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,

\(\displaystyle\overline{{X}}=\frac{{1}}{{10}}{\sum_{{{i}={1}}}^{{10}}}{X}_{{i}}\),

\(\displaystyle\overline{{X}}\) 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

\(\displaystyle{P}{\left({71}\le\overline{{X}}\le{73}\right)}\)

\(\displaystyle={P}{\left(\frac{{{71}-{72}}}{{9}}/{\left(\sqrt{{{10}}}\right)}\le{\left(\frac{{\overline{{X}}-{72}}}{{9}}/{\left(\sqrt{{{10}}}\right)}\le{\left(\frac{{{73}-{72}}}{{9}}/{\left(\sqrt{{{10}}}\right)}\right)},{Z}={\left(\frac{{\overline{{X}}-{72}}}{{9}}/{\left(\sqrt{{{10}}}\right)}\sim{N}{\quad\text{or}\quad}{m}{a}{l}{\left({0},{1}\right)}\right.}\right.}\right.}\)

\(\displaystyle={P}{\left(-{0.3514}\le{Z}\le{0.3514}\right)}\)

\(\displaystyle={P}{\left({Z}\le{0.3514}\right)}-{P}{\left({Z}\le-{0.3514}\right)}\)

\(\displaystyle=\Phi{\left({0.3514}\right)}-\Phi{\left({0.3514}\right)},\Phi{\left({z}\right)}={P}{\left({Z}\le{z}\right)}\)

\(\displaystyle={2}\Phi{\left({0.3514}\right)}-{1},\Phi{\left({z}\right)}={1}-\Phi{\left(-{z}\right)}\)

\(\displaystyle={\left({2}\times{0.6374}\right)}-{1}\)

=0.2748

The values of \(\displaystyle\Phi{\left({0.3514}\right)}\) 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.

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,

\(\displaystyle\overline{{X}}=\frac{{1}}{{10}}{\sum_{{{i}={1}}}^{{10}}}{X}_{{i}}\),

\(\displaystyle\overline{{X}}\) 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

\(\displaystyle{P}{\left({71}\le\overline{{X}}\le{73}\right)}\)

\(\displaystyle={P}{\left(\frac{{{71}-{72}}}{{9}}/{\left(\sqrt{{{10}}}\right)}\le{\left(\frac{{\overline{{X}}-{72}}}{{9}}/{\left(\sqrt{{{10}}}\right)}\le{\left(\frac{{{73}-{72}}}{{9}}/{\left(\sqrt{{{10}}}\right)}\right)},{Z}={\left(\frac{{\overline{{X}}-{72}}}{{9}}/{\left(\sqrt{{{10}}}\right)}\sim{N}{\quad\text{or}\quad}{m}{a}{l}{\left({0},{1}\right)}\right.}\right.}\right.}\)

\(\displaystyle={P}{\left(-{0.3514}\le{Z}\le{0.3514}\right)}\)

\(\displaystyle={P}{\left({Z}\le{0.3514}\right)}-{P}{\left({Z}\le-{0.3514}\right)}\)

\(\displaystyle=\Phi{\left({0.3514}\right)}-\Phi{\left({0.3514}\right)},\Phi{\left({z}\right)}={P}{\left({Z}\le{z}\right)}\)

\(\displaystyle={2}\Phi{\left({0.3514}\right)}-{1},\Phi{\left({z}\right)}={1}-\Phi{\left(-{z}\right)}\)

\(\displaystyle={\left({2}\times{0.6374}\right)}-{1}\)

=0.2748

The values of \(\displaystyle\Phi{\left({0.3514}\right)}\) 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.