\displaystyle \textbf{Question 1: }\text{Find the mean, variance and standard deviation for the following data:}
\displaystyle \textbf{(i)}\;2,4,5,6,8,17\qquad\textbf{(ii)}\;6,7,10,12,13,4,8,12
\displaystyle \textbf{(iii)}\;227,235,255,269,292,299,312,321,333,348\qquad\textbf{(iv)}\;15,22,27,11,9,21,14,9
\displaystyle \text{Answer:}

\displaystyle \textbf{(i)}
\displaystyle \text{Given data: }2,4,5,6,8,17
\displaystyle \overline{X}=\frac{2+4+5+6+8+17}{6}=\frac{42}{6}=7
\displaystyle \text{Calculation of Variance}
\displaystyle \begin{array}{|c|c|c|}\hline x_i&x_i-\overline{X}=x_i-7&(x_i-\overline{X})^2\\\hline2&-5&25\\\hline4&-3&9\\\hline5&-2&4\\\hline6&-1&1\\\hline8&1&1\\\hline17&10&100\\\hline&&\sum\limits_{i=1}^{6}(x_i-\overline{X})^2=140\\\hline\end{array}
\displaystyle \text{Clearly, }n=6\text{ and }\sum(x_i-\overline{X})^2=140
\displaystyle \therefore \mathrm{Var}(X)=\frac{1}{n}\sum(x_i-\overline{X})^2=\frac{140}{6}=23.33
\displaystyle \sigma=\sqrt{\mathrm{Var}(X)}=\sqrt{23.33}=4.83
\displaystyle \\

\displaystyle \textbf{(ii)}
\displaystyle \text{Given data: }6,7,10,12,13,4,8,12
\displaystyle \overline{X}=\frac{72}{8}=9
\displaystyle \text{Calculation of Variance}
\displaystyle \begin{array}{|c|c|c|}\hline x_i&x_i-\overline{X}=x_i-9&(x_i-\overline{X})^2\\\hline6&-3&9\\\hline7&-2&4\\\hline10&1&1\\\hline12&3&9\\\hline13&4&16\\\hline4&-5&25\\\hline8&-1&1\\\hline12&3&9\\\hline&&\sum\limits_{i=1}^{8}(x_i-\overline{X})^2=74\\\hline\end{array}
\displaystyle \text{Clearly, }n=8\text{ and }\sum(x_i-\overline{X})^2=74
\displaystyle \therefore \mathrm{Var}(X)=\frac{74}{8}=9.25
\displaystyle \sigma=\sqrt{9.25}=3.04
\displaystyle \\

\displaystyle \textbf{(iii)}
\displaystyle \text{Given data: }227,235,255,269,292,299,312,321,333,348
\displaystyle \overline{X}=\frac{2891}{10}=289.1
\displaystyle \text{Calculation of Variance}
\displaystyle \begin{array}{|c|c|c|}\hline x_i&x_i-\overline{X}&(x_i-\overline{X})^2\\\hline227&-62.1&3856.41\\\hline235&-54.1&2926.81\\\hline255&-34.1&1162.81\\\hline269&-20.1&404.01\\\hline292&2.9&8.41\\\hline299&9.9&98.01\\\hline312&22.9&524.41\\\hline321&31.9&1017.61\\\hline333&43.9&1927.21\\\hline348&58.9&3469.21\\\hline&&\sum\limits_{i=1}^{10}(x_i-\overline{X})^2=15394.9\\\hline\end{array}
\displaystyle \text{Clearly, }n=10\text{ and }\sum(x_i-\overline{X})^2=15394.9
\displaystyle \therefore \mathrm{Var}(X)=\frac{15394.9}{10}=1539.49
\displaystyle \sigma=\sqrt{1539.49}=39.24
\displaystyle \\

