Statistical Distribution Problem...

Question

The length of time, in tens of minutes, that patients spend waiting at a doctor’s surgery is modelled by the continuous random variable T, with the following cumulative distribution function:

F(t) = (the following)

{ = 0, t < 0,

{ = (1/135)(54t + 9t 2 − 4t 3 ), 0 ≤ t ≤ 3,

{ = 1, t > 3.

(a) Find the probability that a patient waits for more than 20 minutes.

(b) Show that the median waiting time is between 11 and 12 minutes.

(c) Define fully the probability density function f(t) of T.

(d) Find the modal waiting time in minutes.

(e) Give one reason why this model may need to be refined. ...

Answer 1

(a) We are looking for P(T > 2). I say 2, not 20, because t is in tens of minutes. Since we are given the cumulative distribution function (cdf), probably the easiest way to calculate this is to calculate its complement and subtract from 1. All the is needed to find the complement, P(T < 2), is to plug in 2 for t. Using technology or working by hand, we get F(2) = 112/135. The complement of this is 1 - 112/135 = 23/135.

(b) The median is where the cdf = 0.5. Now this cdf is a cubic equation, which is rather nasty to solve by hand. But that's not what we're being asked to do. All we need is to establish that it must be between 11 and 12 minutes, or 1.1 < med < 1.2. This is much more easily done by plugging in both 1.1 and 1.2 into F(t) and showing that F(1.1) < 0.5 and F(1.2 ) > 0.5. When we do this, we see that F(1.1) = 0.4812, and F(1.2) = 0.5248, which satisfies the conditions that it needs to satisfy.

(c) Since a cumulative density function (cdf) is the integral of a probability density function (pdf), then the pdf is the derivative of the cdf that we're given. So, using our derivative rules, we get

f(t) = (1/135)(54 + 18t - 12t^2)

(d) The mode of f(t) is the t value where it hits its maximum f(t) value. This can be found by taking derivative f'(t) and finding out what t value makes it equal 0--in other words, what t value makes f(t) level off.

So f'(t) = (1/135)(18 - 24t)

Setting this equal to 0, we get

(1/135)(18 - 24t) = 0

18 - 24t = 0

18 = 24t

t = 18/24 = 3/4 = 0.75, which is 7.5 minutes.

(e) I've been thinking about this part as I've been answering the earlier parts, and I'm finding it to be the most difficult part of all of them. The best that I can come up with right now is that it does not seem realistic to claim that the probability of waiting 30 minutes or less is 1, or 100%. Stated another way, it does not seem realistic to claim that the probability of waiting more than 30 minutes is 0, that is, that it is impossible that you will be kept waiting more than 30 minutes. There may well be other reasons why this model needs to be refined, but this is what I'm seeing right now.