Respuesta :
Answer:
96.33% probability that everyone who appears for the flight will get a seat
Step-by-step explanation:
I am going to 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:
[tex]E(X) = np[/tex]
The standard deviation of the binomial distribution is:
[tex]\sqrt{V(X)} = \sqrt{np(1-p)}[/tex]
Normal probability distribution
Problems of normally distributed samples can be solved using the z-score formula.
In a set with mean [tex]\mu[/tex] and standard deviation [tex]\sigma[/tex], the zscore of a measure X is given by:
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
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 [tex]\mu = E(X)[/tex], [tex]\sigma = \sqrt{V(X)}[/tex].
In this problem, we have that:
[tex]n = 105, p = 0.9[/tex]
So
[tex]\mu = E(X) = np = 105*0.9 = 94.5[/tex]
[tex]\sigma = \sqrt{V(X)} = \sqrt{np(1-p)} = \sqrt{105*0.9*0.1} = 3.07[/tex]
What is the probability that everyone who appears for the flight will get a seat
100 or less people appearing to the flight, which is the pvalue of Z when X = 100. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{100 - 94.5}{3.07}[/tex]
[tex]Z = 1.79[/tex]
[tex]Z = 1.79[/tex] has a pvalue of 0.9633
96.33% probability that everyone who appears for the flight will get a seat
