Question

I arrive at a bus stop at a time that is normally distributed with mean 07:59 and SD 1.5 minutes. My bus arrives at the stop at an independent normally distributed time with mean 08:03 a.m. and SD 2 minutes. The bus remains at the stop for 1 minute and then leaves.

What is the chance that I miss the bus?

Answer 1

Say you arrive at the bus stop with a normal distribution X~ N(0, 1.5)

The bus arrives at the stop with a normal distribution Y~ N(4, 2)

Let's find the distribution of Z = X - Y

Z will also follow a normal distribution with mean = u1 - u2 = 0 - 4 = -4

The variance will be:

Variance = 1.5*1.5 + 2*2 = 2.25 + 4 = 6.25

s = 2.5

Hence, Z~ N(-4, 2.5)

Now, the bus remains at the stop for 1 minute. If you arrive at the bus stop later than a minute (or Z > 1) when the bus arrives, you will miss the bus. We need to find the probability:

P(Z > 1)

z = (1 - (-4))/2.5 = 5/2.5 = 2

Required probability = P(z > 2) = 1 - P(z < 2)

= 1 - 0.9772 = 0.0228

Hence, there is a 0.0228 probability that you will miss the bus