\displaystyle \textbf{(iv)}
\displaystyle \text{Given data: }15,22,27,11,9,21,14,9
\displaystyle \overline{X}=\frac{128}{8}=16
\displaystyle \text{Calculation of Variance}
\displaystyle \begin{array}{|c|c|c|}\hline x_i&x_i-\overline{X}=x_i-16&(x_i-\overline{X})^2\\\hline15&-1&1\\\hline22&6&36\\\hline27&11&121\\\hline11&-5&25\\\hline9&-7&49\\\hline21&5&25\\\hline14&-2&4\\\hline9&-7&49\\\hline&&\sum\limits_{i=1}^{8}(x_i-\overline{X})^2=310\\\hline\end{array}
\displaystyle \text{Clearly, }n=8\text{ and }\sum(x_i-\overline{X})^2=310
\displaystyle \therefore \mathrm{Var}(X)=\frac{310}{8}=38.75
\displaystyle \sigma=\sqrt{38.75}=6.22
\displaystyle \\

\displaystyle \textbf{Question 2: }\text{The variance of }20\text{ observations is }5.\text{ If each observation} \\ \text{is multiplied by }2,\text{ find the variance of the resulting observations.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }x_1,x_2,x_3,\ldots,x_{20}\text{ be the }20\text{ given observations.}
\displaystyle \text{Their mean is }\overline{X}=\frac{\sum\limits_{i=1}^{20}x_i}{20}.
\displaystyle \text{Given }\mathrm{Var}(X)=5.
\displaystyle \text{We know, }\mathrm{Var}(X)=\frac{1}{20}\sum\limits_{i=1}^{20}(x_i-\overline{X})^2.
\displaystyle \therefore \frac{1}{20}\sum\limits_{i=1}^{20}(x_i-\overline{X})^2=5.
\displaystyle \text{Let }u_1,u_2,u_3,\ldots,u_{20}\text{ be the new observations, where }u_i=2x_i.
\displaystyle \text{Then }\overline{U}=\frac{\sum\limits_{i=1}^{20}u_i}{20}=\frac{\sum\limits_{i=1}^{20}2x_i}{20}=2\overline{X}.
\displaystyle u_i-\overline{U}=2x_i-2\overline{X}=2(x_i-\overline{X}).
\displaystyle \text{Squaring both sides,}
\displaystyle (u_i-\overline{U})^2=4(x_i-\overline{X})^2.
\displaystyle \therefore \frac{1}{20}\sum\limits_{i=1}^{20}(u_i-\overline{U})^2=4\left(\frac{1}{20}\sum\limits_{i=1}^{20}(x_i-\overline{X})^2\right).
\displaystyle \therefore \mathrm{Var}(U)=4\times\mathrm{Var}(X)=4\times5=20.
\displaystyle \therefore \text{The variance of the new observations is }20.
\displaystyle \\

\displaystyle \textbf{Question 3: }\text{The variance of }15\text{ observations is }4.\text{ If each observation} \\ \text{is increased by }9,\text{ find the variance of the resulting observations.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }x_1,x_2,x_3,\ldots,x_{15}\text{ be the }15\text{ given observations.}
\displaystyle \text{Their mean is }\overline{X}=\frac{\sum\limits_{i=1}^{15}x_i}{15}.
\displaystyle \text{Given }\mathrm{Var}(X)=4.
\displaystyle \text{We know, }\mathrm{Var}(X)=\frac{1}{15}\sum\limits_{i=1}^{15}(x_i-\overline{X})^2.
\displaystyle \therefore \frac{1}{15}\sum\limits_{i=1}^{15}(x_i-\overline{X})^2=4.
\displaystyle \text{Let }u_1,u_2,u_3,\ldots,u_{15}\text{ be the new observations, where }u_i=x_i+9.
\displaystyle \text{Then }\overline{U}=\frac{\sum\limits_{i=1}^{15}u_i}{15}=\frac{\sum\limits_{i=1}^{15}(x_i+9)}{15}=\frac{\sum\limits_{i=1}^{15}x_i}{15}+\frac{9\times15}{15}=\overline{X}+9.
\displaystyle u_i-\overline{U}=(x_i+9)-(\overline{X}+9)=x_i-\overline{X}.
\displaystyle \text{Squaring both sides,}
\displaystyle (u_i-\overline{U})^2=(x_i-\overline{X})^2.
\displaystyle \therefore \frac{1}{15}\sum\limits_{i=1}^{15}(u_i-\overline{U})^2=\frac{1}{15}\sum\limits_{i=1}^{15}(x_i-\overline{X})^2.
\displaystyle \therefore \mathrm{Var}(U)=\mathrm{Var}(X)=4.
\displaystyle \therefore \text{The variance of the new observations is }4.
\displaystyle \\

