\displaystyle \textbf{Question 1: }\text{Calculate the mean and standard deviation for the following data:}
\displaystyle \begin{array}{|l|c|c|c|c|c|}\hline \text{Expenditure in Rs.:}&0-10&10-20&20-30&30-40&40-50\\ \hline \text{Frequency:}&14&13&27&21&15\\ \hline \end{array}
\displaystyle \text{Answer:}
\displaystyle \text{Taking assumed mean }A=25\text{ and class width }h=10.
\displaystyle \begin{array}{|c|c|c|c|c|c|c|}\hline \text{Class interval}&f_i&x_i&u_i=\frac{x_i-25}{10}&f_iu_i&u_i^2&f_iu_i^2\\ \hline 0-10&14&5&-2&-28&4&56\\ \hline 10-20&13&15&-1&-13&1&13\\ \hline 20-30&27&25&0&0&0&0\\ \hline 30-40&21&35&1&21&1&21\\ \hline 40-50&15&45&2&30&4&60\\ \hline &N=\sum f_i=90&&&\sum f_iu_i=10&&\sum f_iu_i^2=150\\ \hline \end{array}
\displaystyle \text{Clearly, }N=90,\quad \sum f_iu_i=10,\quad \sum f_iu_i^2=150,\quad A=25\text{ and }h=10.
\displaystyle \overline{X}=A+h\left(\frac{\sum f_iu_i}{N}\right)
\displaystyle =25+10\left(\frac{10}{90}\right)
\displaystyle =25+\frac{10}{9}=26.11\text{ (approx.).}
\displaystyle \mathrm{Var}(X)=h^2\left[\frac{\sum f_iu_i^2}{N}-\left(\frac{\sum f_iu_i}{N}\right)^2\right]
\displaystyle =10^2\left[\frac{150}{90}-\left(\frac{10}{90}\right)^2\right]
\displaystyle =100\left[\frac{5}{3}-\frac{1}{81}\right]
\displaystyle =100\left(\frac{134}{81}\right)=165.43\text{ (approx.).}
\displaystyle \text{Standard deviation}=\sqrt{\mathrm{Var}(X)}
\displaystyle =\sqrt{165.43}=12.86\text{ (approx.).}
\displaystyle \therefore \text{The mean is }26.11\text{ and the standard deviation is }12.86.
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{Calculate the standard deviation for the following data:}
\displaystyle \begin{array}{|l|c|c|c|c|c|c|c|}\hline \text{Class:}&0-30&30-60&60-90&90-120&120-150&150-180&180-210\\ \hline \text{Frequency:}&9&17&43&82&81&44&24\\ \hline \end{array}
\displaystyle \text{Answer:}
\displaystyle \text{Taking assumed mean }A=105\text{ and class width }h=30.
\displaystyle \begin{array}{|c|c|c|c|c|c|c|}\hline \text{Class interval}&f_i&x_i&u_i=\frac{x_i-105}{30}&f_iu_i&u_i^2&f_iu_i^2\\ \hline 0-30&9&15&-3&-27&9&81\\ \hline 30-60&17&45&-2&-34&4&68\\ \hline 60-90&43&75&-1&-43&1&43\\ \hline 90-120&82&105&0&0&0&0\\ \hline 120-150&81&135&1&81&1&81\\ \hline 150-180&44&165&2&88&4&176\\ \hline 180-210&24&195&3&72&9&216\\ \hline &N=\sum f_i=300&&&\sum f_iu_i=137&&\sum f_iu_i^2=665\\ \hline \end{array}
\displaystyle \text{Clearly, }N=300,\quad \sum f_iu_i=137,\quad \sum f_iu_i^2=665,\quad A=105\text{ and }h=30.
\displaystyle \overline{X}=A+h\left(\frac{\sum f_iu_i}{N}\right)
\displaystyle =105+30\left(\frac{137}{300}\right)=118.7.
\displaystyle \mathrm{Var}(X)=h^2\left[\frac{\sum f_iu_i^2}{N}-\left(\frac{\sum f_iu_i}{N}\right)^2\right]
\displaystyle =30^2\left[\frac{665}{300}-\left(\frac{137}{300}\right)^2\right]
\displaystyle =900\left[\frac{665}{300}-\frac{18769}{90000}\right]
\displaystyle =1807.31.
