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.

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* (90)/(100)+0* (10)/(100)\\\\E(x)=3* 0.9+0* 0.10\\\\E(x)=2.7`

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