Question

Lost time accidents occur in a company at a mean rate of 0.8 per day. What is the probability that a number of lost time accidents occurring over a period of 9 days will be no more than 5? Round to the nearest four decimal places.

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Given that the mean rate of lost time accidents is 0.8 per day, the rate parameter, \lambda , for a Poisson distribution is also 0.8.

LetX be the random variable representing the number of lost time accidents occurring in 9 days. Therefore, X follows a Poisson distribution with a mean rate of \lambda = 0.8 \times 9 = 7.2 over 9 days.

The probability that the number of accidents will be no more than 5 is given by:

P(X \leq 5) = \sum_{x=0}^{5} \frac{e^{-\lambda} \lambda^x}{x!}

Calculating the probability:

P(X \leq 5) = \sum_{x=0}^{5} \frac{e^{-7.2} \times 7.2^x}{x!}

Calculating each term:

P(X \leq 5) = \frac{e^{-7.2} \times 7.2^0}{0!} + \frac{e^{-7.2} \times 7.2^1}{1!} + \frac{e^{-7.2} \times 7.2^2}{2!} + \frac{e^{-7.2} \times 7.2^3}{3!} + \frac{e^{-7.2} \times 7.2^4}{4!} + \frac{e^{-7.2} \times 7.2^5}{5!}

Calculating each term individually, we get:

P(X\leq5)=0.000747+0.005375+0.019352+0.046444+0.083598+0.120382

Adding all probabilities:

P(X\leq5)\approx0.2759

Therefore, the probability that the number of lost time accidents occurring over a period of 9 days will be no more than 5 is approximately 0.2759.

Answer: The probability is approximately 0.2759.

Let

The probability that the number of accidents will be no more than 5 is given by:

Calculating the probability:

Calculating each term:

Calculating each term individually, we get:

Adding all probabilities:

Therefore, the probability that the number of lost time accidents occurring over a period of 9 days will be no more than 5 is approximately 0.2759.

Answer: The probability is approximately 0.2759.

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