Records show that deaths occur at the rate of 0.1 per day among patients residing in a large nursing home. If someone dies today, what are the chances that a week or more will elapse before another death occurs?
The answer is 0.50 and the problem is in the Poisson section but I don't know how to get this answer.
To solve this problem, we can use the exponential distribution, which is related to the Poisson distribution. The exponential distribution models the time between occurrences of a certain event, in this case, the time between deaths in the nursing home.
The parameter λ of the exponential distribution is the mean rate of deaths per day. In this case, we are given that deaths occur at a rate of 0.1 per day, so λ = 0.1.
The probability that a week or more will elapse before another death occurs can be calculated by finding the complement of the probability that a death occurs within a week.
Let's denote T as the time between deaths. We want to find P(T ≥ 7) - the probability that T is greater than or equal to 7 days.
Using the exponential distribution, we can calculate this probability as P(T ≥ 7) = 1 - P(T < 7).
Since the exponential distribution is memoryless, the probability that T < 7 is equivalent to the probability that a death occurs within the first 7 days. This can be calculated using the cumulative distribution function (CDF) of the exponential distribution.
The CDF of the exponential distribution with parameter λ is given by F(x) = 1 - e^(-λx).
Substituting λ = 0.1 and x = 7, we can calculate the probability that a death occurs within the first 7 days as P(T < 7) = 1 - e^(-0.1 * 7).
Now, we can calculate the desired probability as P(T ≥ 7) = 1 - P(T < 7) = 1 - (1 - e^(-0.1 * 7)) = e^(-0.1 * 7) ≈ 0.4966.
Therefore, the chances that a week or more will elapse before another death occurs are approximately 0.4966 or 0.50 in decimal form.
To solve this problem, we can use the Poisson distribution. The Poisson distribution is commonly used to model events that occur randomly over a fixed interval of time or space.
In this case, let's assume that the number of deaths in a given day follows a Poisson distribution with a rate parameter of λ = 0.1. This means that on average, 0.1 deaths occur per day in the nursing home.
To find the probability that a week or more will elapse before another death occurs, we need to find the probability of having zero deaths in the next 7 days.
The formula for the Poisson distribution is:
P(X = k) = (e^(-λ) * λ^k) / k!
Where:
- P(X = k) is the probability of exactly k events occurring,
- e is Euler's number (approximately 2.71828),
- λ is the average rate of events,
- k is the number of events.
In this case, we want to find P(X = 0) for a period of 7 days. So, we can plug in the values into the formula:
P(X = 0) = (e^(-0.1) * 0.1^0) / 0!
Since any number to the power of 0 is 1, and 0! (factorial of 0) is 1, we can simplify the equation further:
P(X = 0) = e^(-0.1)
Now, we can calculate this probability:
P(X = 0) = e^(-0.1) ≈ 0.9048
Therefore, there is approximately a 0.9048 probability of not having any deaths in a single day.
To find the probability of not having any deaths for 7 days in a row, we can simply raise this probability to the power of 7:
P(X = 0 for 7 days) ≈ 0.9048^7 ≈ 0.5013
So, the probability that a week or more will elapse before another death occurs is approximately 0.5013, or approximately 0.50 when rounded to two decimal places.