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.

Calculators work out binomial probabilities directly, so why is a normal approximation still worth having?

Because a normal distribution is easier to work with once you have it: the probability of a whole range of values comes out in a single step, where a binomial may need several cumulative probabilities combined. More importantly, there is an inverse normal function and no inverse binomial one on most calculators, so a question giving a probability and asking for the value behind it is far more tractable in the normal setting. The approximation is a convenience rather than a necessity, which is why the answer is only ever close rather than exact.

A LevelMathsEdexcelStatisticsFlashcardsStatistical DistributionsWorking with Distributions

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

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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.
When can a binomial distribution be approximated by a normal distribution?
When $n$ is large and $p$ is close to 0.5.
The condition on $p$ is about shape: a binomial with $p$ close to 0.5 is nearly symmetrical, so a normal curve sits neatly over its bars.
With $p$ far from 0.5 the binomial is skewed and a symmetrical curve cannot follow it, while a large $n$ gives the distribution enough separate values for a smooth curve to be a fair description of discrete bars.
Which normal distribution is used to approximate $X ~ B (n, p)$ , and where do its parameters come from?
The one with the same mean and variance as the binomial, since matching them is what makes the curve sit over the bars in the right place with the right width.
You already know those two results for a binomial distribution, so
$X_{N} ~ N (n p, n p (1 - p))$
So, for example, $X ~ B (1250, 0.4)$ gives $μ = 500$ and $σ^{2} = 300$ ; the second parameter is the variance, so enter $σ = \sqrt{300}$ into a calculator, not 300.
Define continuity correction.
A continuity correction is an adjustment made to the boundaries of an inequality when a discrete distribution is approximated by a continuous one.
A binomial variable takes only whole numbers, while a normal variable takes every value in between, so each whole number $k$ is replaced by the interval from $k - 0.5$ to $k + 0.5$ : every value in that interval rounds to $k$ .
Without the correction the approximation systematically misses part of the probability at each end.
How do you decide whether to add or subtract the 0.5 at a boundary?
Ask whether the boundary value itself is included in the inequality, then move the boundary so that the interval still contains what it should.
- If $k$ is included, move the boundary outwards to take it in: $P (X \leq k)$ becomes $P (X_{N} < k + 0.5)$ , and $P (X \geq k)$ becomes $P (X_{N} > k - 0.5)$
- If $k$ is not included, move the boundary inwards to leave it out: $P (X < k)$ becomes $P (X_{N} < k - 0.5)$ , and $P (X > k)$ becomes $P (X_{N} > k + 0.5)$
Working it out this way is safer than memorising four lines, and it handles a two-sided inequality by treating each end separately.
$X ~ B (1250, 0.4)$ is approximated by a normal distribution in order to find $P (485 \leq X \leq 530)$ . Complete the corrected inequality:
$P (______\leq X_{N} \leq______)$
The completed inequality is:
$P (484.5 \leq X_{N} \leq 530.5)$
Both 485 and 530 are included by the original inequality, so both boundaries move outwards, one down and one up.
They move in opposite directions only because both endpoints are included; if one were excluded they would move the same way, so $P (a < X \leq b)$ becomes $P (a + 0.5 < X_{N} < b + 0.5)$ .
The safe method is to take each end separately: decide whether that endpoint is included, then move the boundary so the interval still holds what it should.
Why is $P (X = k)$ approximated by an interval rather than by a single value?
Because a normal distribution gives zero probability to any single value, so $P (X_{N} = k)$ would come out as 0 whatever $k$ was.
The probability has to be carried by an area, so the single binomial value $k$ becomes
$P (X = k) \approx P (k - 0.5 < X_{N} < k + 0.5)$
Since the endpoints contribute nothing to a continuous distribution, it makes no difference whether those inequalities are strict.
Calculators work out binomial probabilities directly, so why is a normal approximation still worth having?
Because a normal distribution is easier to work with once you have it: the probability of a whole range of values comes out in a single step, where a binomial may need several cumulative probabilities combined.
More importantly, there is an inverse normal function and no inverse binomial one on most calculators, so a question giving a probability and asking for the value behind it is far more tractable in the normal setting.
The approximation is a convenience rather than a necessity, which is why the answer is only ever close rather than exact.

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