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.

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:

In which

x is the number of sucesses

e = 2.71828 is the Euler number

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 and standard deviation , the zscore of a measure X is given by:

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 and standard deviation

In this problem, we have that:

88 days

0.4 accidents per day

Probability of no more than 22 accidents:

pvalue of Z when X = 22. So

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.

33

step-by-step explanation: g=6

unless if u could do my questions, if not i wouldn't want to