\displaystyle \text{Standard deviation}=\sqrt{\mathrm{Var}(X)}
\displaystyle =\sqrt{1807.31}=42.51\text{ (approx.).}
\displaystyle \therefore \text{The standard deviation is }42.51\text{ (approx.).}
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{Calculate the A.M. and standard deviation for the following distribution:}
\displaystyle \begin{array}{|l|c|c|c|c|c|c|c|c|}\hline \text{Class:}&0-10&10-20&20-30&30-40&40-50&50-60&60-70&70-80\\ \hline \text{Frequency:}&18&16&15&12&10&5&2&1\\ \hline \end{array}
\displaystyle \text{Answer:}
\displaystyle \text{Taking assumed mean }A=35\text{ and class width }h=10.
\displaystyle \begin{array}{|c|c|c|c|c|c|c|}\hline \text{Class interval}&f_i&x_i&u_i=\frac{x_i-35}{10}&f_iu_i&u_i^2&f_iu_i^2\\ \hline 0-10&18&5&-3&-54&9&162\\ \hline 10-20&16&15&-2&-32&4&64\\ \hline 20-30&15&25&-1&-15&1&15\\ \hline 30-40&12&35&0&0&0&0\\ \hline 40-50&10&45&1&10&1&10\\ \hline 50-60&5&55&2&10&4&20\\ \hline 60-70&2&65&3&6&9&18\\ \hline 70-80&1&75&4&4&16&16\\ \hline &N=\sum f_i=79&&&\sum f_iu_i=-71&&\sum f_iu_i^2=305\\ \hline \end{array}
\displaystyle \text{Clearly, }N=79,\quad \sum f_iu_i=-71,\quad \sum f_iu_i^2=305,\quad A=35\text{ and }h=10.
\displaystyle \overline{X}=A+h\left(\frac{\sum f_iu_i}{N}\right)
\displaystyle =35+10\left(\frac{-71}{79}\right)
\displaystyle =26.0127\approx26.01.
\displaystyle \mathrm{Var}(X)=h^2\left[\frac{\sum f_iu_i^2}{N}-\left(\frac{\sum f_iu_i}{N}\right)^2\right]
\displaystyle =10^2\left[\frac{305}{79}-\left(\frac{-71}{79}\right)^2\right]
\displaystyle =100\left[\frac{305}{79}-\frac{5041}{6241}\right]
\displaystyle =305.3036\approx305.30.
\displaystyle \text{Standard deviation}=\sqrt{\mathrm{Var}(X)}
\displaystyle =\sqrt{305.3036}=17.4729\approx17.47.
\displaystyle \therefore \text{The A.M. is }26.01\text{ and the standard deviation is }17.47.
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{A student obtained mean and standard deviation of }100\text{ observations}
\displaystyle \text{as }40\text{ and }5.1\text{ respectively. It was later found that one observation}
\displaystyle \text{was wrongly copied as }50,\text{ the correct figure being }40.\text{ Find the correct}
\displaystyle \text{mean and standard deviation.}
\displaystyle \text{Answer:}
\displaystyle \text{Given }N=100,\quad \overline{X}=40,\quad \sigma=5.1.
\displaystyle \sum x_i=N\overline{X}=100\times40=4000.
\displaystyle \sigma^2=5.1^2=26.01.
\displaystyle \text{Also, }\sigma^2=\frac{\sum x_i^2}{N}-\overline{X}^{\,2}.
\displaystyle 26.01=\frac{\sum x_i^2}{100}-40^2.
\displaystyle \frac{\sum x_i^2}{100}=26.01+1600=1626.01.
\displaystyle \therefore \sum x_i^2=162601.
\displaystyle \text{Since }50\text{ should be replaced by }40,
\displaystyle \text{Corrected }\sum x_i=4000-50+40=3990.
\displaystyle \text{Corrected }\sum x_i^2=162601-50^2+40^2=161701.
\displaystyle \text{Correct mean}=\frac{3990}{100}=39.90.
\displaystyle \text{Correct variance}=\frac{161701}{100}-(39.90)^2.
\displaystyle =1617.01-1592.01=25.
\displaystyle \text{Correct standard deviation}=\sqrt{25}=5.
\displaystyle \therefore \text{The correct mean is }39.90\text{ and the correct standard deviation is }5.
