Data from the scores of the 100 Student Quantitative Methodalogy have

Harold Kessler 2021-12-10 Answered
Data from the scores of the 100 Student Quantitative Methodalogy have been compiled in the frequency distribution table below:
\[\begin{array}{|c|c|c|} \hline \text{Value of Quantitative Methodology}&\text{amount} \\ \hline 31-40&5 \\ \hline 41-50&8\\ \hline 51-60&12\\ \hline 61-70&17\\ \hline 71-80&25\\ \hline 81-90&21\\ \hline 91-100&12\\ \hline &100\\ \hline \end{array}\]
Based on the data above, calculate:
a. Mean
b. Median
c. Mode

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Expert Answer

Stuart Rountree
Answered 2021-12-11 Author has 4451 answers
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.
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