Griffiths D. Head First Statistics
Подождите немного. Документ загружается.
you are here 4 311
geometric, binomial, and poisson distributions
Q:
How come we use λ to represent
the mean for the Poisson distribution?
Why not use μ like we do elsewhere?
A: We use λ because for the Poisson
distribution, the parameter of the distribution,
expectation and variance are all the same.
It’s a way of making sure we keep everything
neutral.
Q:
Where does the formula for the
Poisson distribution come from?
A: It can actually be derived from the
other distributions, but the mathematics
are quite involved. In practice it’s best to
just accept the formula, and remember the
situations in which it’s useful.
Q:
What’s the difference between
the Poisson distribution and the other
probability distributions?
A: The key difference is that the Poisson
distribution doesn’t involve a series of
trials. Instead, it models the number of
occurrences in a particular interval.
Q:
Does λ have to be an integer?
A: Not at all. λ can be any non-negative
number. It can’t be negative as it’s the mean
number of occurrences in an interval, and
it doesn’t make sense to have a negative
number of occurrences.
Q:
What’s that “e” in the formula all
about?
A: e is a constant in mathematics that
stands for the number 2.718. So you can
substitute in 2.718 for e in the formula for
calculating Poisson probabilities.
The constant e is used frequently in calculus,
and it also has many other applications
in everything from calculating compound
interest to advanced probability theory.
Further discussion of e is outside the scope
of this book, though.
Q:
I keep getting the wrong answer
when I try to calculate probabilities using
the Poisson distribution. Where am I
going wrong?
A: There are two main areas where it’s
easy to trip up. The first thing is to make
sure you’re using the right formula. It’s easy
to get the r and the λ mixed up, so make
sure you’ve got them the right way round.
The second thing is to make sure you’re
using the e
x
function correctly on your
calculator. One way of doing this is to leave
the e
-λ
calculation until the end. Calculate
everything else first, then multiply by e
-λ
.
Where’s my drink? I want
a drink to go with my popcorn.
Give me my drink now!
The Statsville Cinema has another problem.
It’s not just the popcorn machine that keeps breaking down, now the drinks
machine has begun malfunctioning too. The mean number of breakdowns
per week of the drinks machine is 2.3.
The cinema manager can’t afford for anything to go wrong next week when
the promotion is on. What’s the probability that there will be no breakdowns
next week, either with the popcorn machine nor the drinks machine?
What’s the probability distribution of the drinks
machine? How can we find the probability that
neither the popcorn machine nor the drinks
machine go wrong next week?
312 Chapter 7
So what’s the probability distribution?
Let’s take a closer look at this situation.
We have two machines, a popcorn machine and a drinks machine, and we know
the mean number of breakdowns of each machine in a week. We want to find
the probability that there will be no breakdowns next week.
Here are the distributions of the two machines:
Popcorn machine
X ~ Po(3.4)
Y ~ Po(2.3)
Drinks machine
The mean number of
breakdowns per week of
the popcorn machine is 3.4.
The mean number of
breakdowns per week of
the drinks machine is 2.3.
If X represents the number of breakdowns of the popcorn machine and Y
represents the number of breakdowns of the drinks machine, then both X and
Y follow Poisson distributions. What’s more, X and Y are independent. In other
words, the popcorn machine breaking down has no impact on the probability
that the drinks machine will malfunction, and the drinks machine breaking down
has no impact on the probability that the popcorn machine will malfunction.
We need to find the probability that the total number of malfunctions next week
is 0. In other words, we need to find
P(X + Y = 0)
Think back to the chapter on probabilities. If X and Y are independent variables,
how can we find probabilities for X + Y?
x + y poisson distribution
you are here 4 313
geometric, binomial, and poisson distributions
Combine Poisson variables
You saw in previous chapters that if X and Y are independent random
variables, then
P(X + Y) = P(X) + P(Y)
E(X + Y) = E(X) + E(Y)
This means that if X ~ Po(λ
x
) and Y ~ Po(λ
y
),
X + Y ~ Po(λ
x
+ λ
y
)
This means that if X and Y both follow Poisson distributions, then so does
X + Y. In other words, we can use our knowledge of the way both X and Y
are distributed to find probabilities for X + Y.
