Thread: A few statistics questions

Originally Posted by margaritas

Here are the questions:

1. The petrol stations along a road are located according to a Poisson distribution, with an average of 1 station in 10 km. Because of an oil shortage worldwide, there is a probabililty of 0.2 that a petrol station will be out of petrol.

(i) Find the probability that there is at most 1 petrol station in 15 km of the road.

(ii) Find the probability that the next 3 stations a driver encounters will be out of petrol.

A driver on this road knows that he can go another 15 km before his car runs out of petrol. Find the probability that he will be stranded on the road without petrol. Give your answer correct to 2 decimal places.

For this question, I can't solve the last part (in bold). Also, for (ii) I got the answer which is 0.008 but I don't quite know how I arrived at that so I need someone to explain that to me. Oh I can use the GC to solve this question so there's no need to go through all the formula.

The distribution of the distance to the next petrol station has an exponential
distribution with parameter lambda = mean distance between petrol stations
with fuel.

The mean distance between petrol stations is 10km, and as 20% of them
do not have fuel the mean distance between those with fuel is 12.5km
(in 1000km there are 100 stations of which 80 have petrol so mean distance
between these is 1000/80=12.5km).

RonL

Thanks but I still don't get it!

Originally Posted by margaritas

[snip]

For this question, I can't solve the last part (in bold). Also, for (ii) I got the answer which is 0.008 but I don't quite know how I arrived at that so I need someone to explain that to me. Oh I can use the GC to solve this question so there's no need to go through all the formula.


2. [snip]

Again it's the last part I have problems with, and I can use the GC for this as well.

What is the GC?

RonL

Originally Posted by margaritas

Thanks but I still don't get it!

The wikipedia article on the exponential distribution gives the distribution
functions and describes its relation to the Poisson distribution.

RonL

Originally Posted by CaptainBlack

What is the GC?

RonL

Graphic calculator.

Originally Posted by margaritas

2. Vehicles approaching a T-junction must either turn left or turn right. Observations by traffic engineers showed that on average, for every ten vehicles approaching the T-junction, one will turn left. It is assumed that the driver of each vehicle chooses direction independently. Out of 5 randomly chosen vehicles approaching the T-junction,

(i) find the probability that at least 3 vehicles turn right,

(ii) find the probability that exactly 4 vehicles turn right given that at least 3 vehicles turn right.

On a particular weekend, 40 randomly chosen vehicles approached the T-junction. Using a suitable approximation, find the probability that at least 38 of them turn right.

Again it's the last part I have problems with, and I can use the GC for this as well.

On the last part you are supposed to use the Normal approximation to the
Binomial distribution.

This has mean Np (in this case 40*0.9), and standard deviation sqrt(Np(1-p)
(in this case 40*0.9*0.1).

RonL

Originally Posted by margaritas

Graphic calculator.

Perhaps your Graphic Calculator should take the exam for you

RonL

Originally Posted by CaptainBlack

Perhaps your Graphic Calculator should take the exam for you

RonL

Oh. We are allowed and supposed to use the GC to aid us in the A-Levels examinations.

Originally Posted by margaritas

Oh. We are allowed and supposed to use the GC to aid us in the A-Levels examinations.

Reminds me of a story I heard when Hewlett Packard first put put symbolic
capabilities on their calculators (computer algebra system). I was said that
it would get 80% on a calculus exam if only it did not need an operator.

RonL

Originally Posted by CaptainBlack

On the last part you are supposed to use the Normal approximation to the
Binomial distribution.

This has mean Np (in this case 40*0.9), and standard deviation sqrt(Np(1-p)
(in this case 40*0.9*0.1).

RonL

Thanks I got the answer but using Poisson approximation to Binomial.

Originally Posted by CaptainBlack

The distribution of the distance to the next petrol station has an exponential
distribution with parameter lambda = mean distance between petrol stations
with fuel.

The mean distance between petrol stations is 10km, and as 20% of them
do not have fuel the mean distance between those with fuel is 12.5km
(in 1000km there are 100 stations of which 80 have petrol so mean distance
between these is 1000/80=12.5km).

RonL

I got my answer using Po(1.5)

Probability = P(X=0) + P(X=1)x0.2 + P(X=2)x0.2^2 + P(X=3)x0.2^3 = 0.30 (2 d.p.)

I still can't solve question 3 that is,

X is a binomial random variable, where the number of trials is 5 and the probability of success of each trial is p. Find the values of p if P(X=4)=0.12.

Anybody? Thanks!

Originally Posted by margaritas

I still can't solve question 3 that is,

X is a binomial random variable, where the number of trials is 5 and the probability of success of each trial is p. Find the values of p if P(X=4)=0.12.

Anybody? Thanks!

so we have:



Now I think you need to solve this numerical, there will be two positive
solutions (since Descartes rule of signs tell us there are either 2 or 0
positive solutions, and we know that there must be at least one root).

Now I could tell you what the roots are (at least approximately) but it
would be better if you try to find them yourself.

RonL