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Answer:
0.5438 = 54.38% probability that a seat will be available for every person holding a reservation and planning to fly
Step-by-step 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:
The standard deviation of the binomial distribution is:
Normal probability distribution
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean and standard deviation
, the zscore of a measure X is given by:
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 ,
.
In this problem, we have that:
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:
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
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