If X is the number of times the popcorn machine malfunctions
and Y is the number of times the drinks machine malfunctions,
then X ~ Po(3.4) and Y ~ Po(2.3).
1. What’s the distribution of X + Y?
2. Once you’ve found how X + Y is distributed, you can use it to find probabilities. What’s P(X + Y = 0)?
314 Chapter 7
If X is the number of times the popcorn machine malfunctions
and Y is the number of times the drinks machine malfunctions,
then X ~ Po(3.4) and Y ~ Po(2.3).
1. What’s the distribution of X + Y?
λ
x
+ λ
y
= 3.4 + 2.3
= 5.7
X + Y ~ Po(5.7)
2. Once you’ve found how X + Y is distributed, you can use it to find probabilities. What’s P(X + Y = 0)?
P(X + Y = 0) = e
-
λ
λ
r
r!
= e
-5.7
x 5.7
0
0!
= e
-5.7
x 1
1
= 0.003
Q:
Does that mean that the probability
and expectation shortcuts we saw
earlier in the book work for the Poisson
distribution too?
A: Yes they do. X and Y are independent
random variables, because the popcorn
machine malfunctioning does not affect
the probability that the drinks machine will
malfunction, and vice versa. This means that
we can use all of the shortcuts that apply to
independent variables.
Q:
Why does X + Y follow a Poisson
distribution?
A: X + Y follows a Poisson distribution
because both X and Y are independent, and
they both follow a Poisson distribution.
Both the popcorn machine and drinks
machine each malfunction at random but at
a mean rate. This means that together they
also breakdown at random and at a mean
rate. Together, they still meet the criteria for
the Poisson distribution.
Q:
So can we use the distribution of
X + Y in the same we would any other
Poisson distribution?
A: Yes, we use it in exactly the same way,
so once you know what the parameter λ is,
you can use it to find probabilities.
Only a .003 chance
of no breakdowns next
week? Guess we better
get some new machines
after all.
sharpen your pencil solution
you are here 4 315
geometric, binomial, and poisson distributions
The Case of the Broken Cookies
Kate works at the Statsville cookie factory, and her job is to make sure
that boxes of cookies meet the factory’s strict rules on quality control.
Kate know that the probability that a cookie is broken is 0.1, and her
boss has asked her to find the probability that there will be 15 broken
cookies in a box of 100 cookies. “It’s easy,” he says. “Just use the
binomial distribution where n is 100, and p is 0.1.”
Kate picks up her calculator, but when she tries to calculate 100!,
her calculator displays an error because the number is too big.
“Well,” says her boss, “you’ll just have to calculate it manually.
But I’m going home now, so have a nice night.”
Kate stares at her calculator, wondering what to do. Then she smiles.
“Maybe I can leave early tonight, after all.”
Within a minute, Kate’s calculated the probability. She’s managed
to find the probability and has managed to avoid calculating 100!
altogether. She picks up her coat and walks out the door.
How did Kate find the probability so quickly, and avoid
the error on her calculator?
Five Minute
Mystery
Five Minute
Mystery
316 Chapter 7
The Poisson in disguise
The Poisson distribution has another use too. Under certain
circumstances it can be used to approximate the binomial distribution.
Why should I care? Why
would I want to do that?
Sometimes it’s simpler to use the Poisson
distribution than the binomial.
As an example, imagine if you had to calculate a binomial
probability where n is 3000. At some point you’d need to calculate
3000!, which would be difficult even with a good calculator.
Because of this, it’s useful to know when you can use the Poisson
distribution to accurately approximate the answer instead.
So under what circumstances can we use this, and how?
Imagine we have a variable X where X ~ B(n, p). We want to find
a set of circumstances where B(n, p) is similar to Po(λ).
Let’s start off by looking at the expectation and variance of the
two distributions. We want to find the circumstances in which the
expectation and variance of the Poisson distribution are like those
of the Binomial distribution. In other words, we want
λ np
λ
npq
to be like
to be like
np
npq
to be like
Expectation
Variance
np and npq are close to each other if q is close to 1 and n is large.