\displaystyle \textbf{Question 4: }\text{The mean of }5\text{ observations is }4.4\text{ and their variance is }8.24.
\displaystyle \text{If three of the observations are }1,2\text{ and }6,\text{ find the other two observations.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }x\text{ and }y\text{ be the other two observations.}
\displaystyle \text{Given }\overline X=4.4
\displaystyle \therefore \frac{1+2+6+x+y}{5}=4.4
\displaystyle \Rightarrow 9+x+y=22
\displaystyle \Rightarrow x+y=13\qquad\text{... ... ... ... ... (i)}
\displaystyle \text{Given }\mathrm{Var}(X)=8.24
\displaystyle \text{We know, }\mathrm{Var}(X)=\frac{1}{n}\sum_{i=1}^{n}(x_i-\overline X)^2
\displaystyle \therefore \frac{1}{5}\left[(1-4.4)^2+(2-4.4)^2+(6-4.4)^2+(x-4.4)^2+(y-4.4)^2\right]=8.24
\displaystyle \Rightarrow (1-4.4)^2+(2-4.4)^2+(6-4.4)^2+(x-4.4)^2+(y-4.4)^2=41.2
\displaystyle \Rightarrow 3.4^2+2.4^2+1.6^2+(x-4.4)^2+(y-4.4)^2=41.2
\displaystyle \Rightarrow (x-4.4)^2+(y-4.4)^2=21.32
\displaystyle \Rightarrow x^2+y^2-8.8(x+y)+38.72=21.32
\displaystyle \Rightarrow x^2+y^2-8.8(13)=-17.4
\displaystyle \Rightarrow x^2+y^2=97\qquad\text{... ... ... ... ... (ii)}
\displaystyle \text{Now, }(x+y)^2=x^2+y^2+2xy
\displaystyle \Rightarrow 13^2=97+2xy
\displaystyle \Rightarrow 169=97+2xy
\displaystyle \Rightarrow xy=36
\displaystyle \text{Therefore, }x\text{ and }y\text{ are the roots of the equation}
\displaystyle t^2-(x+y)t+xy=0
\displaystyle \Rightarrow t^2-13t+36=0
\displaystyle \Rightarrow (t-4)(t-9)=0
\displaystyle \Rightarrow t=4\text{ or }t=9
\displaystyle \therefore \text{The other two observations are }4\text{ and }9.
\displaystyle \\

