\displaystyle \textbf{Question 1: }\text{Find the standard deviation for the following distribution:}
\displaystyle \begin{array}{|c|c|c|c|c|c|c|c|}\hline x&4.5&14.5&24.5&34.5&44.5&54.5&64.5\\\hline f&1&5&12&22&17&9&4\\\hline\end{array}
\displaystyle \text{Answer:}
\displaystyle \text{Let the assumed mean be }A=34.5\text{ and }h=10.
\displaystyle \text{Calculation of Variance and Standard Deviation}
\displaystyle \begin{array}{|c|c|c|c|c|c|c|}\hline x_i&f_i&d_i=x_i-34.5&u_i=\frac{x_i-34.5}{10}&f_iu_i&u_i^2&f_iu_i^2\\\hline 4.5&1&-30&-3&-3&9&9\\\hline 14.5&5&-20&-2&-10&4&20\\\hline 24.5&12&-10&-1&-12&1&12\\\hline 34.5&22&0&0&0&0&0\\\hline 44.5&17&10&1&17&1&17\\\hline 54.5&9&20&2&18&4&36\\\hline 64.5&4&30&3&12&9&36\\\hline &N=\sum f_i=70&&&\sum f_iu_i=22&&\sum f_iu_i^2=130\\\hline\end{array}
\displaystyle \text{Here, }N=70,\qquad \sum f_iu_i=22,\qquad \sum f_iu_i^2=130,\qquad h=10
\displaystyle \mathrm{Var}(X)=h^2\left[\frac{1}{N}\sum f_iu_i^2-\left(\frac{1}{N}\sum f_iu_i\right)^2\right]
\displaystyle =10^2\left[\frac{130}{70}-\left(\frac{22}{70}\right)^2\right]
\displaystyle =100\left[\frac{13}{7}-\frac{121}{1225}\right]
\displaystyle =100\left(\frac{2154}{1225}\right)
\displaystyle =\frac{8616}{49}
\displaystyle \approx175.84
\displaystyle \therefore \sigma=\sqrt{\mathrm{Var}(X)}
\displaystyle =\sqrt{175.84}
\displaystyle \approx13.26
\displaystyle \therefore \text{The standard deviation of the distribution is approximately }13.26.
\displaystyle \\

\displaystyle \textbf{Question 2: } \text{Table below shows the frequency }f\text{ with which }x\text{ alpha particles}
\displaystyle \text{were radiated from a diskette. Calculate the mean and variance.}
\displaystyle \begin{array}{|c|c|c|c|}\hline x_i&f_i&f_ix_i&f_ix_i^2\\ \hline 0&51&0&0\\ \hline 1&203&203&203\\ \hline 2&383&766&1532\\ \hline 3&525&1575&4725\\ \hline 4&532&2128&8512\\ \hline 5&408&2040&10200\\ \hline 6&273&1638&9828\\ \hline 7&139&973&6811\\ \hline 8&43&344&2752\\ \hline 9&27&243&2187\\ \hline 10&10&100&1000\\ \hline 11&4&44&484\\ \hline 12&2&24&288\\ \hline &N=\sum f_i=2600&\sum f_ix_i=10078&\sum f_ix_i^2=48522\\ \hline \end{array}
\displaystyle \text{Answer:}
\displaystyle \overline{X}=\frac{\sum f_ix_i}{N}=\frac{10078}{2600}=3.87615\approx3.88.
\displaystyle \text{Variance }=\frac{\sum f_ix_i^2}{N}-\left(\frac{\sum f_ix_i}{N}\right)^2.
\displaystyle =\frac{48522}{2600}-\left(\frac{10078}{2600}\right)^2.
\displaystyle =18.66231-15.02457=3.63774\approx3.64.
\displaystyle \therefore \text{Standard deviation }=\sqrt{3.63774}=1.90729\approx1.91.
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Find the mean and standard deviation for the following data:}
\displaystyle \text{(i) }\begin{array}{|l|c|c|c|c|c|c|}\hline \text{Years under:}&10&20&30&40&50&60\\ \hline \text{Number of persons (cumulative):}&15&32&51&78&97&109\\ \hline \end{array}
\displaystyle \text{(ii) }\begin{array}{|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|c|}\hline \text{Marks:}&2&3&4&5&6&7&8&9&10&11&12&13&14&15&16\\ \hline \text{Frequency:}&1&6&6&8&8&2&2&3&0&2&1&0&0&0&1\\ \hline \end{array}
\displaystyle \text{Answer:}
\displaystyle \text{(i) The ordinary frequencies are obtained by taking successive differences.}
\displaystyle \begin{array}{|c|c|c|c|c|}\hline x_i&\text{Cumulative frequency}&f_i&f_ix_i&f_ix_i^2\\ \hline 10&15&15&150&1500\\ \hline 20&32&17&340&6800\\ \hline 30&51&19&570&17100\\ \hline 40&78&27&1080&43200\\ \hline 50&97&19&950&47500\\ \hline 60&109&12&720&43200\\ \hline &&\sum f_i=109&\sum f_ix_i=3810&\sum f_ix_i^2=159300\\ \hline \end{array}
\displaystyle \text{Clearly, }N=109,\quad \sum f_ix_i=3810\text{ and }\sum f_ix_i^2=159300.
\displaystyle \overline{X}=\frac{\sum f_ix_i}{N}=\frac{3810}{109}=34.9541\approx34.95.
\displaystyle \mathrm{Var}(X)=\frac{\sum f_ix_i^2}{N}-\left(\frac{\sum f_ix_i}{N}\right)^2
\displaystyle =\frac{159300}{109}-\left(\frac{3810}{109}\right)^2
\displaystyle =1461.4679-1221.7911=239.6768\approx239.68.
\displaystyle \text{Standard deviation}=\sqrt{\mathrm{Var}(X)}
\displaystyle =\sqrt{239.6768}=15.4815\approx15.48.
\displaystyle \therefore \text{The mean is }34.95\text{ and the standard deviation is }15.48.
\displaystyle \\

