Griffiths D. Head First Statistics

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geometric, binomial, and poisson distributions

Let’s see if we can find an expression for the variance of the

geometric distribution in the same way that we did for the

expectation. Complete the table below. What do you notice?

Expectation is 1/p

Drawing the chart for the running total of xP(X = x) shows you that as x gets

larger, the running total gets closer and closer to a particular value, 5. In fact, the

running total of xP(X = x) for an infinite number of trials is 5 itself. This means

that

E(X) = 5

This makes intuitive sense. The probability of a successful outcome is 0.2. This is a

bit like saying that 1 in 5 attempts tend to be successful, so we can expect Chad to

make 5 attempts before he is successful.

We can generalize this for any value p. If X ~ Geo(p) then

E(X) =

We’re not just limited to finding the expectation of the geometric distribution,

we can find the variance too.

Remember, the variance is

given by E(X

) - E

(X).

x P(X = x) x

P(X = x) x

P(X ≤ x)

1 0.2

2 0.8

0.2 = 0.16

3 0.8

0.2 = 0.128

4 0.8

0.2 = 0.1024

5 0.8

0.2 = 0.08192

6 0.8

0.2 = 0.065536

7 0.8

0.2 = 0.0524288

8 0.8

0.2 = 0.04194304

9 0.8

0.2 = 0.033554432

10 0.8

0.2 = 0.0268435456

The expectation is 1 divided by

the probability of success.

I can expect to

make it down in 5

tries? Not bad!

282 Chapter 7

Values start off small, get larger,

and then decrease again.

Let’s see if we can find an expression for the variance of the

geometric distribution in the same way that we did for the

expectation. Complete the table below. What do you notice?

x P(X = x) x

P(X = x) x

P(X ≤ x)

1 0.2

0.2 0.2

2 0.8

0.2 = 0.16

0.64 0.84

3 0.8

0.2 = 0.128

1.152 1.992

4 0.8

0.2 = 0.1024

1.6384 3.6304

5 0.8

0.2 = 0.08192

2.048 5.6784

6 0.8

0.2 = 0.065536

2.359296 8.037696

7 0.8

0.2 = 0.0524288

2.5690112 10.6067072

8 0.8

0.2 = 0.04194304

2.68435456 13.29106176

9 0.8

0.2 = 0.033554432

2.717908992 16.00897075

10 0.8

0.2 = 0.0268435456

2.68435456 18.69332531

This time x

P(X = x) increases until x reaches 10. When x reaches 10 it starts to go down again.

I get it, so x

P(X = x) gets larger

for a while, but after that, it gets

smaller and smaller as x gets larger

and larger.

That’s right.

P(X = x) gets larger and larger up until a certain point, and then it starts

decreasing again. Eventually it becomes very close to 0.

P(X = x)

sharpen solution

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geometric, binomial, and poisson distributions

Var(X) =

Finding the variance for our distribution

So how does this help us find the variance of the number of trials it takes

Chad to make a successful run down the slopes?

We find the variance of a probability distribution by calculating

This means that we calculate

P(X = x), and then subtract E(X) squared.

By graphing the resulting values against the values of x, you can see the

pattern of Var(X) as x increases. Here’s the graph of x

P(X x) - E

(X)

Var(X) = E(X

) – E

(X)

P(X ≤ x) - E

(X)

As x gets larger, the value of x

P(X x) – E

(X) gets closer and closer to a

particular value, this time 20.

As with the expectation, we can generalize this. If X ~ Geo(p) then

Even though there’s no fixed number of

trials, you can still work out what the

expectation and variance are.

284 Chapter 7

A quick guide to the geometric distribution

Here’s a quick summary of everything you could possibly need to know about the Geometric distribution

When do I use it?

Use the Geometric distribution if you’re running independent trials, each one can have a success or failure, and

you’re interested in how many trials are needed to get the first successful outcome

How do I calculate probabilities?

Use the following handy formulae. p is the probability of success in a trial, q = 1 - p, and X is the number of

trials needed in order to get the first successful outcome. We say X ~ Geo(p).

P(X = r) = p q

r - 1

P(X ≤ r) = 1 - q

P(X > r) = q

What about the expectation and variance?

Just use the following

E(X) = 1/p

Var(X) = q/p

Can I trust these formulae? Can

I use them any time I need to find

probabilities and expectations?

A: You can use these shortcuts whenever

you’re dealing with the geometric distribution,

as they’re shortcuts for that probability

distribution. If you’re dealing with a situation

that can’t be modelled by the geometric

distribution, don’t use these shortcuts.

