What is the first question to ask when choosing a distribution to model something?

Whether the variable counts something or measures something. Counting points to a discrete distribution, and measuring to a continuous one, which settles half the question immediately. After that it is a matter of which counting situation, or whether the measured data is symmetrical and bell-shaped.

What distinguishes a geometric model from a binomial one?

A geometric variable counts the trials up to and including the first success , so the trials continue until that success happens. A binomial fixes the number of trials in advance and counts how many succeed. The three other conditions are the same for both: independent trials, two outcomes, and a constant probability of success.

True or False? A histogram of real data can tell you whether a normal model is reasonable.

True. If the histogram is roughly symmetrical and bell-shaped, a normal model is worth using. As more data is collected the outline should smooth out and come to resemble the normal curve itself, so a lopsided or two-peaked histogram is telling you to look elsewhere.

Why must you always state what your variables and parameters represent?

Because a question can involve more than one variable measuring quite different things, such as a mass and a count. Writing each one down in words is what stops a mass being fed into a distribution that counts, or a parameter from one model being used in another. It also forces you to decide what is actually being modelled before choosing anything.

Which of the two approximations needs a continuity correction, and why only that one?

Only the normal approximation to a Poisson, because it replaces a discrete variable with a continuous one. A Poisson approximation to a binomial needs none at all, since both distributions are discrete and whole numbers stay whole numbers. Applying a correction where none is needed is as much an error as leaving one out where it is.

You cannot compute a probability from the original distribution. What do you decide first?

Which distribution you are approximating from , since that limits what you can approximate to. A binomial can be approximated by either a Poisson or a normal distribution, depending on which conditions hold. A Poisson can only be approximated by a normal distribution.

A LevelMathsCambridge (CIE)Probability & Statistics 2FlashcardsStatistical DistributionsWorking with Distributions

Working with Distributions (Cambridge (CIE) A Level Maths: Probability & Statistics 2): Flashcards

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What is the first question to ask when choosing a distribution to model something?

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What is the first question to ask when choosing a distribution to model something?
Whether the variable counts something or measures something.
Counting points to a discrete distribution, and measuring to a continuous one, which settles half the question immediately.
After that it is a matter of which counting situation, or whether the measured data is symmetrical and bell-shaped.
Fill in the three conditions for a Poisson model:
the mean number of occurrences, $λ$ , is known and $______$
the events occur at $______$
the events occur singly and $______$
The completed conditions are:
the mean number of occurrences, $λ$ , is known and finite
the events occur at random
the events occur singly and independently
Independence is the one worth checking hardest, because it is the condition a real context most often breaks.
Both a binomial and a Poisson variable count something. What decides between them?
Whether there is a fixed number of trials.
A binomial counts successes among a fixed number $n$ of trials, so it can never exceed $n$ .
A Poisson counts occurrences in an interval of time or space with no fixed number of trials at all, so it has no upper limit.
What distinguishes a geometric model from a binomial one?
A geometric variable counts the trials up to and including the first success, so the trials continue until that success happens.
A binomial fixes the number of trials in advance and counts how many succeed.
The three other conditions are the same for both: independent trials, two outcomes, and a constant probability of success.
True or False?
A histogram of real data can tell you whether a normal model is reasonable.
True.
If the histogram is roughly symmetrical and bell-shaped, a normal model is worth using.
As more data is collected the outline should smooth out and come to resemble the normal curve itself, so a lopsided or two-peaked histogram is telling you to look elsewhere.
Why must you always state what your variables and parameters represent?
Because a question can involve more than one variable measuring quite different things, such as a mass and a count.
Writing each one down in words is what stops a mass being fed into a distribution that counts, or a parameter from one model being used in another.
It also forces you to decide what is actually being modelled before choosing anything.
Fill in the condition for approximating a Poisson distribution by a normal one:
$λ$ must be large, in practice greater than about $______$
The completed condition is:
$λ$ must be large, in practice greater than about 15
The larger $λ$ becomes, the more symmetrical the Poisson distribution is, and the better a bell-shaped curve fits it.
Below that the distribution still has a noticeable tail to the right, and a symmetrical curve fits it badly.
Which normal distribution approximates $X ~ Po (λ)$ ?
It is $N (λ, λ)$ : both the mean and the variance are $λ$ .
That is because a Poisson distribution has its mean and variance exactly equal, so matching them to the normal puts the same number in both places.
The standard deviation you then standardise with is $\sqrt{λ}$ , not $λ$ .
For $X ~ Po (40)$ , how do you approximate $P (X > 50)$ ?
$λ = 40$ is large, so use $X_{N} ~ N (40, 40)$ .
Apply the continuity correction, which turns $X > 50$ into $X_{N} > 50.5$ , then standardise with $\sqrt{40}$ :
$z = \frac{50.5 - 40}{\sqrt{40}} = 1.660$
That gives $1 - Φ (1.660) = 0.0485$ .
Which of the two approximations needs a continuity correction, and why only that one?
Only the normal approximation to a Poisson, because it replaces a discrete variable with a continuous one.
A Poisson approximation to a binomial needs none at all, since both distributions are discrete and whole numbers stay whole numbers.
Applying a correction where none is needed is as much an error as leaving one out where it is.
Fill in the two conditions for approximating a binomial distribution by a Poisson one:
$n$ is large, in practice greater than about $______$
$p$ is small, so that $n p$ is less than about $______$
The completed conditions are:
$n$ is large, in practice greater than about 50
$p$ is small, so that $n p$ is less than about 5
The Poisson distribution arises from the binomial as $n$ grows and $p$ shrinks with their product staying finite, which is why those two pull in opposite directions.
Notice this is the reverse of what a normal approximation to a binomial wants, where $n p$ has to be large.
What is $λ$ when a binomial distribution is approximated by a Poisson one?
It is $n p$ , the mean of the binomial.
The approximating distribution is matched to the original by its mean, and a Poisson has only that one parameter to fix.
So 3000 trials with probability $0.001$ each give $Po (3)$ .
You cannot compute a probability from the original distribution. What do you decide first?
Which distribution you are approximating from, since that limits what you can approximate to.
A binomial can be approximated by either a Poisson or a normal distribution, depending on which conditions hold.
A Poisson can only be approximated by a normal distribution.
True or False?
An approximation is always second best, since the exact distribution would give a better answer.
False.
The exact answer is better only if you can actually reach it, and a cumulative probability spanning dozens of values needs one term for each.
When $n$ or $λ$ is large enough, the approximation is not a compromise but the only practical route, and the conditions exist precisely to keep it accurate.
A binomial has $n = 3000$ and $p = 0.001$ . Which approximation fits, and why not the other one?
A Poisson approximation, with $λ = n p = 3$ , since $n$ is well over 50 and $n p$ is under 5.
A normal approximation would fail here: it needs $n p$ to be greater than 5, and 3 is not.
The two sets of conditions are close to being opposites, so checking $n p$ almost always settles which one is meant.

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