Question

if a football team has a 90% chance of making a field goal for three points and 35% chance of making a touchdown for seven point find the expected value of a touchdown

Answer 1

Answer:

$V(X) = 2.45\ pts$

Step-by-step explanation:

We look for the expected value for a touchdown.

If X represents the number of points scored per touchdown scored, then X is a discrete random variable, and by definition, the expected value V (X) for a discrete random variable is defined as:

$V(X) = \sum{XP(X)}$

Where P(X) is the probability that X will occur.

In the sample space of the random variable X there are two possible values.

$X = 7$ points (1 touchdown) with $P(7) = 0.35$

$X = 0$ points (0 touchdown) with $P(0) = 1-0.35 = 0.65$

Then the expected value V(X) is:

$V(X) = 7P(7) + 0P(0)$

$V(X) = 7(0.35) + 0(0.65)$

$V(X) = 2.45\ pts$