Answer: 11.9%
Explanation:
Given : On a stretch of Interstate-89, car speed is a normally distributed variable with
mph and
mph.
Let x be a random variable that represents the car speed.
Since ,
![z=(x-\mu)/(\sigma)](https://img.qammunity.org/2020/formulas/mathematics/high-school/10fia1p0qwvlz4zhb867kzy3u7bscognwz.png)
z-score corresponds x = 73 ,
![z=(73-69.1)/(3.3)\approx1.18](https://img.qammunity.org/2020/formulas/mathematics/college/926n36pubgdt8kgt91mpovsad8mqlz5k9i.png)
Required probability :
![\text{P-value }: P(x>73)=P(z>1.18)\\\\=1-P(z<1.18)\\\\1-0.8809999\\\\=0.1190001=11.90001\approx11.9\%](https://img.qammunity.org/2020/formulas/mathematics/college/j0nz9tq50vzadkys552u6nqbglixzadv08.png)
[using z-value table.]
Hence, the approximate percentage of cars are traveling faster than you = 11.9%