Question

Suppose that 15% of all sparkplugs produced for a specific model of automobile will require a gap adjustment before they are installed in the engine.  We are about to perform a tune up with new plugs on a V6 engine (6 plugs needed):

What is the probability that during the install of the plugs that 2 of them need to be gapped?  You may assume that each plug was randomly selected (Not from the same run of production)

On average, how many of the spark plugs would we expect to need to be gapped before installing in the engine? (still V6)

What is the probability that we have to gap fewer plugs than we expect before installing them in the engine?

Answer 1

Given 15% of all sparkplugs require a gap adjustment before installation. p = 0.15

For a V6 engine, 6 plugs are needed. n = 6

Let X be the number of sparkplugs that require a gap adjustment before installation, then X is binomially distributed as:

X ~ B(6, 0.15)

What is the probability that during the install of the plugs that 2 of them need to be gapped?

It is simply P(X=2) = 6C2 (0.15)^2 ( 1-0.15)^4 = 0.176

On average, how many of the spark plugs would we expect to need to be gapped before installing in the engine?

It is just the expected value of X.

For a binomial distribution, expected value = np

So spark plugs would we expect to need to be gapped before installing = 6*0.15 = 0.9

But X can take only integer values, hence X = 1.

What is the probability that we have to gap fewer plugs than we expect before installing them in the engine?

Expected plugs = 1.

Fewer than 1 has the only case X=0

So P(X=0) = 6C0(1-0.15)^6 = 0.377