Answer
$7\frac{15}{32} \ in,7\frac{13}{32} \ in, \frac{1}{16}\ in$
Work Step by Step
The upper limit is the measurement + tolerance, that is,
$$
7\frac{7}{16}+\frac{1}{32}=7\frac{15}{32} \ in
.$$
The lower limit is the measurement - tolerance, that is,
$$
7\frac{7}{16}-\frac{1}{32}=7\frac{13}{32} \ in
.$$
The tolerance interval = upper limit-lower limit, that is,
$$
7\frac{15}{32}-7\frac{13}{32}=\frac{2}{32}=\frac{1}{16}\ in
.$$
