To solve this problem, we first need to define the random variable \( X \), which is the number of patrons who do not show up for the performance. Given that 5% of patrons do not show up, we can model \( X \) using the binomial distribution \( B(n, p) \), where \( n = 100 \) (the number of seats) and \( p = 0.05 \) (the probability that a patron does not show up).
Step 1: Calculate the mean and variance
The mean \( \mu \) and variance \( \sigma^2 \) of a binomial distribution are calculated as follows:
\[ \mu = n \cdot p = 100 \cdot 0.05 = 5 \]
\[ \sigma^2 = n \cdot p \cdot (1-p) = 100 \cdot 0.05 \cdot (1 - 0.05) = 100 \cdot 0.05 \cdot 0.95 = 4.75 \]
Thus, the standard deviation \( \sigma \) is:
\[ \sigma = \sqrt{4.75} \approx 2.179 \]
Step 2: Normal approximation to the binomial distribution
Since \( n \) is large, we can use the normal approximation to the binomial distribution. The random variable \( X \) can be approximated by a normal distribution with mean \( \mu = 5 \) and standard deviation \( \sigma \approx 2.179 \).
Step 3: Find the probability that 6 or more patrons do not show up
We are interested in the probability \( P(X \geq 6) \). To use the normal approximation, we apply a continuity correction. Thus, we want to find:
\[ P(X \geq 6) = P(X > 5.5) \]
We standardize this to a standard normal variable \( Z \):
\[ Z = \frac{X - \mu}{\sigma} = \frac{5.5 - 5}{2.179} \approx \frac{0.5}{2.179} \approx 0.229 \]
Step 4: Find the corresponding probability from z-table
Now, we need to find \( P(Z > 0.229) \). We can look this value up in the standard normal distribution table or use a calculator.
Using a standard normal distribution table:
\[ P(Z < 0.229) \approx 0.5910 \]
Thus,
\[ P(Z > 0.229) = 1 - P(Z < 0.229) \approx 1 - 0.5910 = 0.4090 \]
Step 5: Round to the nearest thousandth
Finally, rounding this probability to the nearest thousandth gives:
\[ \boxed{0.409} \]
This means the probability that 6 or more patrons will not show up is approximately 0.409.