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 first need to define the random variable and its distribution, as requested.

Let X be the random variable that represents the number of intervals (out of 1000) that contain the mean value μ = 5. X follows a binomial distribution with parameters n = 1000 (the number of trials) and p (the probability of success).

Now let's consider the probability that at least 920 of the intervals contain the mean value μ = 5. We need to calculate P(X >= 920), where X follows a binomial distribution.

To find the probability, we can use the normal approximation to the binomial distribution, also known as the normal approximation to the binomial distribution.

The conditions for using this approximation are as follows:
1. The number of trials, n, is large (n > 30).
2. Both np and n(1-p) are greater than or equal to 10.

Given that our sample size is 200 (n = 200) and the probability of success is p = P(μ = 5) = e^(-5) * (5^0) / 0! = 0.00673664, we can check if the conditions for using the approximation are satisfied.

np = 200 * 0.00673664 = 1.347328
n(1-p) = 200 * (1 - 0.00673664) = 198.652672

Both np and n(1-p) are greater than 10, so we can proceed with the normal approximation.

Approximating the binomial distribution to a normal distribution, X (the number of intervals containing μ = 5) follows a normal distribution with mean μ_X = n * p and standard deviation σ_X = sqrt(n * p * (1 - p)).

μ_X = 1000 * 0.00673664 = 6.73664
σ_X = sqrt(1000 * 0.00673664 * (1 - 0.00673664)) = 2.58588

Now we need to find the probability P(X >= 920) using the normal approximation. We will use the z-score formula to calculate this probability.

z = (920 - μ_X) / σ_X = (920 - 6.73664) / 2.58588

Now, we can use a standard normal distribution table or a calculator to find the probability of z-score greater than or equal to the calculated z-value. This probability represents the cumulative probability from z to positive infinity.

Using the z-table or a calculator, we find the cumulative probability P(Z >= z) to be approximately 0.9999.

Therefore, the approximate probability that at least 920 of the intervals contain the mean value μ = 5 is approximately 0.9999 or 99.99%.

Note: The normal approximation to the binomial distribution is an approximation, and the accuracy of the result may vary depending on the specific circumstances.