We are tasked with finding the probability that more than 68 people show up for a flight when 70 tickets are sold, and each person has a 97% chance of showing up.

This situation can be modeled using a binomial distribution, as there are two possible outcomes for each passenger (they either show up or they don’t), and each has the same probability.

Steps:

1. Define the binomial distribution parameters:

   • The number of trials (booked passengers) $n = 70$.
   • The probability of success (showing up) $p = 0.97$.

   The random variable $X$, the number of people who show up, follows a binomial distribution $X \sim \text{Binomial}(70, 0.97)$.

2. Find the probability that more than 68 people show up:

   • We want to find $P(X > 68)$, which is $1 - P(X \leq 68)$.

   Using the binomial cumulative distribution function (CDF), we can calculate $P(X \leq 68)$.

3. Approximate using the normal distribution: Since $n$ is large, we can use the normal approximation to the binomial distribution:

   • Mean $\mu = n \times p = 70 \times 0.97 = 67.9$,
   • Standard deviation $\sigma = \sqrt{n \times p \times (1 - p)} = \sqrt{70 \times 0.97 \times 0.03} \approx 1.431$.

   Now, we convert to the standard normal distribution: $Z = \frac{X - \mu}{\sigma}$ To find $P(X > 68)$, we first find the z-score for $X = 68$: $Z = \frac{68 - 67.9}{1.431} \approx 0.07$

4. Find the probability: Using a standard normal table or calculator, the probability of $Z > 0.07$ is approximately $P(Z > 0.07) = 0.4726$.

Final answer:

Thus, the probability that more than 68 people show up is approximately $P(X > 68) \approx 0.4726$.

Would you like more details or have any questions?

Here are some related questions:

1. How is the binomial distribution related to the normal distribution?
2. What is the central limit theorem, and how does it apply here?
3. What would happen if the probability of people showing up was less than 97%?
4. Can you solve the problem without using the normal approximation?
5. What is the impact of booking more passengers than the plane's capacity?

Tip: The normal approximation to the binomial distribution works well when $n$ is large and $p$ is not too close to 0 or 1.