\displaystyle \text{(ii)}
\displaystyle \begin{array}{|c|c|c|c|}\hline x_i&f_i&f_ix_i&f_ix_i^2\\ \hline 2&1&2&4\\ \hline 3&6&18&54\\ \hline 4&6&24&96\\ \hline 5&8&40&200\\ \hline 6&8&48&288\\ \hline 7&2&14&98\\ \hline 8&2&16&128\\ \hline 9&3&27&243\\ \hline 10&0&0&0\\ \hline 11&2&22&242\\ \hline 12&1&12&144\\ \hline 13&0&0&0\\ \hline 14&0&0&0\\ \hline 15&0&0&0\\ \hline 16&1&16&256\\ \hline &\sum f_i=40&\sum f_ix_i=239&\sum f_ix_i^2=1753\\ \hline \end{array}
\displaystyle \text{Clearly, }N=40,\quad \sum f_ix_i=239\text{ and }\sum f_ix_i^2=1753.
\displaystyle \overline{X}=\frac{\sum f_ix_i}{N}=\frac{239}{40}=5.975.
\displaystyle \mathrm{Var}(X)=\frac{\sum f_ix_i^2}{N}-\left(\frac{\sum f_ix_i}{N}\right)^2
\displaystyle =\frac{1753}{40}-\left(\frac{239}{40}\right)^2
\displaystyle =43.825-35.700625=8.124375\approx8.12.
\displaystyle \text{Standard deviation}=\sqrt{\mathrm{Var}(X)}
\displaystyle =\sqrt{8.124375}=2.8503\approx2.85.
\displaystyle \therefore \text{The mean is }5.975\text{ and the standard deviation is }2.85.
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{Find the standard deviation for the following data:}
\displaystyle \text{(i) }\begin{array}{|c|c|c|c|c|c|}\hline x:&3&8&13&18&23\\ \hline f:&7&10&15&10&6\\ \hline \end{array}
\displaystyle \text{(ii) }\begin{array}{|c|c|c|c|c|c|c|}\hline x:&2&3&4&5&6&7\\ \hline f:&4&9&16&14&11&6\\ \hline \end{array}
\displaystyle \text{Answer:}
\displaystyle \text{(i)}
\displaystyle \begin{array}{|c|c|c|c|c|}\hline x_i&f_i&f_ix_i&f_ix_i^2\\ \hline 3&7&21&63\\ \hline 8&10&80&640\\ \hline 13&15&195&2535\\ \hline 18&10&180&3240\\ \hline 23&6&138&3174\\ \hline &\sum f_i=48&\sum f_ix_i=614&\sum f_ix_i^2=9652\\ \hline \end{array}
\displaystyle \text{Clearly, }N=48,\quad \sum f_ix_i=614\text{ and }\sum f_ix_i^2=9652.
\displaystyle \overline{X}=\frac{\sum f_ix_i}{N}=\frac{614}{48}=12.7917.
\displaystyle \mathrm{Var}(X)=\frac{\sum f_ix_i^2}{N}-\left(\frac{\sum f_ix_i}{N}\right)^2
\displaystyle =\frac{9652}{48}-\left(\frac{614}{48}\right)^2
\displaystyle =201.0833-163.6267=37.4566.
\displaystyle \text{Standard deviation}=\sqrt{\mathrm{Var}(X)}
\displaystyle =\sqrt{37.4566}=6.1202\approx6.12.
\displaystyle \therefore \text{The standard deviation is }6.12.
\displaystyle \\

\displaystyle \text{(ii)}
\displaystyle \begin{array}{|c|c|c|c|c|}\hline x_i&f_i&f_ix_i&f_ix_i^2\\ \hline 2&4&8&16\\ \hline 3&9&27&81\\ \hline 4&16&64&256\\ \hline 5&14&70&350\\ \hline 6&11&66&396\\ \hline 7&6&42&294\\ \hline &\sum f_i=60&\sum f_ix_i=277&\sum f_ix_i^2=1393\\ \hline \end{array}
\displaystyle \text{Clearly, }N=60,\quad \sum f_ix_i=277\text{ and }\sum f_ix_i^2=1393.
\displaystyle \overline{X}=\frac{\sum f_ix_i}{N}=\frac{277}{60}=4.6167.
\displaystyle \mathrm{Var}(X)=\frac{\sum f_ix_i^2}{N}-\left(\frac{\sum f_ix_i}{N}\right)^2
\displaystyle =\frac{1393}{60}-\left(\frac{277}{60}\right)^2
\displaystyle =23.2167-21.3136=1.9031.
\displaystyle \text{Standard deviation}=\sqrt{\mathrm{Var}(X)}
\displaystyle =\sqrt{1.9031}=1.3795\approx1.38.
\displaystyle \therefore \text{The standard deviation is }1.38.
\displaystyle \\

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