Question

Suppose that the speed at which cars go on the freeway is normally distributed with mean 68 mph and standard deviation 5 miles per hour. Let X be the speed for a randomly selected car. Round all answers to two decimal places.
A. X ~ N( _____, _______ )
B. If one car is randomly chosen, find the probability that is traveling more than 70 mph.
C. If one of the cars is randomly chosen, find the probability that it is traveling between 65 and 75 mph.
D. 90% of all cars travel at least how fast on the freeway? ________ mph.

Answer 1

Answer:

A)$X \sim N(68 , 25)$

B) the probability that is traveling more than 70 mph is 0.3446

C) the probability that it is traveling between 65 and 75 mph is 0.6449

D) 90% of all cars travel at least 56.85 mph fast on the freeway

Step-by-step explanation:

The speed at which cars go on the freeway is normally distributed with mean 68 mph and standard deviation 5 miles per hour.

Mean = $\mu = 68 mph$

Standard deviation = $\sigma = 5 mph$

A) X ~ N( _____, _______ )

In general $X \sim N( \mu , \sigma^2)$

$\mu = 68 mph$

$\sigma = 5 mph$

$\sigma^2 = 5^2 = 25$

So, $X \sim N(68 , 25)$

B) If one car is randomly chosen, find the probability that is traveling more than 70 mph.i.e.P(X>70)

So,$Z = \frac{x-\mu}{\sigma}\\Z=\frac{70-68}{5}$

Z=0.4

Using Z table

P(Z>70)=1-P(Z<70)=1-0.6554=0.3446

Hence the probability that is traveling more than 70 mph is 0.3446

C) If one of the cars is randomly chosen, find the probability that it is traveling between 65 and 75 mph.

P(65<X<75)

$Z = \frac{x-\mu}{\sigma}$

AT x = 65

$Z=\frac{65-68}{5}$

Z=-0.6

AT x = 75

$Z=\frac{75-68}{5}$

Z=1.4

Using Z table

P(65<X<75)=P(-0.6<Z<1.4)=P(Z<1.4)-P(Z<-0.6)=0.9192-0.2743=0.6449

Hence the probability that it is traveling between 65 and 75 mph is 0.6449

D)90% of all cars travel at least how fast on the freeway?

Since we are supposed to find at least how fast on the freeway

So,P(X>x)=0.9

1-P(X<x)=0.9

1-0.9=P(X<x)

0.1=P(X<x)

Z value at 10% =-2.23

So, $Z=\frac{x-\mu}{\sigma}\\-2.23=\frac{x-68}{5}\\-2.23 \times 5 =x-68\\(-2.23 \times 5)+68=x$

56.85 = x

Hence 90% of all cars travel at least 56.85 mph fast on the freeway