In one year there was 675 deaths due to impaired driving. For any randomly selected day what is the probability of 2 impaired driving related deaths in that one day?
To calculate the probability of 2 impaired driving related deaths in a randomly selected day, we need to know the average number of impaired driving related deaths per day.
Given that there were 675 deaths due to impaired driving in one year, we divide this number by the total number of days in a year to calculate the average number of impaired driving related deaths per day.
Assuming there are 365 days in a year, the average number of impaired driving related deaths per day is:
Average = 675 / 365
Therefore, the average number of impaired driving deaths per day is approximately 1.85.
To calculate the probability of having exactly 2 impaired driving related deaths in one randomly selected day, we can use the Poisson distribution. The Poisson distribution is commonly used to model the probability of events that occur randomly over a certain time period.
The formula for the Poisson probability is:
P(x) = (e^(-λ) * λ^x) / x!
where:
- P(x) is the probability of having x events
- e is the base of the natural logarithm (approximately 2.71828)
- λ is the average number of events per time period (in this case, per day)
- x is the number of events we are interested in (in this case, 2)
Using the average of 1.85 impaired driving related deaths per day as λ, we can calculate the probability of exactly 2 impaired driving related deaths in one randomly selected day using the Poisson probability formula.
P(2) = (e^(-1.85) * 1.85^2) / 2!
Now, you can substitute the values and calculate the probability.