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

Accepted Solution

Answer:[tex]V(X) = 2.45\ pts[/tex]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: [tex]V(X) = \sum{XP(X)}[/tex]Where P(X) is the probability that X will occur. In the sample space of the random variable X there are two possible values. [tex]X = 7[/tex] points (1 touchdown) with [tex]P(7) = 0.35[/tex] [tex]X = 0[/tex] points (0 touchdown) with [tex]P(0) = 1-0.35 = 0.65[/tex]Then the expected value V(X) is: [tex]V(X) = 7P(7) + 0P(0)[/tex][tex]V(X) = 7(0.35) + 0(0.65)[/tex] [tex]V(X) = 2.45\ pts[/tex]