Here's the sketch of the normal distribution with mean $\mu = 155.31$ and standard deviation $\sigma = 3.15$. The shaded area represents the probability $P(X \geq 155)$, which is approximately 60.26%.

This indicates that there's about a 60.26% chance that at least 155 passengers will show up, making it likely that all seats may be occupied, and potentially someone may get bumped.

Would you like further details or have any questions?

Here are five related questions to expand your understanding:

1. How does the probability change if the no-show rate decreases?
2. What if the airline sold only 160 tickets instead of 167?
3. How does the choice of continuity correction affect our approximation?
4. How would you interpret the result if the standard deviation were larger?
5. Can we apply this approach to situations other than airlines?

Tip: When working with binomial distributions and high sample sizes, using a normal approximation is a common and efficient technique for calculating probabilities.