Probability & stats questions

Brentmeister

Hi guys,

I have a couple of questions that i'm stuck on:

1. On average, three cars arrive (independently of each other) at a carpark per hour. Find the probability that, in a given hour, (i) four cars or less will arrive, (ii) two or more cars will arrive
If more than four cars arrive in any hour, congestion occurs. What is the probability that, in a working day of six hours, congestion will occur twice?

2. In 1997, 45% of all households in Dublin used the internet. At the end of 2007, in a random sample of 400 households, 204 used the internet. Is there significant evidence that the proportion had increased from the 1997 level. Use alpha=0.02
How large a sample would have been required to ensure, with 99% confidence, that the sample proportion was within 0.01 of the true proportion?

Thanks guys, i dont know where to start with these really!

LeixlipRed

This looks like a homework assignment so let's give pointers not complete solutions please. The first one is a poisson distribution as it's discrete with an observed average. And the second one involves the normal distribution, confidence intervals and hypothesis testing.

Brentmeister

thanks for the help leixlipred!

there past exam questions that im trying to get my head round, so not trying to get someone to do my homework.

with your pointers though i have a starting point to understanding them thanks!

LeixlipRed

No, not accusing you of anything. Do you have a full set of notes? There should definitely be sections on Poisson and Normal Distributions!

Brentmeister

LeixlipRed wrote: »

No, not accusing you of anything. Do you have a full set of notes? There should definitely be sections on Poisson and Normal Distributions!

its one section of an exam that im not strong in, so im aiming just to get by in this part and pick up more marks in the theory section. i have notes but havent been concentrating on this area at all. and the exam is in a couple of weeks:eek:

from your pointers i've found this...http://stattrek.com/Lesson2/Poisson.aspx. thanks!

i have other questions on least squares, moving average and chi square but know where to start on them.

one other question where i'm stumped for a starting point is

The IQ of a population is normally distributed with mean 100 and standard deviation 25. What proportion of the population has an IQ a) greater than 150 b) between 80 and 130?
The government decides to provide special educational facilities for the bottom 5% of the population (in terms of IQ). What should the cut off point be?

Some questions i just dont know where to start

LeixlipRed

You need to look at Confidence Intervals for proportions for the that question. Check out the Confidence Interval wiki page. You can post your workings up here if you want and we can tell you were you're going wrong.

MathsManiac

LeixlipRed wrote: »

You need to look at Confidence Intervals for proportions for the that question. Check out the Confidence Interval wiki page. You can post your workings up here if you want and we can tell you were you're going wrong.

Actually I wouldn't say that that the confidence interval for a proportion is particularly relevant to that last question. It's really just about transforming a normal distribution to and from the standardised distribution. The transformation equation is z = (x - mu) / sigma, or you can do it by just counting standard deviations above and below the mean.

The first part asks what proportion of the distribution is over 150. This is the same as asking what proportion of a normal distribution is more than two standard deviations above the mean. You look up your z-tables and Bob's your uncle.

The second part is similar: what proportion of a normal distribution is between ... and ...
The last part is the same idea but working the opposite way: the z-tables will tell you what value of z will cut off the bottom 5% of a standardised normal distribution, and you've to transform that back to a normal distribution with mean 100 and s.d. 25.

LeixlipRed

Right you are. I just seen proportion and jumped.

Brentmeister

thanks for the offer of checking my work leixlipred, here's my shot at the first one....

Brentmeister wrote: »

1. On average, three cars arrive (independently of each other) at a carpark per hour. Find the probability that, in a given hour, (i) four cars or less will arrive,


P(x; μ) = (e-μ) (μx) / x!
P(4; 3) = (2.71828-3) (81) / 4!
P(4; 3) = (0.13534) (8) / 6
P(4; 3) = 0.180

P(4 or less) = P(4) + P(3)....P(0)
=0.814


(ii) two or more cars will arrive

1-P(1) +P(0)
1-0.149+0.049
=0.9
If more than four cars arrive in any hour, congestion occurs. What is the probability that, in a working day of six hours, congestion will occur twice?

this i'm still stuck on

Mathsmaniac, that definitely makes sense for the z scores method, i hadnt looked at it like that but can see how it would work, thanks!

LeixlipRed

I'm not sure you're using the right formula there. I'm tired so maybe you are.

For Poisson, pmf is P(X=k)=[e^(-u).u^k]/k!

The second part you've definitely made a mistake in. P(X>=2) = 1 - (P(X<=1) = 1- (P(X=0) + P(X=1)).

Concentrate on getting this part done correctly first. Last part of that is a little different.

LeixlipRed

No, you done the 1st one right. It's just the way you typed it had me confused. Second part is just simple mistake.

LeixlipRed

The last part of that you can do using the binomial distribution I reckon. You want to know what is the probability that in 2 out of 6 hours you'll get congestion (i.e. 4 or more cars). So you have the probability of that happening and the probability of that not happening. So two possible outcomes. So you can the binomial distribution. Read up on it and see if you can come up with anything.

Brentmeister

thanks a mil for your help leixlipred, think i'll leave it till the morning though, gonna hit the hay!

TommyGunne

LeixlipRed wrote: »

The last part of that you can do using the binomial distribution I reckon. You want to know what is the probability that in 2 out of 6 hours you'll get congestion (i.e. 4 or more cars). So you have the probability of that happening and the probability of that not happening. So two possible outcomes. So you can the binomial distribution. Read up on it and see if you can come up with anything.

Question is worded quite indeterminately. If congestion occurs once, then the probability of it occurring again in the following few minutes (up to an hour), asa you already have a good density of events for the previous x number of minutes (where x is less than 60). ie hours are not independent. Thus, I'm not sure if what you suggest is entirely appropriate. I may be wrong though. Question seems to be worded horribly to me either way.

Brentmeister

If more than four cars arrive in any hour, congestion occurs. What is the probability that, in a working day of six hours, congestion will occur twice?

This is a binomial distribution as there is the probability of congestion happening and the probability of it not happening. So two possible outcomes.

Binomial Formula:

n = number of trials
k = number of successes
n – k = number of failures
p = probability of success in one trial
q = 1 – p = probability of failure in one trial
In this case ‘success’ is congestion occurring twice k = 2
Number of trials is each hour, n = 6
Probability of success in one trial
P( 4) = 1 – P ( 4)
= 1- 0.814 (from part i)
= 0.186
q = 1- p = 1 – 0.186 = 0.814

Plug these values into formula answer = 0.228
Probability of 23% that congestion occurs twice in six hours.
Note: assumption made that congestion can only happen once in an hour.


am i on the right track?? TommyGunne...ya its strangely worded, even though its an NUI exam.

LeixlipRed

TommyGunne wrote: »

Question is worded quite indeterminately. If congestion occurs once, then the probability of it occurring again in the following few minutes (up to an hour), asa you already have a good density of events for the previous x number of minutes (where x is less than 60). ie hours are not independent. Thus, I'm not sure if what you suggest is entirely appropriate. I may be wrong though. Question seems to be worded horribly to me either way.

Yeh, if you look at it that way it's not quite so simple. It's a bit ambiguous alright.