Answer
$6\frac{21}{32} \ in, 6\frac{19}{32} \ in ,\frac{1}{16} \ in$
Work Step by Step
The upper limit is the measurement + tolerance, that is,
$$
6\frac{5}{8}+\frac{1}{32}=6\frac{21}{32} \ in
.$$
The lower limit is the measurement - tolerance, that is,
$$
6\frac{5}{8}-\frac{1}{32}=6\frac{19}{32} \ in
.$$
The tolerance interval = upper limit-lower limit, that is,
$$
6\frac{21}{32} \ in-6\frac{19}{32} \ in=\frac{2}{32} \ in=\frac{1}{16} \ in
.$$
