Answer:
a) P(27<x<57)=0.68
b) P(X>57)=0.18
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.
![z_1=(X_1-\mu)/(\sigma)=(27-43)/(15)=(-16)/(15)=-1.0667 \\\\\\z_2=(X_2-\mu)/(\sigma)=(57-43)/(15)=(14)/(15)=0.9333](https://img.qammunity.org/2021/formulas/mathematics/college/zwbc7sp8d4kxd5vp0hlp8umdbay5iv4z1u.png)
![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](https://img.qammunity.org/2021/formulas/mathematics/college/jnoktlz2ihkm3h77m3c2ngsgst23pq1zyq.png)
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:
![P(x>57)=P(z>0.9333)=1-P(z<0.9333)=1-0.82467=0.17533](https://img.qammunity.org/2021/formulas/mathematics/college/2c7nlgaf0vbi93tp5doerm4oigba6diodv.png)