Answer:
0.9826 = 98.26% probability that a seat will be available for every person holding a reservation and planning to fly.
Step-by-step explanation:
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](https://tex.z-dn.net/?f=E%28X%29%20%3D%20np)
The standard deviation of the binomial distribution is:
![\sqrt{V(X)} = \sqrt{np(1-p)}](https://tex.z-dn.net/?f=%5Csqrt%7BV%28X%29%7D%20%3D%20%5Csqrt%7Bnp%281-p%29%7D)
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:
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
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:
6% of the people who make reservations on a certain flight do not show up for the flight. So 100-6 = 94% show up, which means that ![p = 0.94](https://tex.z-dn.net/?f=p%20%3D%200.94)
190 tickets, so ![n = 190](https://tex.z-dn.net/?f=n%20%3D%20190)
So
![\mu = E(X) = np = 190*0.94 = 178.6](https://tex.z-dn.net/?f=%5Cmu%20%3D%20E%28X%29%20%3D%20np%20%3D%20190%2A0.94%20%3D%20178.6)
![\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{190*0.94*0.06} = 3.27](https://tex.z-dn.net/?f=%5Csigma%20%3D%20%5Csqrt%7BV%28X%29%7D%20%3D%20%5Csqrt%7Bnp%281-p%29%7D%20%3D%20%5Csqrt%7B190%2A0.94%2A0.06%7D%20%3D%203.27)
Find the probability that a seat will be available for every person holding a reservation and planning to fly.
At most 185 people show up.
Using continuity correction, we have to find
, which is the pvalue of Z when X = 185.5. So
![Z = \frac{X - \mu}{\sigma}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7BX%20-%20%5Cmu%7D%7B%5Csigma%7D)
![Z = \frac{185.5 - 178.6}{3.27}](https://tex.z-dn.net/?f=Z%20%3D%20%5Cfrac%7B185.5%20-%20178.6%7D%7B3.27%7D)
![Z = 2.11](https://tex.z-dn.net/?f=Z%20%3D%202.11)
has a pvalue of 0.9826
0.9826 = 98.26% probability that a seat will be available for every person holding a reservation and planning to fly.