\displaystyle \textbf{Question 5: }\text{The mean and standard deviation of }6\text{ observations are }8\text{ and }4\text{ respectively.}
\displaystyle \text{If each observation is multiplied by }3,\text{ find the new mean and new standard} \\ \text{deviation of the resulting observations.}
\displaystyle \text{Answer:}
\displaystyle \text{Given: Mean }=\overline{X}=8,\qquad n=6,\qquad \sigma_X=4
\displaystyle \text{Let }x_1,x_2,\ldots,x_6\text{ be the given observations.}
\displaystyle \therefore \overline{X}=\frac{\sum\limits_{i=1}^{6}x_i}{6}=8
\displaystyle \text{Let }u_1,u_2,\ldots,u_6\text{ be the new observations, where }u_i=3x_i.
\displaystyle \therefore \overline{U}=\frac{\sum\limits_{i=1}^{6}u_i}{6}=\frac{\sum\limits_{i=1}^{6}3x_i}{6}=3\left(\frac{\sum\limits_{i=1}^{6}x_i}{6}\right)=3\times8=24
\displaystyle \text{Given }\sigma_X=4
\displaystyle \therefore \mathrm{Var}(X)=\sigma_X^2=4^2=16
\displaystyle \therefore \frac{1}{6}\sum\limits_{i=1}^{6}(x_i-\overline{X})^2=16
\displaystyle \mathrm{Var}(U)=\frac{1}{6}\sum\limits_{i=1}^{6}(u_i-\overline{U})^2
\displaystyle =\frac{1}{6}\sum\limits_{i=1}^{6}(3x_i-3\overline{X})^2
\displaystyle =\frac{1}{6}\sum\limits_{i=1}^{6}9(x_i-\overline{X})^2
\displaystyle =9\left(\frac{1}{6}\sum\limits_{i=1}^{6}(x_i-\overline{X})^2\right)
\displaystyle =9\times16=144
\displaystyle \therefore \sigma_U=\sqrt{\mathrm{Var}(U)}=\sqrt{144}=12
\displaystyle \therefore \text{The new mean and the new standard deviation are }24\text{ and }12\text{ respectively.}
\displaystyle \\

\displaystyle \textbf{Question 6: }\text{The mean and the variance of }8\text{ observations are }9\text{ and }9.25\text{ respectively.}
\displaystyle \text{If six of the observations are }6,7,10,12,12\text{ and }13,\text{ find the remaining two observations.}
\displaystyle \text{Answer:}
\displaystyle \text{Let }x\text{ and }y\text{ be the remaining two observations.}
\displaystyle \text{Given }\overline X=9
\displaystyle \therefore \frac{6+7+10+12+12+13+x+y}{8}=9
\displaystyle \Rightarrow 60+x+y=72
\displaystyle \Rightarrow x+y=12\qquad\text{... ... ... ... ... (i)}
\displaystyle \text{Given }\mathrm{Var}(X)=9.25
\displaystyle \text{We know, }\mathrm{Var}(X)=\frac{1}{n}\sum_{i=1}^{n}(x_i-\overline X)^2
\displaystyle \therefore \frac{1}{8}\left[(6-9)^2+(7-9)^2+(10-9)^2+(12-9)^2+(12-9)^2+(13-9)^2+(x-9)^2+(y-9)^2\right]=9.25
\displaystyle \Rightarrow (6-9)^2+(7-9)^2+(10-9)^2+(12-9)^2+(12-9)^2+(13-9)^2+(x-9)^2+(y-9)^2=74
\displaystyle \Rightarrow 3^2+2^2+1^2+3^2+3^2+4^2+(x-9)^2+(y-9)^2=74
\displaystyle \Rightarrow (x-9)^2+(y-9)^2=26
\displaystyle \Rightarrow x^2+y^2-18(x+y)+162=26
\displaystyle \Rightarrow x^2+y^2-18(12)=-136
\displaystyle \Rightarrow x^2+y^2=80\qquad\text{... ... ... ... ... (ii)}
\displaystyle \text{Now, }(x+y)^2=x^2+y^2+2xy
\displaystyle \Rightarrow 12^2=80+2xy
\displaystyle \Rightarrow 144=80+2xy
\displaystyle \Rightarrow xy=32
\displaystyle \text{Therefore, }x\text{ and }y\text{ are the roots of the equation}
\displaystyle t^2-(x+y)t+xy=0
\displaystyle \Rightarrow t^2-12t+32=0
\displaystyle \Rightarrow (t-4)(t-8)=0
\displaystyle \Rightarrow t=4\text{ or }t=8
\displaystyle \therefore \text{The remaining two observations are }4\text{ and }8.
\displaystyle \\

