Step 1

To find the median we have to arrange the data set first.

That is: 15 17 18 19 20 31 33 35 37 39

Since there are 10 values in the data set we take the average of \(\displaystyle{\left({\frac{{{10}}}{{{2}}}}\right)}={5}-{t}{h}\ {\quad\text{and}\quad}\ {\left({\frac{{{10}}}{{{2}}}}\right)}+{1}={6}-{t}{h}\) value

\(\displaystyle{5}-{t}{h}\ {v}{a}{l}{u}{e}={20}\)

\(\displaystyle{6}-{t}{h}\ {v}{a}{l}{u}{e}={31}\)

So median \(\displaystyle={\frac{{{20}+{31}}}{{{2}}}}={\frac{{{51}}}{{{2}}}}={25.5}\)

Answer: \(\displaystyle{M}{e}{d}{i}{a}{n}={25.5}\)

Step 2

To find the standard deviation we need to find mean first

Mean is:

\(\displaystyle{\frac{{{15}+{19}+{17}+{39}+{33}+{20}+{18}+{37}+{31}+{35}}}{{{10}}}}={26.4}\)

Then we make a table

\[\begin{array}{|c|c|c|} \hline data&data-mean&(data-mean)^{2}\\ \hline 15&-11.4&129.96\\ \hline 19&-7.4&54.76\\ \hline 17&-9.4&88.36\\ \hline 39&12.6&158.76\\ \hline 33&6.6&43.56\\ \hline 20&-6.4&40.96\\ \hline 18&-8.4&70.56\\ \hline 37&10.6&112.36\\ \hline 31&4.6&21.16\\ \hline 35&8.6&73.96\\ \hline \end{array}\]

Then we add the values in the last column

\(\displaystyle\sum{\left({x}_{{{i}}}-\overline{{{X}}}\right)}^{{{2}}}={794.4}\)

Then use it to find the standard deviation

\(\displaystyle\sqrt{{{\frac{{\sum{\left({x}_{{{i}}}-\overline{{{X}}}\right)}^{{{2}}}}}{{{n}-{1}}}}}}=\sqrt{{{\frac{{{794.4}}}{{{10}-{1}}}}}}={9.39503414931}\)

Answer: \(\displaystyle{S}{D}={9.39503414931}\)

To find the median we have to arrange the data set first.

That is: 15 17 18 19 20 31 33 35 37 39

Since there are 10 values in the data set we take the average of \(\displaystyle{\left({\frac{{{10}}}{{{2}}}}\right)}={5}-{t}{h}\ {\quad\text{and}\quad}\ {\left({\frac{{{10}}}{{{2}}}}\right)}+{1}={6}-{t}{h}\) value

\(\displaystyle{5}-{t}{h}\ {v}{a}{l}{u}{e}={20}\)

\(\displaystyle{6}-{t}{h}\ {v}{a}{l}{u}{e}={31}\)

So median \(\displaystyle={\frac{{{20}+{31}}}{{{2}}}}={\frac{{{51}}}{{{2}}}}={25.5}\)

Answer: \(\displaystyle{M}{e}{d}{i}{a}{n}={25.5}\)

Step 2

To find the standard deviation we need to find mean first

Mean is:

\(\displaystyle{\frac{{{15}+{19}+{17}+{39}+{33}+{20}+{18}+{37}+{31}+{35}}}{{{10}}}}={26.4}\)

Then we make a table

\[\begin{array}{|c|c|c|} \hline data&data-mean&(data-mean)^{2}\\ \hline 15&-11.4&129.96\\ \hline 19&-7.4&54.76\\ \hline 17&-9.4&88.36\\ \hline 39&12.6&158.76\\ \hline 33&6.6&43.56\\ \hline 20&-6.4&40.96\\ \hline 18&-8.4&70.56\\ \hline 37&10.6&112.36\\ \hline 31&4.6&21.16\\ \hline 35&8.6&73.96\\ \hline \end{array}\]

Then we add the values in the last column

\(\displaystyle\sum{\left({x}_{{{i}}}-\overline{{{X}}}\right)}^{{{2}}}={794.4}\)

Then use it to find the standard deviation

\(\displaystyle\sqrt{{{\frac{{\sum{\left({x}_{{{i}}}-\overline{{{X}}}\right)}^{{{2}}}}}{{{n}-{1}}}}}}=\sqrt{{{\frac{{{794.4}}}{{{10}-{1}}}}}}={9.39503414931}\)

Answer: \(\displaystyle{S}{D}={9.39503414931}\)