In other words:
X ~ B(n, p) can be approximated by X ~ Po(np)
if n is large and p is small
The approximation is typically very close if n is larger than 50, and p is less
than 0.1.
approximating with the poisson distribution
you are here 4 317
geometric, binomial, and poisson distributions
A student needs to take an exam, but hasn’t done any revision for it. He needs to guess the
answer to each question, and the probability of getting a question right is 0.05. There are 50
questions on the exam paper. What’s the probability he’ll get 5 questions right? Use the Poisson
approximation to the binomial distribution to find out.
Q:
Why would I ever want to use the
Poisson distribution to approximate the
binomial distribution?
A: When n is very large, it can be difficult
to calculate
n
C
r
. Some calculators run out
of memory, and the results can be so large
they’re just unwieldy. Using the Poisson
distribution in this way is a way round this
sort of problem.
Q:
So when can I use this
approximation?
A: You can use it when n is large (say
over 50) and p is small (say less than
0.1). When this is the case, the binomial
distribution and the Poisson distribution are
approximately the same.
Q:
Why do we use np as the
parameter for the Poisson distribution?
A: The Poisson distribution takes one
parameter, λ, and E(X) = λ. This means that
if we have use the Poisson approximation
of the binomial distribution, we can
substitute in the expectation of the binomial
distribution, np.
318 Chapter 7
Solved: The Case of the Broken Cookies
How did Kate find the probability so quickly, and avoid
the Out of Memory error on her calculator?
Kate spotted that even though she needed to use the
binomial distribution, her values of n and p were
such that she could approximate the probability using
the Poisson distribution instead.
A lot of calculators can’t cope with high factorials,
and this can sometimes make the binomial distribution
unwieldy. Knowing how to approximate it with the Poisson distribution
can sometimes save you quite a bit of time.
Five Minute
Mystery
Solved
Five Minute
Mystery
Solved
A student needs to take an exam, but hasn’t done any revision for it. He needs to guess the
answer to each question, and the probability of getting a question right is 0.05. There are 50
questions on the exam paper. What’s the probability he’ll get 5 questions right? Use the Poisson
approximation to the binomial distribution to find out.
Let’s use X to represent the number of questions the student gets right. In this problem, n = 50
and p = 0.05, np = 2.5. This means we can use X ~ Po(2.5) to approximate the probability.
P(X = 5) = e
-
λ
λ
r
r!
= e
-2.5
x 2.5
5
5!
= e
-2.5
x 97.65625
120
= e
-2.5
x 0.8138
= 0.067
exercise and mystery solutions
you are here 4 319
geometric, binomial, and poisson distributions
Anyone for popcorn?
You’ve covered a lot of ground in this chapter. You’ve built on your
existing knowledge of probability and statistics by tackling three of the
most important discrete probability distributions. Moreover, you’ve
gained a deeper understanding of how probability distributions work
and the sort of shortcuts you can make to save yourself time and
produce reliable results, skills that will come in useful in the rest of the
book.
So sit back and enjoy the popcorn — you’ve earned it.
Your quick guide to the Poisson distribution
Here’s a quick summary of everything you could possibly need to know about the Poisson distribution
When do I use it?
Use the Poisson distribution if you have independent events such as malfunctions occurring in a given interval,
and you know λ, the mean number of occurrences in a given interval. You’re interested in the number of
occurrences in one particular interval.
How do I calculate probabilities, and the expectation and variance?
Use
How do I combine independent random variables?
If X ~ Po(λ
x
) and Y ~ Po(λ
y
), then
E(X) = λ Var(X) = λ
P(X = r) = e
-λ
λ
r
r!
X + Y ~ Po(λ
x
+ λ
y
)
What connection does it have to the binomial distribution?
If X ~ B(n, p), where n is large and p is small, then X can be approximated using
X ~ Po(np)
320 Chapter 7
Here are some scenarios. Your job is to say which distribution each of them follows, say what
the expectation and variance are, and find any required probabilities.
1. A man is bowling. The probability of him knocking all the pins over is 0.3. If he has 10 shots, what’s the probability he’ll
knock all the pins over less than three times?
long exercise