Respuesta :
Answer:
1.32% probability that the number of lost-time accidents occurring over a period of 88 days will be no more than 22.
Step-by-step explanation:
To solve this question, i am going to approximate the Poisson distribution to the normal to solve this question.
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. The variance is the same as the mean.
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.
The poisson distribution can be approximated to the normal with mean [tex]\mu[/tex] and standard deviation [tex]\sigma = \sqrt{\mu}[/tex]
In this problem, we have that:
88 days
0.4 accidents per day
[tex]\mu = 88*0.4 = 35.2[/tex]
[tex]\sigma = \sqrt{35.2} = 5.933[/tex]
Probability of no more than 22 accidents:
pvalue of Z when X = 22. So
[tex]Z = \frac{X - \mu}{\sigma}[/tex]
[tex]Z = \frac{22 - 35.2}{5.933}[/tex]
[tex]Z = -2.22[/tex]
[tex]Z = -2.22[/tex] has a pvalue of 0.0132
1.32% probability that the number of lost-time accidents occurring over a period of 88 days will be no more than 22.
