Respuesta :
Answer:
15.87% probability of 36 or more accidents in a 2 year period
Step-by-step explanation:
I am going to approximate the normal probability distribution to the Poisson to solve this question.
Normal probability distribution:
Problems of normally distributed samples are 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.
Poisson probability distribution:
In a Poisson distribution, the probability that X represents the number of successes of a random variable is given by the following formula:
[tex]P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}[/tex]
In which
x is the number of sucesses
e = 2.71828 is the Euler number
[tex]\mu[/tex] is the mean in the given interval. The variance has the same value as the mean.
An average of 15 aircraft accidents occur each year
So for 2 years, [tex]\mu = 2*15 = 30[/tex]
Approximation to the normal
[tex]\mu = 30, \sigma = \sqrt{30} = 5.4772[/tex]
What is the probability of 36 or more accidents in a 2 year period?
[tex]P(X \geq 36)[/tex], which, using continuity correction, is [tex]P(X \geq 36 - 0.5) = P(X \geq 35.5)[/tex], which is 1 subtracted by the pvalue of Z when X = 35.5.
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{35.5 - 30}{5.4772}[/tex]
[tex]Z = 1[/tex]
[tex]Z = 1[/tex] has a pvalue of 0.8413
1 - 0.8413 = 0.1587
15.87% probability of 36 or more accidents in a 2 year period