Remember, the geometric distribution is

used for situations where you’re running

independent trials (so the probability stays

the same for each one), each trial ends in

either success or failure, and the thing you’re

interested in is how many trials are needed

to get the first successful outcome.

What about if my circumstances

are different? What if I have a fixed

number of trials and I want to find the

number of successful outcomes?

A: You can’t use the geometric

distribution to model this sort of situation, but

don’t worry, there are other methods.

Do I have to learn all of these

shortcuts?

A: If you have to deal with the geometric

distribution, knowing the formulae will

save you a lot of time. If you’re sitting for a

statistics exam, check whether your exam

syllabus covers it.

Why does the distribution use the

letters p and q?

A: The letter p stands for probability. In

this case, it’s the probability of getting a

successful outcome in one trial.

The letter q is often used in statistics to

represent 1 - p, or p

. You’ll see quite a lot

of it through the rest of this chapter and the

rest of the book.

The probability of the first

success being in the r’th trial

The probability you’ll need more than

r trials to get your first success

The probability you’ll need r trials

or less to get your first success

geometric distribution cheat sheet

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geometric, binomial, and poisson distributions

BE the snowboarder

The probability that another

snowboarder will make it down

the slope without falling over is

0.4. Your job is to play like you’re

the snowboarder and work out the

following probabilities for

your slope success.

1. The probability that you will be successful on your second attempt, while failing on your first.

2. The probability that you will be successful in 4 attempts or fewer.

3. The probability that you will need more than 4 attempts to be successful.

4. The number of attempts you expect you’ll need to make before being successful.

5. The variance of the number of attempts.

286 Chapter 7

BE the snowboarder solution

The probability that another

snowboarder will make it down

the slope without falling over is

0.4. Your job is to play like you’re

the snowboarder and work out the

following probabilities for

your slope success.

1. The probability that you will be successful on your second attempt, while failing on your first.

P(X = 2) = p x q

= 0.4 x 0.6

= 0.24

Let’s use X ~ Geo(0.4), where X

is the number of trials needed by

this second snowboarder to make a

clean run down the slope.

2. The probability that you will be successful in 4 attempts or fewer.

P(X ≤ 4) = 1 - q

= 1 - 0.6

= 1 - 0.1296

= 0.8704

3. The probability that you will need more than 4 attempts to be successful.

P(X > 4) = q

= 0.6

= 0.1296

4. The number of attempts you expect you’ll need to make before being successful.

E(X) = 1/p

= 1/0.4

= 2.5

4. The variance of the number of attempts.

Var(X) = q/p

= 0.6/0.4

= 0.6/0.16

= 3.75

Or you could have found this by

using P(X> 4) = 1 - P(X ≤ 4)

= 1 - 0.8704 = 0.1296

be the snowboarder solution

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geometric, binomial, and poisson distributions

You’ve mastered the geometric

distribution

Thanks to your skills with the geometric distribution, Chad not only

knows the probability of him making a clear run down the slopes after

any number of tries, but also how many times he can expect it to take to

get down the hill successfully, and how much variability there is.

With an expectation of 5 tries to make it down the slopes, and a

variance of 20, he feels much more confident he can impress the ladies

without serious bodily harm.

Now let’s move on to…

Ladies And GentleMen,

We Interrupt This Chapter

To Bring You

An Exciting Installment

of Statsville’s Favorite

Quiz Show:

Who Wants To Win A

Swivel Chair!

Let’s hit the slopes!

Hello, and welcome to Who Wants To

Win A Swivel Chair, Statville’s favorite

quiz show. We’ve got some fiendishly

difficult questions on the show tonight.

I hope you’re feeling lucky.

290 WHO WANTS TO WIN A SWIVEL CHAIR

D: Intelligence

B: Charm

C: Sense of Humor

3. What do people like most about him?

A: Good looks

D: April

B: February

C: March

2. In what month is his birthday?

A: January

Here are the questions for Round One. The

questions are all about the game show host.

Put a check mark next to the correct answer.

We’ve got some great questions for you today,

so let’s get started. In Round One I’m going to ask you

three questions, and for each question there are four possible

answers. You can quit now and walk away with the consolation

prize, but if you play on and beat your competitors, you’ll move

on to the next round and be one step closer to winning a swivel

chair. The title of Round One is “All About Me.” Good luck!

What’s a quiz show doing in the middle of my chapter? I

thought we were talking about probability distributions.

A: We still are. This situation is ideal for another sort of probability

distribution. Keep reading and everything will become clear.

I don’t know the answers to these questions. What should

I do?

A: If you don’t know the answers you’ll have to answer them at

random. Give it your best shot - you might win a swivel chair.

D: Yellow

B: Blue

C: Green

1. What’s his favorite color?

A: Red