From the given question:
[tex]\mu=\$2.715 \\ \\ \sigma=\$0.049[/tex]
The probability of a normally distributed data between two values (a, b) is given by:
[tex]P(a\ \textless \ \bar{x}\ \textless \ b)=P\left( \frac{b-\mu}{\sigma/\sqrt{n}} \right)-P\left( \frac{a-\mu}{\sigma/\sqrt{n}} \right)[/tex]
Thus,
[tex]P(\$2.691\ \textless \ \bar{x}\ \textless \ \$2.732)=P\left( \frac{\$2.732-\$2.715}{\$0.049/\sqrt{32}} \right)-P\left( \frac{\$2.691-\$2.715}{\$0.049/\sqrt{32}} \right) \\ \\ =P(1.963)-P(-2.771)=0.97515-0.0028=\bold{0.97235}[/tex]