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{Calculate the mean, median and standard deviation for the following}
\displaystyle \text{distribution:}
\displaystyle \begin{array}{|l|c|c|c|c|c|c|c|c|}\hline \text{Class interval:}&31-35&36-40&41-45&46-50&51-55&56-60&61-65&66-70\\ \hline \text{Frequency:}&2&3&8&12&16&5&2&3\\ \hline \end{array}
\displaystyle \text{Answer:}
\displaystyle \text{Taking assumed mean }A=53\text{ and class width }h=5.
\displaystyle \begin{array}{|c|c|c|c|c|c|c|}\hline \text{Class interval}&f_i&x_i&u_i=\frac{x_i-53}{5}&f_iu_i&u_i^2&f_iu_i^2\\ \hline 31-35&2&33&-4&-8&16&32\\ \hline 36-40&3&38&-3&-9&9&27\\ \hline 41-45&8&43&-2&-16&4&32\\ \hline 46-50&12&48&-1&-12&1&12\\ \hline 51-55&16&53&0&0&0&0\\ \hline 56-60&5&58&1&5&1&5\\ \hline 61-65&2&63&2&4&4&8\\ \hline 66-70&3&68&3&9&9&27\\ \hline &N=\sum f_i=51&&&\sum f_iu_i=-27&&\sum f_iu_i^2=143\\ \hline \end{array}
\displaystyle \text{Clearly, }N=51,\quad \sum f_iu_i=-27,\quad \sum f_iu_i^2=143,\quad A=53\text{ and }h=5.
\displaystyle \overline{X}=A+h\left(\frac{\sum f_iu_i}{N}\right)
\displaystyle =53+5\left(\frac{-27}{51}\right)
\displaystyle =50.3529\approx50.35.
\displaystyle \mathrm{Var}(X)=h^2\left[\frac{\sum f_iu_i^2}{N}-\left(\frac{\sum f_iu_i}{N}\right)^2\right]
\displaystyle =5^2\left[\frac{143}{51}-\left(\frac{-27}{51}\right)^2\right]
\displaystyle =25\left[\frac{143}{51}-\frac{729}{2601}\right]
\displaystyle =63.0911\approx63.09.
\displaystyle \text{Standard deviation}=\sqrt{\mathrm{Var}(X)}
\displaystyle =\sqrt{63.0911}=7.9430\approx7.94.
\displaystyle \text{For calculating the median, the inclusive classes are made continuous.}
\displaystyle \begin{array}{|c|c|c|}\hline \text{Continuous class interval}&f_i&\text{Cumulative frequency}\\ \hline 30.5-35.5&2&2\\ \hline 35.5-40.5&3&5\\ \hline 40.5-45.5&8&13\\ \hline 45.5-50.5&12&25\\ \hline 50.5-55.5&16&41\\ \hline 55.5-60.5&5&46\\ \hline 60.5-65.5&2&48\\ \hline 65.5-70.5&3&51\\ \hline \end{array}
\displaystyle N=51\quad\Rightarrow\quad \frac{N}{2}=25.5.
\displaystyle \text{The cumulative frequency just greater than }25.5\text{ is }41.
\displaystyle \therefore \text{The median class is }50.5-55.5.
\displaystyle l=50.5,\quad F=25,\quad f=16,\quad h=5.
\displaystyle \text{Median}=l+\frac{\frac{N}{2}-F}{f}\times h
\displaystyle =50.5+\frac{25.5-25}{16}\times5
\displaystyle =50.5+\frac{2.5}{16}
\displaystyle =50.65625\approx50.66.
\displaystyle \therefore \text{The mean is }50.35,\text{ the median is }50.66\text{ and the standard deviation is }7.94.
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{Calculate the mean and variance of the frequency distribution given below:}
\displaystyle \begin{array}{|l|c|c|c|c|}\hline x_i:&1\leq x<3&3\leq x<5&5\leq x<7&7\leq x<10\\ \hline f_i:&6&4&5&1\\ \hline \end{array}
\displaystyle \text{Answer:}
\displaystyle \text{The mid-value of each class interval is used as }x_i.
\displaystyle \begin{array}{|c|c|c|c|c|}\hline \text{Class interval}&f_i&x_i&f_ix_i&f_ix_i^2\\ \hline 1\leq x<3&6&2&12&24\\ \hline 3\leq x<5&4&4&16&64\\ \hline 5\leq x<7&5&6&30&180\\ \hline 7\leq x<10&1&8.5&8.5&72.25\\ \hline &N=\sum f_i=16&&\sum f_ix_i=66.5&\sum f_ix_i^2=340.25\\ \hline \end{array}
\displaystyle \text{Clearly, }N=16,\quad \sum f_ix_i=66.5\text{ and }\sum f_ix_i^2=340.25.
