Question

if a football team has a 90% chance of making a field goal for 3 points and a 35% chance of making a touchdown for 7 points, find the expected value of a field goal.

Answer 1

The expected value of a set of outcomes can be found by taking the weighted sum of the outcomes, with the probabilities of the weights:

$E([X])= \sum_{i=1}^{n}x_{i}p_{i}$

$E(fieldgoal)=(x_{1}p_{1})+(x_{2}p_{2})$

$E(fieldgoal)=((3pts)(0.90))+((0pts)(0.10))$

$E(fieldgoal)=2.7pts+0pts$

$E(fieldgoal)=2.7pts$

Answer 2

Answer: The expected value of a field goal is 2.7.

Step-by-step explanation:

Since we have given that

Probability of making a field goal for 3 points = 90%

Probability of making a field goal for 7 points = 35%

So, Expected value of a field goal is given by

$E(x)=\sum xp(x)$

X= 3 points ( make a field goal) and P(3)= 0.90

and

X = 0 points ( make no field goal) and P(7)= 1-0.90=0.10

$E(x)=3\times \dfrac{90}{100}+0\times \dfrac{10}{100}\\\\E(x)=3\times 0.9+0\times 0.10\\\\E(x)=2.7$

Hence, the expected value of a field goal is 2.7.