Question

The speeds of vehicles on a highway with speed limit 100 km/h are normally distributed with mean 114 km/h and standard deviation 8 km/h. (Round your answers to two decimal places.) ((a) What is the probability that a randomly chosen vehicle is traveling at a legal speed?

_________%

(b) If police are instructed to ticket motorists driving 125 km/h or more, what percentage of motorist are targeted?


_______%

Answer 1

Answer:

a) $P(X<100)=P(\frac{X-\mu}{\sigma}<\frac{100-\mu}{\sigma})=P(Z<\frac{100-114}{8})=P(z<-1.75)$

And we can find this probabiliy using the normal standard table or excel and we got:

$P(z<-1.75)=0.04$

So then we expect about 4%

b) $P(X>125)=P(\frac{X-\mu}{\sigma}>\frac{125-\mu}{\sigma})=P(Z>\frac{125-114}{8})=P(z>1.375)$

And we can find this probabiliy using the complement rule and the normal standard table or excel and we got:

$P(z>1.375)=1-P(Z<1.375) = 1-0.915=0.085$

So then we expect about 8.5%

Step-by-step explanation:

Part a

Let X the random variable that represent the speeds of vehicles, and for this case we know the distribution for X is given by:

$X \sim N(114,8)$  

Where $\mu=114$ and $\sigma=8$

We want to find this probability  (legal speed)

$P(X<100)$

We can use the z score given by:

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

If we apply this formula to our probability we got this:

$P(X<100)=P(\frac{X-\mu}{\sigma}<\frac{100-\mu}{\sigma})=P(Z<\frac{100-114}{8})=P(z<-1.75)$

And we can find this probabiliy using the normal standard table or excel and we got:

$P(z<-1.75)=0.04$

So then we expect about 4%

Part b

$P(X>125)=P(\frac{X-\mu}{\sigma}>\frac{125-\mu}{\sigma})=P(Z>\frac{125-114}{8})=P(z>1.375)$

And we can find this probabiliy using the complement rule and the normal standard table or excel and we got:

$P(z>1.375)=1-P(Z<1.375) = 1-0.915=0.085$

So then we expect about 8.50%