Question

Lost-time accidents occur in a company at a mean rate of 0.40.4 per day. What is the probability that the number of lost-time accidents occurring over a period of 88 days will be no more than 22.

Answer 1

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:

$P(X = x) = \frac{e^{-\mu}*\mu^{x}}{(x)!}$

In which

x is the number of sucesses

e = 2.71828 is the Euler number

$\mu$ 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 $\mu$ and standard deviation $\sigma$, the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

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 $\mu$ and standard deviation $\sigma = \sqrt{\mu}$

In this problem, we have that:

88 days

0.4 accidents per day

$\mu = 88*0.4 = 35.2$

$\sigma = \sqrt{35.2} = 5.933$

Probability of no more than 22 accidents:

pvalue of Z when X = 22. So

$Z = \frac{X - \mu}{\sigma}$

$Z = \frac{22 - 35.2}{5.933}$

$Z = -2.22$

$Z = -2.22$ 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.