Statistics Help

[baseline bum]

At a particular crossroads in the middle of the desert, vehicles do not drive by very often. If on average, only 36 vehicles drive by over a 24 hour period, what is the probability that exactly one vehicle will drive by in the next 10 minutes? (24 hours=144 ten minute intervals)

X <- number of vehicles to come in a period

X is Poisson (just look up the prob density fcn for it, at apply it at 1 with mean 36/144)

The pdf for Poisson is EX^k * e^(-EX) / k!, where EX is the mean num of vechicles coming in a period, ie. the expected value of X for the period

EX = 36/144 = 0.25 gives the mean number of vehicles to come in a period

P(X=1) = (EX)^1 * (e^(-EX)) /1!

where exp(x) = (2.71828..)^x


Therefore,

P(X=1) = 0.25 * e^-0.25 / 1 = 0.321/1 = 0.321 = 32.1%

[CubanMustGo]

Math+CS major

Calculus (3 quarters)
Linear Algebra (2 quarters)
Abstract Algebra (2 quarters)
Real Analysis (2 quarters)
Complex Analysis
Probability (2 quarters)
Partial Differential Equations
Ordinary Differential Equations
Topology
Numerical Analysis (2 quarters)
Queuing Theory
Algorithms and Complexity
Discrete Math
Mathematical Modelling
Stochastic Processes
Computer Graphics

prob some other too

Beats the out of 4 semesters of calculus and 1 semester each of stat and numerical methods 28 (!) years ago. Thanks, don't feel so bad now.

[1369]

I played lacrosse and drank beer instead of going to my stats class, but reading this thread makes me want to read this even more...

[E20]

Math+CS major

Calculus (3 quarters)
Linear Algebra (2 quarters)
Abstract Algebra (2 quarters)
Real Analysis (2 quarters)
Complex Analysis
Probability (2 quarters)
Partial Differential Equations
Ordinary Differential Equations
Topology
Numerical Analysis (2 quarters)
Queuing Theory
Algorithms and Complexity
Discrete Math
Mathematical Modelling
Stochastic Processes
Computer Graphics

prob some other too

No wonder.

[spurster]

I disagree with BB on

4. What is the probability that exactly two of the three cards will be the same?

My answer is:

(52*51*50 - 52*48*44 - 52*3*2) / (52*51*50)

which can be simplified, but here is the story:

There are 52*51*50 ways to draw three cards.
There are 52*48*44 ways to draw three different ranks.
There are 52*3*2 ways to draw three of a kind.
Everything else left must be a pair.

[CubanMustGo]

nevermind

[mrsmaalox]

I took "Statistics for Healthcare Professionals" aka "Weenie Stats" and our prof made every test open book, because if we ever had to do any stats for our jobs, he figured we'd immediately find a book to work it out of! He was realistic.

[baseline bum]

I disagree with BB on

4. What is the probability that exactly two of the three cards will be the same?

My answer is:

(52*51*50 - 52*48*44 - 52*3*2) / (52*51*50)

which can be simplified, but here is the story:

There are 52*51*50 ways to draw three cards.
There are 52*48*44 ways to draw three different ranks.
There are 52*3*2 ways to draw three of a kind.
Everything else left must be a pair.

F*ck! Probability was one of two math classes I didn't get A or better in (Got an A- in the 1st quarter of Probability and in Discrete). Not sure where I'm screwing up using the indicator R.V.s there.

[baseline bum]

I took "Statistics for Healthcare Professionals" aka "Weenie Stats" and our prof made every test open book, because if we ever had to do any stats for our jobs, he figured we'd immediately find a book to work it out of! He was realistic.

My prof had just got out of Cambridge and killed everyone with the first exam in Probability part A. I got a 60 on it and was freaked out, until I found out the mean score was like 15.

[baseline bum]

Eh, crap. I see where I screwed up on that problem. I made everything have to match up with the first card chosen, when another equally likely case was that the second and third card matched. That's why my answer was a factor of 3/2 off from Spurster's (correct) answer.

[TheSanityAnnex]

thanks to all.

[mrsmaalox]

thanks to all.

I just want to know who is getting the money/handjob/fingering!

[Phenomanul]

X <- number of vehicles to come in a period

X is Poisson (just look up the prob density fcn for it, at apply it at 1 with mean 36/144)

The pdf for Poisson is EX^k * e^(-EX) / k!, where EX is the mean num of vechicles coming in a period, ie. the expected value of X for the period

EX = 36/144 = 0.25 gives the mean number of vehicles to come in a period

P(X=1) = (EX)^1 * (e^(-EX)) /1!

where exp(x) = (2.71828..)^x


Therefore,

P(X=1) = 0.25 * e^-0.25 / 1 = 0.321/1 = 0.321 = 32.1%

Too bad we can't ask timvp to embed mathtype fonts in the forum.