\displaystyle \textbf{Question 7: }\text{For a group of }200\text{ candidates, the mean and standard deviation of the}
\displaystyle \text{scores were found to be }40\text{ and }15\text{ respectively. Later, it was discovered that}
\displaystyle \text{the scores }43\text{ and }35\text{ were misread as }34\text{ and 53 respectively. Find the correct mean}
\displaystyle \text{and standard deviation.}
\displaystyle \text{Answer:}
\displaystyle \text{Given }\overline X=40,\qquad n=200,\qquad \sigma=15
\displaystyle \overline X=\frac{\sum\limits_{i=1}^{200}x_i}{200}
\displaystyle \Rightarrow 40=\frac{\sum\limits_{i=1}^{200}x_i}{200}
\displaystyle \Rightarrow \sum\limits_{i=1}^{200}x_i=8000
\displaystyle \text{Since }34\text{ and }53\text{ were used instead of }43\text{ and }35,\text{ the corrected sum is}
\displaystyle \sum\limits_{i=1}^{200}x_i=8000-34-53+43+35=7991
\displaystyle \therefore \text{Correct mean}=\frac{7991}{200}=39.955
\displaystyle \text{Now, }\sigma^2=\frac{\sum\limits_{i=1}^{200}x_i^2}{200}-\overline X^{\,2}
\displaystyle \Rightarrow 15^2=\frac{\sum\limits_{i=1}^{200}x_i^2}{200}-40^2
\displaystyle \Rightarrow 225=\frac{\sum\limits_{i=1}^{200}x_i^2}{200}-1600
\displaystyle \Rightarrow \frac{\sum\limits_{i=1}^{200}x_i^2}{200}=1825
\displaystyle \Rightarrow \sum\limits_{i=1}^{200}x_i^2=365000
\displaystyle \text{Therefore, the corrected sum of squares is}
\displaystyle \sum\limits_{i=1}^{200}x_i^2=365000-34^2-53^2+43^2+35^2
\displaystyle =365000-1156-2809+1849+1225
\displaystyle =364109
\displaystyle \therefore \text{Correct variance}=\frac{364109}{200}-\left(\frac{7991}{200}\right)^2
\displaystyle =1820.545-1596.402025
\displaystyle =224.142975
\displaystyle \therefore \text{Correct standard deviation}=\sqrt{224.142975}
\displaystyle \approx 14.97
\displaystyle \therefore \text{The correct mean and standard deviation are }39.955\text{ and }14.97\text{ respectively.}
\displaystyle \\

\displaystyle \textbf{Question 8: }\text{The mean and standard deviation of }100\text{ observations were calculated as }
\displaystyle 40\text{ and }5.1\text{ respectively.}  \ \text{A student mistakenly took }50\text{ instead of }40
\displaystyle \text{ for one observation.} \ \text{What are the correct mean and standard deviation?}
\displaystyle \text{Answer:}
\displaystyle \text{Given }\overline X=40,\qquad n=100,\qquad \sigma=5.1
\displaystyle \overline X=\frac{\sum\limits_{i=1}^{100}x_i}{100}
\displaystyle \Rightarrow 40=\frac{\sum\limits_{i=1}^{100}x_i}{100}
\displaystyle \Rightarrow \sum\limits_{i=1}^{100}x_i=4000
\displaystyle \text{Since }50\text{ was used instead of }40,\text{ the corrected sum is}
\displaystyle \sum\limits_{i=1}^{100}x_i=4000-50+40=3990
\displaystyle \therefore \text{Correct mean}=\frac{3990}{100}=39.9
\displaystyle \text{Now, }\sigma^2=\frac{1}{n}\sum\limits_{i=1}^{n}x_i^2-\overline X^{\,2}
\displaystyle \Rightarrow 5.1^2=\frac{1}{100}\sum\limits_{i=1}^{100}x_i^2-40^2
\displaystyle \Rightarrow 26.01=\frac{1}{100}\sum\limits_{i=1}^{100}x_i^2-1600
\displaystyle \Rightarrow \frac{1}{100}\sum\limits_{i=1}^{100}x_i^2=1626.01
\displaystyle \Rightarrow \sum\limits_{i=1}^{100}x_i^2=162601
\displaystyle \text{Therefore, the corrected sum of squares is}
\displaystyle \sum\limits_{i=1}^{100}x_i^2=162601-50^2+40^2
\displaystyle =162601-2500+1600
\displaystyle =161701
\displaystyle \therefore \text{Correct variance}=\frac{161701}{100}-\left(\frac{3990}{100}\right)^2
\displaystyle =1617.01-(39.9)^2
\displaystyle =1617.01-1592.01
\displaystyle =25
\displaystyle \therefore \text{Correct standard deviation}=\sqrt{25}=5
\displaystyle \therefore \text{The correct mean and standard deviation are }39.9\text{ and }5\text{ respectively.}
\displaystyle \\

