Question

The time I spend waiting for the bus on any given day has a distribution with mean 4 min- utes and variance off 0.5 minutes. What is the probability that I spend more than 2 hours and 10 minutes waiting for the bus in one month (30 days)? You may assume that waiting times on different days are independent of each other. HINT: Is there a sum of random variables somewhere in here?

Answer 1

0.0049 = 0.49% probability that I spend more than 2 hours and 10 minutes waiting for the bus in one month.

Explanation:

Normal Probability Distribution

Problems of normal distributions can be solved using the z-score formula.

In a set with mean
`\mu` and standard deviation
`\sigma`, the z-score of a measure X is given by:

`Z = (X - \mu)/(\sigma)`

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the p-value, we get the probability that the value of the measure is greater than X.

n instances of a normal variable:

For n instances of a normal variable, the mean is:

`M = n\mu`

`s = \sigma√(n)`

Mean of 4 minutes, standard deviation of 0.5 minutes:

This means that
`\mu = 4, \sigma = √(0.5)`

30 days:

`M = 30(4) = 120`

`s = √(0.5)√(30) = √(0.5*30) = √(15)`

What is the probability that I spend more than 2 hours and 10 minutes waiting for the bus in one month (30 days)?

2 hours and 10 minutes is 2*60 + 10 = 130 minutes, so this probability is 1 subtracted by the p-value of Z when X = 130. So

`Z = (X - \mu)/(\sigma)`

In this context, due to the 30 instances of the normal variable:

`Z = (X - M)/(s)`

`Z = (130 - 120)/(√(15))`

`Z = 2.58`

`Z = 2.58` has a p-value of 0.9951.

1 - 0.9951 = 0.0049

0.0049 = 0.49% probability that I spend more than 2 hours and 10 minutes waiting for the bus in one month.