Answer
$9\frac{7}{32} \ mi,9\frac{5}{32} \ mi,\frac{1}{16} \ mi$
Work Step by Step
The upper limit is the measurement + tolerance, that is,
$$
9\frac{3}{16}+\frac{1}{32}=9\frac{7}{32} \ mi
.$$
The lower limit is the measurement - tolerance, that is,
$$
9\frac{3}{16}-\frac{1}{32}=9\frac{5}{32} \ mi
.$$
The tolerance interval = upper limit-lower limit, that is,
$$
9\frac{7}{32} \ mi-9\frac{5}{32} \ mi=\frac{2}{32} \ mi=\frac{1}{16} \ mi
.$$
