Answer :
Answer:
2.87% probability that on a given day some tankers have to be turned away
Step-by-step explanation:
To solve this question, we need to understand the poisson distribution and the normal distribution.
Poisson 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 time interval, which is the same as the variance.
Normal 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.
The Poisson distribution can be approximated to the normal with mean [tex]\mu[/tex] and standard deviation [tex]\sigma = \sqrt{\mu}[/tex].
The average number of oil tankers arriving each day at a certainport city is 10.
This means that [tex]\mu = 10[/tex]
The facilities at the port can handle at most15 tankers per day. What is the probability that on a given daysome tankers have to be turned away
So at least 16 tankers, which is 1 subtracted by the pvalue of Z when X = 16.
[tex]\sigma = \sqrt{10} = 3.1623[/tex]
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{16 - 10}{3.1623}[/tex]
[tex]Z = 1.90[/tex]
[tex]Z = 1.90[/tex] has a pvalue of 0.9713
1 - 0.9713 = 0.0287
2.87% probability that on a given day some tankers have to be turned away