Lost-time accidents occur in a company at a mean rate of 0.3 per day. What is the probability that the number of lost-time accidents occurring over a period of 9 days will be no more than 3? Round your answer to four decimal places.

Probabilities are used to determine the chances of events.

The probability that accidents occur no more than 3 days in 9 days is 0.7296

The given parameters are:

--- the probability that an accident occurs

The probability that an accident does not occur (q) is then calculated using the following complement rule

The probability that accidents occur no more than 3 days in 9 days is

Each probability is calculated using:

So, we have:

This gives

Hence, the probability that accidents occur no more than 3 days in 9 days is 0.7296

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0.7295

Step-by-step explanation:

Given that :

Mean rate of lost time accident = 0.3 per day

probability that the number of lost-time accidents occurring over a period of 9 days will be no more than 3?

n = 9

Using binomial distribution formula :

P(x ≤3)

Probability of success (p) = 0.3

1 - p = 1- 0.3 = 0.7

nCx * p^x * (1 - p)^(n-x)

Using the binomial probability distribution calculator to save computation time :

P(x ≤3) = P(x = 0) + p(x = 1) + p(x = 2) + p(x = 3)

P(x ≤3) = 0.0403 + 0.1556 + 0.2668 + 0.2668

P(x ≤3) = 0.7295

a box 3 3/4 feet long or 4+ feet.

step-by-step explanation:

4th option is correct xy=-10

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