Answer
The shortest braking distance of a sedan that can be in the top 10% is $131.88$ feet
Work Step by Step
A braking distance in the top 10% is any score above the 90th percentile. Thus, we have to find the z score that corresponds to the cumulative area 0.90. Using the standard normal table or technology we find that a z score of 1.28 corresponds to an area of 0.90.
To find the corresponding $x$ value:
z = $\frac{x - \mu}{\sigma}$
$x$ = $\mu$ + $z\sigma$
$x$ = $127 + (3.81 \times 1.28)$
$x$ = $131.88$ feet
