 0.0023

0.23% probability that not enough seats will be available.