Griffiths D. Head First Statistics
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you are here 4 401
What’s the probability of you winning the jackpot on today’s edition of Who Wants to Win a
Swivel Chair? See if you can find the probability of getting at least 30 questions correct out of 40,
where each question has a choice of 4 possible answers.
402 WHO WANTS TO WIN A SWIVEL CHAIR
What’s the probability of you winning the jackpot on today’s edition of Who Wants to Win a
Swivel Chair? See if you can find the probability of getting at least 30 questions correct out of 40,
where each question has a choice of 4 possible answers.
If X is the number of questions we get right, then we want to find P(X > 30) where X ~ B(40, 0.25).
As np and nq are both greater than 5, it’s appropriate for us to use the normal distribution to approximate
this probability. np = 10 and npq = 30, which means we need to find P(X > 29.5) where X ~ N(10, 30).
Let’s start by finding the standard score.
z = x - μ
= 29.5 - 10
30
= 19.5
30
= 0.65
Looking up 0.65 in probability tables gives us a probability of 0.7422. This means that
P(X > 29.5) = 1 - 0.7422
= 0.2578
So, looks like
you’ve only got about a
26% chance at that swivel chair.
If you lose, you’ll miss out on our
great consolation prize. Why don’t
you take the prize and run?
σ
you are here 4 403
Sorry to see you go. It’s been great
having you back as a contestant on the show,
but we’ve just had an urgent email from
someone called Dexter...
Here are the first five questions for Round Two.
The questions are all about the game show host.
D: $4,000
B: $2,000
C: $3,000
3. On average, how much does he spend on suits each month
A: $1,000
D: Bird on a Wire
B: Curse of the Were-Rabbit
C: Mousehunt
2. What is the favorite film of his cat
A: A Fish Called Wanda
D: All the President’s Men
B: The Italian Job
C: Lawrence of Arabia
1. What is his favorite film
A: The Day of The Jackal
D: Four times a month
B: Twice a month
C: Three times a month
4. How often does he have his hair cut?
A: Once a month
D: www.starbuzzcoffee.com
B: www.gregs-list.net
C: www.you-cube.net
5. What is his favorite web site
A: www.fatdanscasino.com
404 Chapter 9
Head First: Hey, Normal, glad you could make it
on the show.
Normal: Thanks for inviting me, Head First.
Head First: Now, my first question is about your
name. Why are you called Normal?
Normal: It’s really because I’m so representative
of a lot of types of data. They have a probability
distribution that has a distinctive shape and a
smooth, bell-curved shape, and that’s me. I’m
something of an ideal.
Head First: Can you give me an example?
Normal: Sure. Imagine you have a baker’s shop that
sells loaves of bread. Now, each loaf of a particular
sort of bread should theoretically weigh about the
same, but in practice, the actual weight of each loaf
of bread will vary.
Head First: But surely they’ll all weigh about the
same?
Normal: More or less, but with variation. I model
that variation.
Head First: So why’s that so important?
Normal: Well, it means that you can use me to work
out probabilities. Say you want to find the probability
of a randomly chosen loaf of bread being below a
particular weight. That sounds like something that
could be quite difficult, but with me, it’s easy.
Head First: Easy? How do you mean?
Normal: With a lot of the other probability
distributions, there can be lots of complicated
calculations involved. With Binomial you have
factorials, and with Poisson you have to work with
exponentials. With me there’s none of that. Just look
me up in a table and away you go.
Head First: Surely it’s not quite as simple as that?
Normal: Well, you do have to convert me to a
standard score first, but that’s nothing, not in the
grand scheme of things.
Head First: So tell me, do you think you’re better
than the other probability distributions?
Normal: I wouldn’t say that I’m better as such,
but I’m a lot more flexible, and I’m useful in lots
of situations. I’m also a lot more robust. When
the numbers get high for Poisson and Binomial
distributions, they run into trouble. Mind you, I do
what I can to help out.
Head First: You do? How?
Normal: Well under certain circumstances both
Binomial and Poisson look like me. It’s uncanny;
they’re often stopped at parties by people asking
them if they’re Normal. I tell them to take it as a
compliment.
Head First: So how does that help?
Normal: Well, because they look like me, it means
that you can actually use my probability tables to
work out their probabilities. How cool is that? No
more late nights slaving over a calculator; just look it
up.
Head First: I’m afraid that’s all we’ve got time for
tonight. Normal, thanks for coming along, it’s been a
pleasure.
Normal: You’re welcome, Head First.
The Normal Distribution Exposed
This week’s interview:
Why Being Normal Isn’t Dull
interview with the normal distribution
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using the normal distribution ii
All aboard the Love Train
Remember Dexter’s Love Train? He’s started running trials of the ride, and everyone
who’s given it a trial run thinks it’s great. There’s just one problem: sometimes the ride
breaks down and causes delays, and delays cost money.
Dexter’s found some statistics on the Internet about the model of roller coaster he’s been
trying out, and according to one site, you can expect the ride to break down 40 times a
year.
