Answer to Question #127871 in Statistics and Probability for Sourabh

Question

CRASH 4.102 NHTSA crash safety tests. Refer to Exercise 4.21 (p. 195) and the NHTSA crash test data for new cars. One of the variables saved in the accompanying file is the severity of a driver's head injury when the car is in a head-on collision with a fixed barrier while traveling at 35 miles per hour. The more points assigned to the head-injury rating, the more severe the injury. The head-injury ratings can be shown to be approximately normally distributed with a mean of 605 points and a standard deviation of 185 points. One of the crash-tested cars is randomly selected from the data, and the driver's head-injury rating is observed. a. Find the probability that the rating will fall between 500 and 700 points. b. Find the probability that the rating will fall between 400 and 500 points. c. Find the probability that the rating will be less than 850 points. d. Find the probability that the rating will exceed 1,000 points. e. Find the 10th percentile. f. Find the 95th percentile. Please do it step by step

Answer 1

Let $X$ be the head-injury rating: $X\sim N(\mu, \sigma^2).$ Then $Z=\dfrac{X-\mu}{\sigma}\sim N(0, 1)$

Given $N=605, \sigma=185.$

a.

$$P(500<X<700)=P(700)-P(X\leq500)=$$

$$=P(Z<\dfrac{700-605}{185})-P(Z\leq\dfrac{500-605}{185})\approx$$

$$\approx P(Z<0.513514)-P(Z\leq-0.567568)\approx$$

$$\approx0.6962-0.2852=0.4110$$

b.

$$P(400<X<500)=P(500)-P(X\leq400)=$$

$$=P(Z<\dfrac{500-605}{185})-P(Z\leq\dfrac{400-605}{185})\approx$$

$$\approx P(Z<-0.567568)-P(Z\leq-1.108108)\approx$$

$$\approx0.2852-0.1339=0.1513$$

c.

$$P(X<850)=P(Z<\dfrac{850-605}{185})\approx$$

$$\approx P(Z<1.324324)\approx0.9073$$

d.

$$P(X>1000)=1-P(X\leq1000)=$$

$$=1-P(Z<\dfrac{1000-605}{185})\approx1-P(Z\leq2.135135)\approx$$

$$\approx1-0.983625\approx0.0164$$

e.

$$Pr(Z<z_p)=0.1$$

$$z_p\approx-1.28155$$

$$P_{10}=\mu+z_p\times \sigma$$

$$P_{10}=605+(-1.28155)\times185\approx368$$

f.

$$Pr(Z<z_p)=0.95$$

$$z_p\approx1.645$$

$$P_{95}=\mu+z_p\times \sigma$$

$$P_{95}=605+1.645\times185\approx909$$