Given that the waiting time $X$ follows an exponential probability distribution with an average waiting time of 42 minutes, the probability density function (PDF) is given by:

$f(x) = \lambda e^{-\lambda x}, \quad x \geq 0$

where $\lambda$ is the rate parameter, which is the reciprocal of the mean. Since the mean waiting time is 42 minutes, we can find $\lambda$ as:

$\lambda = \frac{1}{42}$

The cumulative distribution function (CDF) of the exponential distribution is:

$F(x) = 1 - e^{-\lambda x}$

We are asked to find the value of $x$ such that 55% of the time, you will have to wait at most that many minutes. Mathematically, we want to find $x$ such that:

$F(x) = 0.55$

Substituting the CDF formula:

$0.55 = 1 - e^{-\lambda x}$

Solving for $x$:

$e^{-\lambda x} = 1 - 0.55 = 0.45$

Taking the natural logarithm on both sides:

$-\lambda x = \ln(0.45)$

Now, solving for $x$:

$x = \frac{-\ln(0.45)}{\lambda}$

Substituting $\lambda = \frac{1}{42}$:

$x = -42 \ln(0.45)$

Let me calculate that.The amount of time you will have to wait at most, 55% of the time, is approximately $33.54$ minutes.

Would you like more details on the solution? Here are some related questions for further exploration:

1. What is the probability of waiting more than 50 minutes?
2. How would the answer change if the average waiting time were reduced to 30 minutes?
3. What is the median waiting time in this distribution?
4. How do you derive the CDF for an exponential distribution?
5. How can we model multiple pubs with different average waiting times?

Tip: The exponential distribution is memoryless, meaning the waiting time until the next event is independent of how much time has already passed.