Answer:
a) P(27<x<57)=0.68
b) P(X>57)=0.18
Step-by-step explanation:
We have a sample of n=5000 accidents, where the sample mean is 43 mph and the sample standard deviation is 15 mph, with a distribution shape approximately normal.
a) We have to calculate what proportion of vehicle speeds were between 27 and 57 mph.
If the distribution is approximately normal, we can calculate a z-score and then the probability.
[tex]z_1=\dfrac{X_1-\mu}{\sigma}=\dfrac{27-43}{15}=\dfrac{-16}{15}=-1.0667 \\\\\\z_2=\dfrac{X_2-\mu}{\sigma}=\dfrac{57-43}{15}=\dfrac{14}{15}=0.9333[/tex]
[tex]P(27<X<57)=P(-1.0667<z<0.9333)\\\\P(27<X<57)=P(z<0.9333)-P(z<-1.0667)\\\\P(27<X<57)=0.82467-0.14305=0.68162[/tex]
b) We have to calculate what proportion of vehicle speeds were 57 mph or more.
We use the same z=0.9333 for X=57, so we can calculate the probability as:
[tex]P(x>57)=P(z>0.9333)=1-P(z<0.9333)=1-0.82467=0.17533[/tex]