40 times a year?! If the ride
breaks down on someone’s
wedding day, they’ll sue!
Given the huge profit the Love Train is bound to make, Dexter thinks that
it’s still worth going ahead with the ride if there’s a high probability of it
breaking down less than 52 times a year.
So how do we work out that probability?
What sort of probability distribution does this follow? How would
you work out the probability of the ride breaking down less than
52 times in a year?
406 Chapter 9
What sort of probability distribution does this follow? How would
you work out the probability of the ride breaking down less than
52 times in a year?
Situations where you’re dealing with things breaking down at a mean rate follow a Poisson distribution,
taking a parameter of the mean. If X represents the number of breakdowns in a year, then X ~ Po(40).
We need to find P(X < 52). To find this, we’d need to find each individual probability for all values of X
up to 52 .
Working out that probability
is gonna be tricky and time-
consuming. I wonder if we can take
a shortcut like we did with the
binomial.
Under certain circumstances, the shape of the
Poisson distribution resembles that of the normal.
The advantage of this is we can use standard normal probability tables
to work out whole ranges of probabilities. This means that we don’t
have to calculate lots of individual probabilities in order to find what
we need.
Approximating the Poisson distribution with the normal is very similar
to when you use the normal in place of the binomial. Once you have
the right set of circumstances, you take the Poisson mean and variance,
and use them as parameters in a normal distribution.
If X ~ Po(λ), this means that the corresponding normal
approximation is X ~ N(λ, λ). But when is this true?
It all comes down to the shape of the distribution.
sharpen solution
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using the normal distribution ii
We can use the normal distribution to approximate the Poisson whenever the
Poisson distribution adopts a shape that’s like the normal, but when does this
happen? Let’s take a look.
When λ is small...
When λ is small, the shape of the Poisson distribution is different from
that of the normal distribution. The shape isn’t symmetrical, and the
curve looks as though it’s “pulled” over to the right.
As the Poisson distribution doesn’t resemble the normal for small values
of λ, the normal distribution isn’t a suitable approximation for the
poisson distribution where λ is small.
When λ is large...
As λ gets larger, the shape of the Poisson distribution looks increasingly
like that of the normal distribution. The main part of the shape is
reasonably symmetrical, and it forms a smooth curve that’s just like the
one for the normal.
This means that as λ gets larger, the normal distribution can be used to
give a better and better approximation of it.
So how large is large enough?
We’ve seen that the Poisson resembles the normal distribution when λ is large, but how
big does it have to be before we can use the normal?
λ actually gets sufficiently large when λ is greater than 15. This means that if X ~ Po(λ)
and λ > 15, we can approximate this using X ~ N(λ, λ).
The shape of the
distribution is skewed,
so we can’t use a normal
approximation
Approximating the Poisson Distribution
If X ~ Po(λ) and λ > 15, you can use X ~ N (λ, λ) to
approximate it.
Vital StatisticsVital Statistics
When to approximate the binomial distribution with the normal
The shape here is
more like the normal,
so we can use the
normal distribution
to approximate it.
408 Chapter 9
The number of breakdowns on Dexter’s Love Train follows a Poisson distribution where λ = 40.
What’s the probability that there will be fewer than 52 breakdowns in the first year?
Hint: Use a normal approximation,
and remember your continuity
corrections.
approximation exercise
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using the normal distribution ii
It’s time to test your statistical knowledge. Complete the table below, saying what normal
distribution suits each situation, and what conditions there are.
Situation Distribution Condition
X + Y
X ~ N(μ
x
, σ
2
x
), Y ~ (μ
y
, σ
2
y
)
X + Y ~ N(μ
x
+ μ
y
, σ
2
x
+ σ
2
y
)
X, Y are independent
X - Y
X ~ N(μ
x
, σ
2
x
), Y ~ (μ
y
, σ
2
y
)
aX + b
X ~ N(μ, σ
2
)
X
1
+ X
2
+ ... + X
n
X ~ N(μ, σ
2
)
Normal approximation of X
X ~ B(n, p)
Normal approximation of X
X ~ Po(λ)
410 Chapter 9
The number of breakdowns on Dexter’s Love Train follows a Poisson distribution where λ = 40.
What’s the probability that there will be fewer than 52 breakdowns in the first year?
If we use X to represent the number of breakdowns in a year, then X ~ Po(40)
As λ is large, we can use the normal distribution to approximate this. In other words, we use
X ~ N(40, 40)
We need to find the probability that there are fewer than 52 breakdowns. As we’re approximating a discrete
probability distribution with a continuous one, we have to apply a continuity correction. We don’t want to
include 52, so we need to find P(X ≤ 51.5).
Before we can find the probability using standard normal tables, we need to calculate the standard score.
z = x - μ
σ
= 51.5 - 40
6.32
= 1.82 (to 2 decima
l places)
Looking this up in probability tables gives us 0.9656. This means that the probability of there being fewer
than 52 breakdowns in a year is 0.9656.
exercise solution