91Ó°ÊÓ

Let X denotes the head injury rating.

X follows a normal distribution with the parameter $\mathrm{μ}$and ${\mathrm{σ}}^2$

$z=\frac{X-\mathrm{μ}}{\mathrm{σ}}$follows normal distribution with mean 0 and a variance 1

Also, given$mean=605SD=185$

We have to calculate the probability that the rating will fall between 500 and 70 points.

That is,$P\left(500<X<700\right)$

role="math" localid="1660714332516" $$\begin{gathered} P\left(500<X<700\right)=P\left(700\right)-P\left(X-≤500\right) \\ =P\left(Z<\frac{X-\mathrm{μ}}{\mathrm{σ}}\right) \\ =P\left(Z<\frac{700-605}{185}\right)-P\left(Z≤\frac{700-605}{185}\right) \\ P\left(500<X<700\right)≈P\left(Z<0.513\right)-P\left(Z≤-0.567\right) \\ P\left(500<X<700\right)≈0.6962-0.2852 \\ =0.411 \end{gathered}$$

Hence, the probability that rating will fall between 500 and 70 points. That is,$P\left(500<X<700\right)$ is 0.411.

For calculating the probability that rating will fall between 400 and 500,that is$P\left(400<X<500\right)$ .

$$\begin{gathered} P\left(400<X<500\right)=P\left(500\right)-P\left(X-≤400\right) \\ =P\left(Z<\frac{X-\mathrm{μ}}{\mathrm{σ}}\right) \\ =P\left(Z<\frac{500-605}{185}\right)-P\left(Z≤\frac{400-605}{185}\right) \\ P\left(400<X<500\right)≈P\left(Z<0.567\right)-P\left(Z≤-1.108\right) \\ ≈0.2852-0,1339 \\ =0.1513 \end{gathered}$$

Hence, the probability that rating will fall between 400 and 500,that is$P\left(400<X<500\right)$ .is 0.1513.

c.

We have to calculate the probability that the rating will be less than 850, that is,,

$P\left(X<850\right)$

Hence,

$$\begin{gathered} P\left(X<850\right)=P\left(Z<\frac{X-\mathrm{μ}}{\mathrm{σ}}\right) \\ P\left(z<\frac{850-605}{185}\right) \\ ≈P\left(z<1.3243\right) \\ P\left(X<850\right)≈0.9073 \end{gathered}$$

Hence, the probability that the rating will be less than 850, that is$P\left(X<850\right)$ is 0.9073.

We will find the probability that rating will exceed 1000 points that is$P\left(X>1000\right)$

$$\begin{gathered} P\left(X>1000\right)=1-P\left(X>1000\right) \\ =1-P\left(Z<\frac{X-\mathrm{μ}}{\mathrm{σ}}\right) \\ =1-P\left(Z<\frac{1000-605}{185}\right) \\ P\left(X>1000\right)≈1-P\left(Z>2.1351\right) \\ ≈1-0.9836 \\ ≈0.0164 \end{gathered}$$

Hence, the probability that rating will exceed 1000 points that is$P\left(X>1000\right)$ is 0.0164 .