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R Tatauranga, Ryan and Miriam's lecturer at Down-Under University, points out that if you add their 99 waiting times at the bus stops during a term, you will see that Ryan and Miriam spend hours and hours waiting at the bus stop over a term!

Using the approximation mentioned in Question 1, what is the probability that Ryan's total waiting time at the bus stop over the course of a term will exceed 13 hours?

Answer :

Final answer:

The total waiting time for Ryan over the term could be considered as the sum of random variables from a probability distribution.

Explanation:

In order to solve this problem, we should first understand that the total waiting time at the bus stop follows a probability distribution. If the average waiting time for each bus stop event is uniform or normally distributed, we can use the Central Limit Theorem, as the number of waiting times is large (99) - which states that the sum of a large number of independent and identically distributed random variables will approximate a normal distribution.

From the given data, we do not know the exact average or standard deviation of their waiting times. However, if it were given, say the average waiting time per stop was 'a' minutes with a standard deviation of 'b' minutes, the sum for 99 stops will be 99*a minutes with a standard deviation of sqrt(99)*b minutes.

Then, we would convert 13 hours into minutes (780 minutes assuming 1 hour = 60 minutes). Finally, by computing the z-score ((780-99*a)/(sqrt(99)*b)), we can find the probability that Ryan's total waiting time exceeds 13 hours from the standard normal distribution table.

Unfortunately, without specific values for the average waiting time and standard deviation, we can't provide a numerical answer for this question. This general approach should work with adequate data.

Learn more about Probability here:

https://brainly.com/question/33594301

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