\displaystyle \overline X=\frac{\sum f_ix_i}{N}
\displaystyle =\frac{66.5}{16}=4.15625.
\displaystyle \mathrm{Var}(X)=\frac{\sum f_ix_i^2}{N}-\left(\frac{\sum f_ix_i}{N}\right)^2
\displaystyle =\frac{340.25}{16}-\left(\frac{66.5}{16}\right)^2
\displaystyle =21.265625-17.2744140625
\displaystyle =3.9912109375\approx3.9912.
\displaystyle \therefore \text{The mean is }4.15625\text{ and the variance is }3.9912\text{ (approx.).}
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{The weight of coffee in }70\text{ jars is shown in the following table:}
\displaystyle \begin{array}{|l|c|c|c|c|c|c|}\hline \text{Weight in grams:}&200-201&201-202&202-203&203-204&204-205&205-206\\ \hline \text{Frequency:}&13&27&18&10&1&1\\ \hline \end{array}
\displaystyle \text{Answer:}
\displaystyle \text{Taking assumed mean }A=203.5\text{ and class width }h=1.
\displaystyle \begin{array}{|c|c|c|c|c|c|c|}\hline \text{Class interval}&f_i&x_i&u_i=\frac{x_i-203.5}{1}&f_iu_i&u_i^2&f_iu_i^2\\ \hline 200-201&13&200.5&-3&-39&9&117\\ \hline 201-202&27&201.5&-2&-54&4&108\\ \hline 202-203&18&202.5&-1&-18&1&18\\ \hline 203-204&10&203.5&0&0&0&0\\ \hline 204-205&1&204.5&1&1&1&1\\ \hline 205-206&1&205.5&2&2&4&4\\ \hline &N=\sum f_i=70&&&\sum f_iu_i=-108&&\sum f_iu_i^2=248\\ \hline \end{array}
\displaystyle \text{Clearly, }N=70,\quad \sum f_iu_i=-108,\quad \sum f_iu_i^2=248,\quad A=203.5\text{ and }h=1.
\displaystyle \overline{X}=A+h\left(\frac{\sum f_iu_i}{N}\right)
\displaystyle =203.5+\frac{-108}{70}=201.9571\approx201.96.
\displaystyle \mathrm{Var}(X)=h^2\left[\frac{\sum f_iu_i^2}{N}-\left(\frac{\sum f_iu_i}{N}\right)^2\right]
\displaystyle =1^2\left[\frac{248}{70}-\left(\frac{-108}{70}\right)^2\right]
\displaystyle =3.542857-2.380408=1.162449\approx1.162.
\displaystyle \text{Standard deviation}=\sqrt{\mathrm{Var}(X)}
\displaystyle =\sqrt{1.162449}=1.0782\approx1.078.
\displaystyle \therefore \text{The variance is }1.162\text{ and the standard deviation is }1.078\text{ (approx.).}
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{The mean and standard deviation of }100\text{ observations were found}
\displaystyle \text{to be }40\text{ and }10\text{ respectively. If, at the time of calculation, two}
\displaystyle \text{observations were wrongly taken as }30\text{ and }70\text{ in place of }3\text{ and }27
\displaystyle \text{respectively, find the correct standard deviation.}
\displaystyle \text{Answer:}
\displaystyle \text{Given }N=100,\quad \overline{X}=40,\quad \sigma=10.
\displaystyle \sum x_i=N\overline{X}=100\times40=4000.
\displaystyle \sigma^2=10^2=100.
\displaystyle \text{Also, }\sigma^2=\frac{\sum x_i^2}{N}-\overline{X}^{\,2}.
\displaystyle 100=\frac{\sum x_i^2}{100}-40^2.
\displaystyle \frac{\sum x_i^2}{100}=1700.
\displaystyle \therefore \sum x_i^2=170000.
\displaystyle \text{These totals are based on the incorrect observations.}
\displaystyle \text{Corrected }\sum x_i=4000-30-70+3+27=3930.
\displaystyle \text{Corrected }\sum x_i^2=170000-30^2-70^2+3^2+27^2.
\displaystyle =170000-900-4900+9+729=164938.
