Question

Lost-time accidents occur in a company at a mean rate of 0.6 per day. What is the probability that the number of lost-time accidents occurring over a period of 10 days will be no more than 2? Round your answer to four decimal places.

Answer 1

The probability of having no more than 2 accidents in a 10 day period, with a daily mean of 0.6, is found using the Poisson distribution and rounding the result to four decimal places.

Step-by-step explanation:

To determine the probability of there being no more than 2 lost-time accidents in a 10 day period, given the mean rate is 0.6 accidents per day, we use the Poisson distribution formula. The Poisson distribution helps us find the probability of a given number of events happening in a fixed interval of time when these events happen with a known constant rate and independently of the time since the last event.

The formula for the Poisson probability is P(X=k) = (e^-λ * λ^k) / k!, where λ is the average number of events in the interval, e is the base of the natural logarithm, k is the number of occurrences of the event, and k! is the factorial of k. In this case, λ for 10 days is 0.6 accidents/day * 10 days = 6 accidents.

We want the probability of X being not more than 2, so we calculate P(X=0), P(X=1), and P(X=2) and then sum these probabilities. After calculating each using the formula, we add them together to get the total probability, which we round to four decimal places to provide our answer.

Answer 2

- This is a Binomial theorem question. The following are the parameters given:

`\begin{gathered} n=10 \\ r=2 \\ p=0.6 \\ q=1-p=0.4 \\ \\ \text{ The formula to use is:} \\ P(r)=\sum_r^nnC_rp^rq^(n-r) \end{gathered}`

- Thus, we have:

`\begin{gathered} P(r\le2)=P(0)+P(1)+P(2) \\ P(0)=^(10)C_00.6^0*0.4^(10)=0.0001048576 \\ P(1)=^(10)C_10.6^1*0.4^9=0.001572864 \\ P(2)=^(10)C_20.6^2*0.4^8=0.010616832 \\ \\ \therefore P(r\le2)=0.0001048576+0.001572864+0.010616832 \\ \\ P(r\le2)=0.0122945536\approx0.0123 \end{gathered}`