\displaystyle \textbf{Question 9: }\text{The mean and standard deviation of }20\text{ observations are found to be }10\text{ and }
\displaystyle 2\text{ respectively.}  \ \text{On rechecking, it was found that an observation }8\text{ was incorrect.}
\displaystyle \text{Calculate the correct mean and standard deviation in each of the following cases:}
\displaystyle \text{(i) If the wrong item is omitted}\qquad\text{(ii) If it is replaced by }12
\displaystyle \text{Answer:}
\displaystyle \text{Given }\overline X=10,\qquad n=20,\qquad \sigma=2
\displaystyle \overline X=\frac{\sum\limits_{i=1}^{20}x_i}{20}
\displaystyle \Rightarrow 10=\frac{\sum\limits_{i=1}^{20}x_i}{20}
\displaystyle \Rightarrow \sum\limits_{i=1}^{20}x_i=200
\displaystyle \text{Also, }\mathrm{Var}(X)=\sigma^2=2^2=4
\displaystyle \mathrm{Var}(X)=\frac{1}{20}\sum\limits_{i=1}^{20}x_i^2-\overline X^{\,2}
\displaystyle \Rightarrow 4=\frac{1}{20}\sum\limits_{i=1}^{20}x_i^2-10^2
\displaystyle \Rightarrow \frac{1}{20}\sum\limits_{i=1}^{20}x_i^2=104
\displaystyle \Rightarrow \sum\limits_{i=1}^{20}x_i^2=2080
\displaystyle \text{(i) If the wrong item is omitted}
\displaystyle \text{The corrected number of observations is }19.
\displaystyle \text{Corrected }\sum x_i=200-8=192
\displaystyle \text{Corrected }\sum x_i^2=2080-8^2=2016
\displaystyle \therefore \text{Correct mean}=\frac{192}{19}\approx 10.11
\displaystyle \text{Correct variance}=\frac{2016}{19}-\left(\frac{192}{19}\right)^2
\displaystyle =\frac{38304-36864}{361}
\displaystyle =\frac{1440}{361}
\displaystyle \therefore \text{Correct standard deviation}=\sqrt{\frac{1440}{361}}
\displaystyle =\frac{12\sqrt{10}}{19}\approx 1.997
\displaystyle \therefore \text{The correct mean and standard deviation are approximately }10.11\text{ and }2.00\text{ respectively.}
\displaystyle \text{(ii) If the wrong item is replaced by }12
\displaystyle \text{Corrected }\sum x_i=200-8+12=204
\displaystyle \text{Corrected }\sum x_i^2=2080-8^2+12^2=2160
\displaystyle \therefore \text{Correct mean}=\frac{204}{20}=10.2
\displaystyle \text{Correct variance}=\frac{2160}{20}-\left(\frac{204}{20}\right)^2
\displaystyle =108-(10.2)^2
\displaystyle =108-104.04
\displaystyle =3.96=\frac{99}{25}
\displaystyle \therefore \text{Correct standard deviation}=\sqrt{\frac{99}{25}}
\displaystyle =\frac{\sqrt{99}}{5}\approx 1.990
\displaystyle \therefore \text{The correct mean and standard deviation are }10.2\text{ and approximately }1.99\text{ respectively.}
\displaystyle \\