\displaystyle \text{Correct mean}=\frac{3930}{100}=39.30.
\displaystyle \text{Correct variance}=\frac{164938}{100}-(39.30)^2.
\displaystyle =1649.38-1544.49=104.89.
\displaystyle \text{Correct standard deviation}=\sqrt{104.89}=10.24\text{ (approx.).}
\displaystyle \therefore \text{The correct standard deviation is }10.24\text{ (approx.).}
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{While calculating the mean and variance of }10\text{ readings, a student}
\displaystyle \text{wrongly used the reading }52\text{ for the correct reading }25.\text{ He obtained the}
\displaystyle \text{mean and variance as }45\text{ and }16\text{ respectively. Find the correct mean}
\displaystyle \text{and variance.}
\displaystyle \text{Answer:}
\displaystyle \text{Given }N=10,\quad \overline{X}=45,\quad \mathrm{Var}(X)=16.
\displaystyle \sum x_i=N\overline{X}=10\times45=450.
\displaystyle \mathrm{Var}(X)=\frac{\sum x_i^2}{N}-\overline{X}^{\,2}.
\displaystyle 16=\frac{\sum x_i^2}{10}-45^2.
\displaystyle \frac{\sum x_i^2}{10}=2041.
\displaystyle \therefore \sum x_i^2=20410.
\displaystyle \text{These totals are based on the incorrect reading.}
\displaystyle \text{Corrected }\sum x_i=450-52+25=423.
\displaystyle \text{Corrected }\sum x_i^2=20410-52^2+25^2.
\displaystyle =20410-2704+625=18331.
\displaystyle \text{Correct mean}=\frac{423}{10}=42.3.
\displaystyle \text{Correct variance}=\frac{18331}{10}-(42.3)^2.
\displaystyle =1833.1-1789.29=43.81.
\displaystyle \therefore \text{The correct mean is }42.3\text{ and the correct variance is }43.81.
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{Calculate the mean, variance and standard deviation of the following}
\displaystyle \text{frequency distribution:}
\displaystyle \begin{array}{|l|c|c|c|c|c|c|}\hline \text{Class:}&0-10&10-20&20-30&30-40&40-50&50-60\\ \hline \text{Frequency:}&11&29&18&4&5&3\\ \hline \end{array}
\displaystyle \text{Answer:}
\displaystyle \text{Taking assumed mean }A=35\text{ and class width }h=10.
\displaystyle \begin{array}{|c|c|c|c|c|c|c|}\hline \text{Class interval}&f_i&x_i&u_i=\frac{x_i-35}{10}&f_iu_i&u_i^2&f_iu_i^2\\ \hline 0-10&11&5&-3&-33&9&99\\ \hline 10-20&29&15&-2&-58&4&116\\ \hline 20-30&18&25&-1&-18&1&18\\ \hline 30-40&4&35&0&0&0&0\\ \hline 40-50&5&45&1&5&1&5\\ \hline 50-60&3&55&2&6&4&12\\ \hline &N=\sum f_i=70&&&\sum f_iu_i=-98&&\sum f_iu_i^2=250\\ \hline \end{array}
\displaystyle \text{Clearly, }N=70,\quad \sum f_iu_i=-98,\quad \sum f_iu_i^2=250,\quad A=35\text{ and }h=10.
\displaystyle \overline X=A+h\left(\frac{\sum f_iu_i}{N}\right)
\displaystyle =35+10\left(\frac{-98}{70}\right)
\displaystyle =35-14=21.
\displaystyle \mathrm{Var}(X)=h^2\left[\frac{\sum f_iu_i^2}{N}-\left(\frac{\sum f_iu_i}{N}\right)^2\right]
\displaystyle =10^2\left[\frac{250}{70}-\left(\frac{-98}{70}\right)^2\right]
\displaystyle =100\left[\frac{25}{7}-\left(\frac{-7}{5}\right)^2\right]
\displaystyle =100\left[\frac{25}{7}-\frac{49}{25}\right]
\displaystyle =161.1429\approx161.14.
\displaystyle \text{Standard deviation}=\sqrt{\mathrm{Var}(X)}
\displaystyle =\sqrt{161.1429}=12.6942\approx12.69.
\displaystyle \therefore \text{The mean is }21,\text{ the variance is }161.14\text{ and the standard deviation is }12.69\text{ (approx.).}
\displaystyle \\

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