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The accuracy x of a professional soccer player鈥檚 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鈥� shots, which have normal distribution .

Mean$渭$ and standard deviation$蟽$ 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,

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

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鈥檚 score is the value of x greater than 9 or less than -9.

Hence

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

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,

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

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鈥檚 reach, the probability that he will score is0.383.