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 places.)

Answer 1

0.5438 = 54.38% probability that a seat will be available for every person holding a reservation and planning to fly

Explanation:

We use the binomial approximation to the normal 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:

`√(V(X)) = √(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 = (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 = √(V(X))`.

In this problem, we have that:

`n = 105, p = 1 - 0.05 = 0.95`

I use p = 0.95 because i consider a success a person showing up to the flight. 5% probability that a person misses the flight, so 100-5 = 95% probability that a person shows up to the flight.

For the approximation:

`\mu = E(X) = np = 105*0.95 = 99.75`

`\sigma = √(V(X)) = √(np(1-p)) = √(105*0.95*0.05) = 2.23`

What is the probability that a seat will be available for every person holding a reservation and planning to fly?

Probability of 100 or less people showing up, which is the pvalue of Z when X = 100. So

`Z = (X - \mu)/(\sigma)`

`Z = (100 - 99.75)/(2.23)`

`Z = 0.11`

`Z = 0.11` has a pvalue of 0.5438

0.5438 = 54.38% probability that a seat will be available for every person holding a reservation and planning to fly