# How can i calculate the probability of waiting for a certain bus to come? Difficult but interesting

## Can anyone explain to me how to solve this question: The waiting timein minutes for a bus in a certain station(from getting to the station until the bus arrival) is a random variable with a mean of 10. But if the the traffic is busy, then itsmean is 20. The probavility that the traffic will be busy when waiting for the bus is 0.18. 1)what is the probability that a man that arrives in time will wait longer than 15 minutes? 2)if a man waits longer than 15 minutes,what is the probability that the traffic is occupied? 3)if a man comes in a random day and after 8 minutes still waiting, what is the probability he'll have to wait for notmlonger than 15 minutes? Please explain how you did, as i am trying to learn from such complicated questions. Thank you so much for teaching me!

May 5, 2018

1) .20
2)$\approx .81$
3) $\approx .34$

#### Explanation:

1) So to begin with you have two scenarios one with mean 10 and the other with mean 20 with some probability of occurrence. The questions are centered around some probability of minutes.

With normal random variables we can think of normal distributions which are symmetrical around the mean. In other words as we go further from the mean the less likely something is to occur. We need to figure out what that mean and spread is aka variance. The variance can be used to determine the probability of observing values further away from the mean.

The normal distribution has some nice theoretical properties and the branch of statistics has found some formulas that tend to hold. One of which is called Expectation or the long run average.

Expected mean is defined as $\setminus \sum p \left(x\right) \cdot x$

Expected Variance as $\setminus \sum p \left(x\right) \cdot {\left(x - \setminus \mu\right)}^{2}$

First the mean is $.18 \cdot 20 + .82 \cdot 10 = 11.8$
now lest solve for the variance is $.18 \cdot {\left(20 - 11.8\right)}^{2} + .82 \cdot {\left(10 - 11.8\right)}^{2} = 14.76$

The normal distribution can compute the probability of observing values away from the mean using the standard deviation (the square root of the variance). everything within two standard deviations account for 95% of all point that you would observe.
$\setminus \sqrt{14.76} \approx 3.84$

so I would expect that most values have a 95% probability of seeing some value between 4.12 and 19.48.

We can compute a Z score which helps quantify the spread of the standard deviation and compute a probability as a result. There is a precomputed table in the back of most statistic books that allow you to convert a Z score into a probability. I will use this approach to answer the questions. the Z score is defined as $\frac{x - \setminus \mu}{\setminus} \sigma$ and the final probability equates to $.2$

2) The second question is geared toward determining in the traffic is busy or not. Here we actually are interested in determining the likelihood of it being late. There is a nice formula for that too. Bayes decomposed joint probability as $p \left(a | b\right) = \frac{p \left(b | a\right) p \left(a\right)}{p \left(b\right)}$

Here we can think of $b = x > 15$ and $a = l a t e$ so we have $p \left(b\right) = .2$, $p \left(a\right) = .18$ for $p \left(b | a\right)$ i use the same standard deviation from above and mean as 20 for my distribution. Here I make a loose assumption that this is reasonable estimate
$\frac{.9 \cdot .18}{.20} = .81$

It makes sense that there is traffic because its past the expected time of 10 minutes. Also I know that its at least .5 since anything greater than 20 accounts for 50% of the probability.

3) for this question we need to compute the difference from 11.8 to 8 and then see what percent of the probability from the zscore is equal to this thus.

$11.8 - 8 = 3.8$

This is actually close to 1 standard deviation based on earlier calculation from problem 1. I know statistically that $\pm 3.84$ is about 68% so i can take about half of that to approximate roughly this amount. Thus 34% of it coming between now and 15 minutes.

I know I used some approximation for answering 2 and 3. I personally am okay with it because the nature of probability.