Consider the marks of all 1st-year students on a statistics test. If the marks have a normal distribution with a mean of 72 and a standard deviation of 9, then the probability that a random sample of 10 students from this group have a sample mean between 71 and 73 is?

Step 1
Suppose, X be the marks of statistics students, X follows Normal distribution with mean 72 and standard deviation 9.
The average marks of 10 students,
${\displaystyle \bar{X}=\frac{1}{10}\sum _{i=1}^{10}{X}_i}$,
${\displaystyle \bar{X}}$ follows Normal distributions with mean 72 and standard deviation ${\displaystyle \frac{9}{\sqrt{10}}}$.
Step 2
The probability that the sample mean between 71 and 73 is
${\displaystyle P(71\le \bar{X}\le 73)}$
${\displaystyle =P(\frac{71-72}{9}/\left(\sqrt{10}\right)\le (\frac{\bar{X}-72}{9}/\left(\sqrt{10}\right)\le \left(\frac{73-72}{9}/\left(\sqrt{10}\right)\right),Z=(\frac{\bar{X}-72}{9}/\left(\sqrt{10}\right)\sim N\text{or}mal(0,1)}$
${\displaystyle =P(-0.3514\le Z\le 0.3514)}$
${\displaystyle =P(Z\le 0.3514)-P(Z\le -0.3514)}$
${\displaystyle =\mathrm{\Phi }\left(0.3514\right)-\mathrm{\Phi }\left(0.3514\right),\mathrm{\Phi }\left(z\right)=P(Z\le z)}$
${\displaystyle =2\mathrm{\Phi }\left(0.3514\right)-1,\mathrm{\Phi }\left(z\right)=1-\mathrm{\Phi }(-z)}$
${\displaystyle =(2\times 0.6374)-1}$
=0.2748
The values of ${\displaystyle \mathrm{\Phi }\left(0.3514\right)}$ is taken from Normal distribution.
Therefore, the probability that a random sample of 10 students from this group have a sample mean between 71 and 73 is 0.2758.