Answer to Question #128088 in Statistics and Probability for Sourabh pol

Question

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

Given : mean = 605

Sd= 185

a. Find the probability that the rating will fall between 500 and 700 points.

P($500 \leq x\leq700)$ =

= P($(\frac{500 - mean}{sd}\leq\frac{X-mean}{sd}\leq\frac{700- mean}{sd})$

=$P(\frac{500 - 605}{185}\leq Z\leq\frac{700- 605}{185})$

=P($−0.5676≤Z≤0.5135)$

=$P(Z≤0.5135)−Pr(Z≤−0.5676)$

=0.6962−0.2852=0.411

b. Find the probability that the rating will fall between 400 and 500 points.

P( $400\leq X \leq 500)$ = P$(\frac{400 - mean}{sd}\leq\frac{X-mean}{sd}\leq\frac{500- mean}{sd})$

= P($(\frac{400 - 605}{185}\leq z\leq\frac{500- 605}{185})$

$=P(Z≤−0.5676)−P(Z≤−1.1081)$

=$P(−1.1081≤Z≤−0.5676)$ =0.2852−0.1339=0.1513

c. Find the probability that the rating will be less than 850 points.

P(X<850)= $P(\frac{X-mean}{sd}< \frac{850-mean}{sd})$

=$P(Z< \frac{850-605}{185})$

$=P(Z<1.3243)$ =0.9073

d. Find the probability that the rating will exceed 1,000 points.

P(Z>1000)= $P(\frac{X-mean}{sd} > \frac{1000- mean}{sd})$

=$=P(Z>2.1351)$ $=P(Z> \frac{1000-650}{185})$ =1−0.9836=0.0164

e. Find the 10th percentile.

In terms of the information that is provided, we have to compute the 10-th percentile.

First, we need to find the z-score associated to this percentile. we need to find $z_{p}$

​that solves the equation below.

$P(Z <z_p ) = 0.1$

The value of $z_p$ that solves the equation above cannot be made directly, it is solved either by looking at a standard normal distribution table .Based on this, we find that that the solution is $z_p = -1.282$ , because from the normal table we see that $P(Z<−1.282)=0.1$

Therefore, the percentile we are looking for is computed using the following formula:

P10= mean+($Z_p*sd)$

=605+(-1.282*185)

= 367.913

f. Find the 95th percentile. in the same way we need to find $Z_p$ such that $P(Z <z_p ) = 0.95$

Based on this, we find that that the solution is $z_p = 1.645$

because from the normal table we see that $P(Z<1.645)=0.95$

Therefore, the percentile we are looking for is computed using the following formula:

P95 = $mean+(Z_p*sd)$

=605+(1.645*185)

=909.298