Optimal goal target in soccer. When attempting to score a goal in soccer, where should you aim your shot? Should you aim for a goalpost (as some soccer coaches teach), the middle of the goal, or some other target? To answer these questions, Chance (Fall 2009) utilized the normal probability distribution. Suppose the accuracy x of a professional soccer player’s shots follows a normal distribution with a mean of 0 feet and a standard deviation of 3 feet. (For example, if the player hits his target,x=0; if he misses his target 2 feet to the right, x=2; and if he misses 1 foot to the left,x=-1.) Now, a regulation soccer goal is 24 feet wide. Assume that a goalkeeper will stop (save) all shots within 9 feet of where he is standing; all other shots on goal will score. Consider a goalkeeper who stands in the middle of the goal.

a. If the player aims for the right goalpost, what is the probability that he will score?

b. If the player aims for the center of the goal, what is the probability that he will score?

c. If the player aims for halfway between the right goal post and the outer limit of the goalkeeper’s reach, what is the probability that he will score?

The accuracy x of a professional soccer player’s shots follows a normal distribution with a mean of 0 feet and a standard deviation of 3 feet.

Let x be the random variable defined as the accuracy of the professional soccer players’ shots, which have normal distribution .

Mean$\mu$ and standard deviation$\sigma$ are given 0 feet and 3 feet.

Also, if the goalkeeper is in the center of the goal and can stop all the balls within 9 feet the only way to score the ball is within 3 feet of either goal post.

Hence, if player aims for the right goalpost, the player will score is the value of x lies between -2 and 0

Therefore,

$$\begin{gathered} P\left(-3<x<0\right)=P\left(\frac{x-\mu }{\sigma }<z<\frac{x+\mu }{\sigma }\right) \\ =P\left(\frac{-3-0}{3}<z<\frac{0+0}{3}\right) \\ =P\left(-1<z<0\right) \\ =P\left(0<z<1\right) \end{gathered}$$

From the table 2,Normal Curve areas, we find that p =0.3413

Hence, the probability that the player will score whenhe aims for the right goalpost, is 0.3413.

When player aims for the center goal, then the player’s score is the value of x greater than 9 or less than -9.

Hence

$$\begin{gathered} P\left(x<-9\right)+P\left(x<-9\right)=P\left(\frac{x-\mu }{\sigma }<z<\frac{x+\mu }{\sigma }\right) \\ =P\left(z<\frac{-9-0}{3}\right)+P\left(z<\frac{9-0}{3}\right) \\ =P\left(z<-3\right)+P\left(z>3\right) \end{gathered}$$

From table 2, Normal Curve Areas, we find that

$\left(0.5-0.4987+0.5-0.4987\right)=0.0026$

Therefore, the probability that player will score when he aims for the center of the goal is 0.0026.

We have to find

$P\left(-1.5<x<1.5\right)$

Thus,

$$\begin{gathered} P\left(-1.5<x<1.5\right)=P\left(\frac{x-\mu }{\sigma }<z<\frac{x+\mu }{\sigma }\right) \\ =P\left(\frac{-1.5-0}{3}<x<\frac{1.5+0}{3}\right) \\ =P\left(-0.5<z<0.5\right) \end{gathered}$$

From table 2, Normal Curve Areas, we find that

$\left(0.1915+0.191\right)=0.0383$

Hence,If the player aims for halfway between the right goal post and the outer limit of the goalkeeper’s reach, the probability that he will score is0.383.