What is the first question to ask when deciding between a binomial and a normal model?

Whether the variable counts or measures . A variable that counts is discrete , and counting successes is what a binomial distribution does; a variable that measures is continuous , such as a mass, a time or a length, and that points to a normal distribution. That settles which family you are in, but the specific conditions for each model then have to be checked separately, because being a count does not by itself make something binomial.

You have real data rather than a description. How can you tell whether a normal model is reasonable?

Draw a histogram of the data and look at its shape: if the outline is roughly symmetrical and bell-shaped , a normal model is reasonable. If the variable really is normally distributed, then the more data you collect the smoother that outline should become, settling towards the shape of a normal distribution curve. A histogram that is clearly lopsided, or that has two humps, is evidence against the model, whatever the context suggests.

True or False? A binomial variable can only take whole-number values, but the normal distribution approximating it can take any value.

True. That is the sense in which it is only an approximation : a discrete set of separate probabilities, one for each whole number, is being replaced by a smooth curve that is defined everywhere in between them. It is also why a normal approximation is never exact, however well the two conditions are met.

A LevelMathsAQAStatisticsFlashcardsStatistical DistributionsWorking with Distributions

Working with Distributions (AQA A Level Maths: Statistics): Flashcards

Exam code: 7357

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FrontModelling with Distributions
What is the first question to ask when deciding between a binomial and a normal model?
BackModelling with Distributions
Whether the variable counts or measures.
A variable that counts is discrete, and counting successes is what a binomial distribution does; a variable that measures is continuous, such as a mass, a time or a length, and that points to a normal distribution.
That settles which family you are in, but the specific conditions for each model then have to be checked separately, because being a count does not by itself make something binomial.
BackModelling with Distributions
Draw a histogram of the data and look at its shape: if the outline is roughly symmetrical and bell-shaped, a normal model is reasonable.
If the variable really is normally distributed, then the more data you collect the smoother that outline should become, settling towards the shape of a normal distribution curve.
A histogram that is clearly lopsided, or that has two humps, is evidence against the model, whatever the context suggests.
BackModelling with Distributions
With two distributions, one feeding the other, starting with the normal distribution for the probability that a single cow is beefy:
$p = P (M > 700) = 0.030396 \dots$
That probability then becomes the $p$ of a binomial distribution for the sample, because each of the 10 cows either is beefy or is not:
$X ~ B (10, 0.030396 \dots), P (X \leq 1) = 0.965$
Carry plenty of decimal places in $p$ : rounding it to 0.03 before the second stage shifts the final answer.

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What is the first question to ask when deciding between a binomial and a normal model?
Whether the variable counts or measures.
A variable that counts is discrete, and counting successes is what a binomial distribution does; a variable that measures is continuous, such as a mass, a time or a length, and that points to a normal distribution.
That settles which family you are in, but the specific conditions for each model then have to be checked separately, because being a count does not by itself make something binomial.
You have real data rather than a description. How can you tell whether a normal model is reasonable?
Draw a histogram of the data and look at its shape: if the outline is roughly symmetrical and bell-shaped, a normal model is reasonable.
If the variable really is normally distributed, then the more data you collect the smoother that outline should become, settling towards the shape of a normal distribution curve.
A histogram that is clearly lopsided, or that has two humps, is evidence against the model, whatever the context suggests.
Cow masses are modelled by $N (550, 80^{2})$ , and a cow is called beefy if it weighs more than 700 kg. A random sample of 10 cows is taken. How do you find the probability that at most one is beefy?
With two distributions, one feeding the other, starting with the normal distribution for the probability that a single cow is beefy:
$p = P (M > 700) = 0.030396 \dots$
That probability then becomes the $p$ of a binomial distribution for the sample, because each of the 10 cows either is beefy or is not:
$X ~ B (10, 0.030396 \dots), P (X \leq 1) = 0.965$
Carry plenty of decimal places in $p$ : rounding it to 0.03 before the second stage shifts the final answer.
In a question that uses two distributions, what must you write down before calculating anything?
Exactly what each variable and each parameter stands for, in words, and then its distribution.
So, for example:
- let $M$ be the mass of a cow, with $M ~ N (550, 80^{2})$
- let $X$ be the number of beefy cows in the sample, with $X ~ B (10, p)$ , where $p$ is the probability that a cow is beefy
Without that the two variables are easy to confuse, and the 10 in the binomial gets mixed up with quantities from the normal distribution; saying what $p$ means also tells you it must be calculated rather than read off.
Complete the two conditions under which a binomial distribution can be approximated by a normal distribution:
$n$ must be $______$
$p$ must be close to $______$
The completed conditions are:
$n$ must be large
$p$ must be close to 0.5
Both are needed, and they do different jobs, so a binomial meeting only one of them should not be approximated this way.
Why must $n$ be large before a normal approximation is reasonable?
Because a binomial with only a few trials has only a few possible values, so its outline is too coarse for a smooth curve to follow.
As $n$ grows the possible values become more numerous and each individual probability smaller, and the outline of the distribution settles towards a smooth bell shape.
Why must $p$ be close to 0.5 before a normal approximation is reasonable?
Because a normal distribution is symmetrical, and a binomial distribution is only roughly symmetrical when $p$ is near 0.5.
With $p$ near 0 the binomial has a long tail to the right, and with $p$ near 1 a long tail to the left, so a symmetrical curve laid over either of them would fit badly at both ends.
At $p = 0.5$ the binomial is exactly symmetrical, which is the value the condition is measured against.
A binomial $B (n, p)$ is approximated by a normal distribution. What are its mean and variance?
They are the mean and variance of the binomial itself: $μ = n p$ and $σ^{2} = n p (1 - p)$ .
So $B (1250, 0.4)$ is approximated by $N (500, 300)$ , since $1250 \times 0.4 = 500$ and $1250 \times 0.4 \times 0.6 = 300$ .
Matching those two quantities is what ties the approximation to that particular binomial rather than to any other.
True or False?
A binomial variable can only take whole-number values, but the normal distribution approximating it can take any value.
True.
That is the sense in which it is only an approximation: a discrete set of separate probabilities, one for each whole number, is being replaced by a smooth curve that is defined everywhere in between them.
It is also why a normal approximation is never exact, however well the two conditions are met.

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