Optimal goal target in soccer. When attempting to score a

Step 1 of 3

(a)

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 by 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 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.

we are asked to find the probability that he will score if the player aims for the right goalpost.

If the goalkeeper stands in the middle of the goal and can any ball within 9 feet then the only way a player can score if and only if he shoots the ball within the 3 feet of either goal post.

If a player aims at the right goal post, then the player will score if x is between -3 and 0.

Hence we need to find the \(P(-3<x<0)\).

The probabilities for all normal distributions are computed by using the standard normal distribution. we can use the standard normal probability table and the appropriate z values to find the desired probabilities.

The formula used to convert any normal random variable x with mean ? and standard deviation ? to the standard normal random variable z follows.

\(z=\frac{x-\mu}{\sigma}\)

We can rewrite \(P(-3<x<0)\) as,

\(P(-3<x<0)=P(x \leq 0)-P(x \leq-3)\)

Therefore, the z - score for x = 0 is,

\(z=\frac{0-0}{3}=0\)

[mean \(\mu=0\), and standard deviation \(\sigma=3\)]

Hence the area to the left of z = 0 is 0.50.

[Please refer the standard normal table II in Appendix D]

\(P(x \leq 0)=0.50\)

the z - score for x = -3 is,

\(z=\frac{-3-0}{3}=-1\)

Hence the area to the left of z = -1 is 0.1587

\(\begin{array}{c}
P(x \leq-3)=0.1587 \\
P(-3<x<0)=0.50-0.1587=0.3413
\end{array}\)

Hence the probability that he will score  if the player aims for the right goalpost is 0.3413