Suppose you took 1000 random samples of size 200 from the Poisson distribution
with u = 5 and computed a 90% confidence interval for each sample. Approximate the probability that at
least 920 of these intervals would contain the mean value u = 5. Be sure to define any random variable you
use, as well as its distribution, and justify any procedure that you use.
the u represent mu sign
To solve this problem, we need to use the properties of the Poisson distribution and confidence intervals.
First, let's define some important terms:
- X: A random variable representing the number of intervals (out of the 1000 samples) that contain the mean value μ = 5.
- X follows a binomial distribution because it represents the number of successes (intervals containing μ) in a fixed number of trials (1000 samples).
Now, let's determine the parameters of the binomial distribution:
- n = 1000: The number of trials (samples taken).
- p: The probability of success (an interval containing μ). We need to calculate this using the confidence interval approach.
In order to calculate p, we need to find the width of the confidence interval assuming a 90% confidence level. Here's how we can do that:
1. Calculate the standard deviation (σ) of the Poisson distribution:
σ = sqrt(μ) = sqrt(5) ≈ 2.236
2. Calculate the margin of error (E) for the confidence interval:
E = z * (σ / sqrt(n))
Here, z represents the critical value from the standard normal distribution that corresponds to the desired confidence level. For a 90% confidence interval, z ≈ 1.645 (obtained from a standard normal distribution table).
3. Calculate the width of the confidence interval (C):
C = 2 * E
Now we can compute p, the probability of an interval containing μ, as the proportion of the Poisson distribution within a width of C:
4. p = P(X > 920) = 1 - P(X ≤ 920)
We subtract the cumulative probability up to 920 from 1 to get the probability of having more than 920 intervals containing μ.
Finally, we can compute p by using the binomial cumulative distribution function (CDF) or, alternatively, approximating it with the normal distribution:
5. p = 1 - binomcdf(n, p, 920) (using a binomial CDF function)
or
p = 1 - P(Z ≤ (920 - n * p) / sqrt(n * p * (1 - p))) (using the standard normal distribution)
By substituting the values into equation 5, you can calculate the approximate probability that at least 920 of the 1000 intervals contain the mean value μ = 5.