An event organiser finds that 6% of people who purchased a ticket to an event will not show up. If the event organiser sold 215 tickets for an event that can only seat 200 people, what is the approximate probability that every ticket holder who shows up will have a seat? Please use the normal approximation.  

To solve this problem, we can use the normal approximation to the binomial distribution. Here's how we approach it:

### Step 1: Define the problem
- Let \( n = 215 \) be the total number of tickets sold.
- Let \( p = 0.94 \) be the probability that a person shows up (since 6% don't show up, 94% do).
- Let \( X \) be the random variable representing the number of people who show up.

\( X \) follows a binomial distribution, \( X \sim \text{Binomial}(n=215, p=0.94) \).

We want to find the probability that the number of people who show up is less than or equal to 200, i.e., \( P(X \leq 200) \).

### Step 2: Use the normal approximation
For a binomial distribution with large \( n \), we can approximate it with a normal distribution \( N(\mu, \sigma^2) \), where:
- \( \mu = np = 215 \times 0.94 = 202.1 \)
- \( \sigma^2 = np(1-p) = 215 \times 0.94 \times 0.06 \approx 12.141 \)
- \( \sigma = \sqrt{12.141} \approx 3.484 \)

Thus, \( X \) can be approximated by \( N(202.1, 3.484^2) \).

### Step 3: Apply the continuity correction
We need to calculate \( P(X \leq 200) \). Using the continuity correction, this becomes:
\[
P(X \leq 200) \approx P\left( Z \leq \frac{200 + 0.5 - 202.1}{3.484} \right)
\]
Where \( Z \) is a standard normal variable.

### Step 4: Calculate the Z-score
\[
Z = \frac{200.5 - 202.1}{3.484} \approx \frac{-1.6}{3.484} \approx -0.459
\]

### Step 5: Find the probability using the Z-score
Using standard normal distribution tables or a calculator, we find:
\[
P(Z \leq -0.459) \approx 0.3236
\]

### Conclusion
The approximate probability that every ticket holder who shows up will have a seat is **0.3236** or **32.36%**.

Would you like more details or have any questions?

Here are five related questions you might find interesting:
1. How would the probability change if the event organizer sold 220 tickets instead?
2. What is the probability that exactly 200 people show up?
3. How would the result change if the no-show percentage was 10%?
4. What is the expected number of people who will not show up?
5. How does the accuracy of the normal approximation compare with exact binomial calculations for this scenario?

**Tip:** The normal approximation to the binomial distribution is most accurate when \( np \) and \( n(1-p) \) are both greater than 5.

Learn how to approximate the probability that every ticket holder who shows up will have a seat at an event. This problem involves using the normal approximation to the binomial distribution, with specific calculations and steps provided.

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