A population of values has a normal distribution with μ=77 and σ=32.2. You intend to...

From the given information, ${\displaystyle \mu =77}$, ${\displaystyle \sigma =32.2}$ and the sample size=15.
Consider,
${\displaystyle P\left(59.5<\bar{X}<98.6\right)=P\left(\frac{59.5-\mu }{\left(\frac{\sigma }{\sqrt{n}}\right)}<\frac{\bar{X}-\mu }{\left(\frac{\sigma }{\sqrt{n}}\right)}<\frac{98.6-\mu }{\left(\frac{\sigma }{\sqrt{n}}\right)}\right)}$
${\displaystyle =P\left(\frac{59.5-\mu }{\left(\frac{32.2}{\sqrt{15}}\right)}<z<\frac{98.6-77}{\left(\frac{32.2}{\sqrt{15}}\right)}\right)}$
${\displaystyle =P(-2.10488<z<2.59803)}$
${\displaystyle =P\left(z<2.59803\right)-P(z<-2.10488)}$
$=0.748828-0.2934=0.4554\left[\begin{array}{c}FromtheExcelfunction\\ =NORM.DIST(0.6770807,0,1,TRUE)\\ =NORM.DIST(-0.54348,0,1,TRUE)\end{array}\right]$
Thus, the probability that a sample of size ${\displaystyle n=15}$ is randomly selected with a mean between 59.5 and 98.6 is 0.9777.