From the given information, $\mu =77$, $\sigma =32.2$ and the sample size=15.

Consider,

$P\left(59.5<\stackrel{\u2015}{X}<98.6\right)=P\left(\frac{59.5-\mu}{\left(\frac{\sigma}{\sqrt{n}}\right)}<\frac{\stackrel{\u2015}{X}-\mu}{\left(\frac{\sigma}{\sqrt{n}}\right)}<\frac{98.6-\mu}{\left(\frac{\sigma}{\sqrt{n}}\right)}\right)$

$=P\left(\frac{59.5-\mu}{\left(\frac{32.2}{\sqrt{15}}\right)}<z<\frac{98.6-77}{\left(\frac{32.2}{\sqrt{15}}\right)}\right)$

$=P(-2.10488<z<2.59803)$

$=P\left(z<2.59803\right)-P(z<-2.10488)$

$=0.748828-0.2934=0.4554\left[\begin{array}{c}From\text{}the\text{}Excel\text{}function\\ =NORM.DIST(0.6770807,0,1,TRUE)\\ =NORM.DIST(-0.54348,0,1,TRUE)\end{array}\right]$

Thus, the probability that a sample of size $n=15$ is randomly selected with a mean between 59.5 and 98.6 is 0.9777.