Step 1:Given

The table shows the distribution of various classes.

Step 2:Objective

To calculate mean ,median and mode of this table.

Step 3: Solution

The table is as shown below,

\[\begin{array}{|c|c|c|} \hline Quantity&Amount(f)&Cumulative\ amount&x&fx\\ \hline 31-40&5&5&35.5&177.5\\ \hline 41-50&8&13&45.5&364\\ \hline 51-60&12&25&55.5&666\\ \hline 61-70&17&42&65.5&1113.5\\ \hline 71-80&25&67&75.5&1887.5\\ \hline 81-90&21&88&85.5&1795.5\\ \hline 91-100&12&100&95.5&1196\\ \hline &\sum A=100\\ \hline \end{array}\]

Summation of fx is given by,

\(\displaystyle\sum{f}_{{{i}}}{x}_{{{i}}}={7150}\)

The mean is given by,

\(\displaystyle\overline{{{x}}}={\frac{{\sum{f}_{{{i}}}{x}_{{{i}}}}}{{\sum{f}}}}\)

\(\displaystyle={\frac{{{7150}}}{{{100}}}}\)

\(\displaystyle={71.5}\)

Median is given by,

\(\displaystyle={L}+{\left[{\frac{{{\left({\frac{{{N}+{1}}}{{{2}}}}\right)}-{\left({F}+{1}\right)}}}{{{f}_{{{m}}}}}}\right]}\cdot{h}\)

Where,

\(\displaystyle{L}={71}\)

\(\displaystyle{N}={100}\)

\(\displaystyle{F}={42}\)

\(\displaystyle{h}={9}\)

\(\displaystyle{f}_{{{m}}}={25}\)

\(\displaystyle{M}{e}{d}{i}{a}{n}={71}+{\frac{{{55.5}-{43}}}{{{25}}}}\cdot{9}\)

\(\displaystyle={75.5}\)

Mode is given by,

\(\displaystyle={L}+{\frac{{{f}_{{{0}}}-{f}_{{{1}}}}}{{{2}{f}_{{{0}}}-{f}_{{{1}}}-{f}_{{{2}}}}}}\cdot{h}\)

\(\displaystyle{f}_{{{0}}}={25}\)

\(\displaystyle{f}_{{{1}}}={17}\)

\(\displaystyle{f}_{{{2}}}={21}\)

\(\displaystyle{M}{o}{d}{e}={71}+{\frac{{{25}-{17}}}{{{2}\times{25}-{17}-{21}}}}\cdot{9}\)

\(\displaystyle={77}\)

So mean, median and mode of this grouped data is 71.5, 75.5 and 77 respectively.

The table shows the distribution of various classes.

Step 2:Objective

To calculate mean ,median and mode of this table.

Step 3: Solution

The table is as shown below,

\[\begin{array}{|c|c|c|} \hline Quantity&Amount(f)&Cumulative\ amount&x&fx\\ \hline 31-40&5&5&35.5&177.5\\ \hline 41-50&8&13&45.5&364\\ \hline 51-60&12&25&55.5&666\\ \hline 61-70&17&42&65.5&1113.5\\ \hline 71-80&25&67&75.5&1887.5\\ \hline 81-90&21&88&85.5&1795.5\\ \hline 91-100&12&100&95.5&1196\\ \hline &\sum A=100\\ \hline \end{array}\]

Summation of fx is given by,

\(\displaystyle\sum{f}_{{{i}}}{x}_{{{i}}}={7150}\)

The mean is given by,

\(\displaystyle\overline{{{x}}}={\frac{{\sum{f}_{{{i}}}{x}_{{{i}}}}}{{\sum{f}}}}\)

\(\displaystyle={\frac{{{7150}}}{{{100}}}}\)

\(\displaystyle={71.5}\)

Median is given by,

\(\displaystyle={L}+{\left[{\frac{{{\left({\frac{{{N}+{1}}}{{{2}}}}\right)}-{\left({F}+{1}\right)}}}{{{f}_{{{m}}}}}}\right]}\cdot{h}\)

Where,

\(\displaystyle{L}={71}\)

\(\displaystyle{N}={100}\)

\(\displaystyle{F}={42}\)

\(\displaystyle{h}={9}\)

\(\displaystyle{f}_{{{m}}}={25}\)

\(\displaystyle{M}{e}{d}{i}{a}{n}={71}+{\frac{{{55.5}-{43}}}{{{25}}}}\cdot{9}\)

\(\displaystyle={75.5}\)

Mode is given by,

\(\displaystyle={L}+{\frac{{{f}_{{{0}}}-{f}_{{{1}}}}}{{{2}{f}_{{{0}}}-{f}_{{{1}}}-{f}_{{{2}}}}}}\cdot{h}\)

\(\displaystyle{f}_{{{0}}}={25}\)

\(\displaystyle{f}_{{{1}}}={17}\)

\(\displaystyle{f}_{{{2}}}={21}\)

\(\displaystyle{M}{o}{d}{e}={71}+{\frac{{{25}-{17}}}{{{2}\times{25}-{17}-{21}}}}\cdot{9}\)

\(\displaystyle={77}\)

So mean, median and mode of this grouped data is 71.5, 75.5 and 77 respectively.