Normal approximation to binomial

Question

What do I do when the normal approximation is not valid?

Here's the question I'm trying to answer:

A student guesses all 15 answers on a multiple-choice test. There are 5 choices for each of the questions. Find the probability (correct to 6 decimal places) that the student passes the test.

thanks

Answer 1

What constitutes "valid" depends on many things (though I bet you've been given one blanket 'rule' which in many cases will be too strict and in some cases too weak, depending on your needs). In this case the requirement of 6dp accuracy preclude using approximations unless the sample sizes get very large indeed.

So obviously the idea is that one should apply the binomial directly.

You'd need to make some assumptions - which are necessary to apply the binomial.

Then you simply compute the actual binomial probabilities in question.

http://en.wikipedia.org/wiki/Binomial_distribution

That is, there's no need to try to apply the normal approximation itself, you can just compute the required probability. Which marks on the test correspond to passing? What is the probability of getting a result among those values?

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Some comments about the form of the question (not the OP's question, the one set for the OP to do):

Seriously - who needs probability to 6dp accuracy?

Even if you did need it, when are the assumptions (like independence, constant probability of success) ever satisfied closely enough to actually get that level of accuracy?

We're talking about an extreme tail probability. Even small amounts of dependence or heterogeneity of probability and the calculation could be out by orders of magnitude (i.e. not even close to one significant figure of accuracy)

This sort of question implies a ludicrously false level of accuracy to our models.

On a real-world problem of this kind we'd usually be lucky to get more than a couple of figures of accuracy, because we don't really have independence, etc. The shadow of George Box's famous maxim (in abbreviated form, 'all models are wrong; some are useful') is ever present.

Answer 2

A tricky question requires a trick to answer it.

Let's say a passing grade equals 60%, that is 9 or more correct answers are needed to pass. The program that I have for the "Binomial Probability Sums" gives the Probability of getting r or less, also true for all the tables that I've seen, so a little algebra is needed.
Since $P(\leq 8)+P(X\geq 9)=1$
Then $P(X\geq 9)=1-P(X\leq 8)$. This is the trick. And our problem becomes to find $1-P(X\leq 8)$ for $\mathrm{Binomial}(x;15,.2)$
For $P(X\leq 8)$, I get $0.999215$
And $P(X\geq 9)=1-0.999215=0.000785$ This is the accurate answer
So less than 4 out of 5000 chances to pass the test on luck alone!

Now if the solution requires the use of the normal approximation then let's look at a table for "Area under the Normal Curve". One with a culmulative area of 0 at the top left and increases to 0.5 at the center and ends up at 1 at the bottom right. For this Normal Approximation to the Binomial problem, the x-value goes from 0 to 15 correct test answers. And because a continuity correction is needed, the culmulative area increments at x-values of 0.5, 1.5, 2.5, etc. Since tables give the cumulative area for the Standard Normal Curve, x-values have to be transformed to z-values. So, we need :
$x=8.5$ (With continuity correction)
$u=p*n=.2*15=3$ Expected x-value, occurs at curve peak
$s=\sqrt{(n\cdot p\cdot (1-p))}=\sqrt{(15\cdot .2\cdot .8)}=1.5491933$ Deviation for this normal curve-

$z=(x-u)/s=(8.5-3)/1.5491933=3.5502348$ Transformed x-value
And our problem to find $1-P(X\leq8)$ for $\mathcal{N}(x;3,1.5491933)$ The trick again
Transforms to $1-P(Z\leq3.5502348)$ for $\mathcal{N}(z;0,1)$
Now $P(X\leq 8)=P(Z\leq 3.5502348)=0.999808$ By computer program
And $P(X\leq 9)=P(Z > 3.5502348)=1-0.999808=0.000192$ This is the approximate answer.
So, less than 1 out of 5000 chances to pass the test on luck alone!

I guess you could say they agree that there is not much chance chance of passing the test in 1000 tries on luck alone!
Now, if in the unlikely event, the passing grade for this test had been 81% or more , the probability of passing on luck alone by both methods is the same: $0.000000$.