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.

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.

Let X 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.