Question

As of the end of the 2010 NFL season, Indianapolis Colts quarterback Peyton Manning, throughout his 13-year career, completed 65% of all of his pass attempts. Suppose the probability each pass attempted in the next season has probability 0.65 of being completed. a. Does this mean that if we watch Manning throw 100 times in the upcoming season, he would complete exactly 65 passes? Explain.

Answer 1

Answer:

This does not mean that he would complete exactly 65 passes.

The probability that he completes exactly 65 passes is only 8.34%

65% is the number of passes he would be expected to complete in a very large sample(like 13 NFL seasons, not just a single season).

Step-by-step explanation:

There is not a 100% probability that he completes exactly 65 passes. So, it does not mean that he would complete exactly 65 passes.

How I arrived at this answer?

For each pass that he throws, either it is complete, or it is not. Since there are only two outcomes, we have a binomial probability distribution.

Binomial probability distribution

The binomial probability is the probability of exactly x successes on n repeated trials, and X can only have two outcomes.

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

In which $C_{n,x}$ is the number of different combinatios of x objects from a set of n elements, given by the following formula.

$C_{n,x} = \frac{n!}{x!(n-x)!}$

And $\pi$ is the probability of X happening.

In this problem, we have that:

There are 100 passes, so $n = 100$

Each pass a 65% probability of being completed, so $\pi = 0.65$

So

$P(X = x) = C_{n,x}.\pi^{x}.(1-\pi)^{n-x}$

$P(X = 65) = C_{100,65}.(0.65)^{65}.(0.35)^{35} = 0.0834$