\displaystyle \textbf{Question 10: }\text{The mean and standard deviation of }100\text{ observations are found to be }20\text{ and }3\text{ respectively.}
\displaystyle \text{Later on, it was found that three observations were incorrect, which were recorded as }21,21\text{ and }18.
\displaystyle \text{Find the mean and standard deviation if the incorrect observations were omitted.}
\displaystyle \text{Answer:}
\displaystyle \text{Given }\overline{X}=20,\qquad n=100,\qquad \sigma=3
\displaystyle \overline{X}=\frac{\sum\limits_{i=1}^{100}x_i}{100}
\displaystyle \Rightarrow \sum\limits_{i=1}^{100}x_i=2000
\displaystyle \text{Also, }\mathrm{Var}(X)=\sigma^2=9
\displaystyle \mathrm{Var}(X)=\frac{1}{100}\sum\limits_{i=1}^{100}x_i^2-\overline{X}^{\,2}
\displaystyle \Rightarrow 9=\frac{1}{100}\sum\limits_{i=1}^{100}x_i^2-20^2
\displaystyle \Rightarrow \sum\limits_{i=1}^{100}x_i^2=40900
\displaystyle \text{When the incorrect observations }21,21\text{ and }18\text{ are omitted, }n=97.
\displaystyle \text{Corrected }\sum x_i=2000-21-21-18=1940
\displaystyle \therefore \text{Correct mean}=\frac{1940}{97}=20
\displaystyle \text{Corrected }\sum x_i^2=40900-21^2-21^2-18^2
\displaystyle =40900-441-441-324=39694
\displaystyle \therefore \text{Correct variance}=\frac{39694}{97}-\left(\frac{1940}{97}\right)^2 = 9.216
\displaystyle \therefore \text{Correct standard deviation}=\sqrt{9.216}\approx3.035
\displaystyle \therefore \text{The correct mean and standard deviation are }20\text{ and approximately }3.035\text{ respectively.}
\displaystyle \\

\displaystyle \textbf{Question 11: }\text{Show that the two formulae for the standard deviation of ungrouped data}
\displaystyle \sigma=\sqrt{\frac{1}{n}\sum\limits_{i=1}^{n}(x_i-\overline{X})^2}\text{ and }\sigma=\sqrt{\frac{1}{n}\sum\limits_{i=1}^{n}x_i^2-\overline{X}^{\,2}}\text{ are equivalent, where }
\displaystyle \overline{X}=\frac{1}{n}\sum\limits_{i=1}^{n}x_i.
\displaystyle \text{Answer:}
\displaystyle \sigma=\sqrt{\frac{1}{n}\sum\limits_{i=1}^{n}(x_i-\overline{X})^2}
\displaystyle =\sqrt{\frac{1}{n}\sum\limits_{i=1}^{n}(x_i^2-2x_i\overline{X}+\overline{X}^{\,2})}
\displaystyle =\sqrt{\frac{1}{n}\sum\limits_{i=1}^{n}x_i^2-\frac{2\overline{X}}{n}\sum\limits_{i=1}^{n}x_i+\frac{\overline{X}^{\,2}}{n}\sum\limits_{i=1}^{n}1}
\displaystyle =\sqrt{\frac{1}{n}\sum\limits_{i=1}^{n}x_i^2-\frac{2\overline{X}}{n}(n\overline{X})+\frac{\overline{X}^{\,2}}{n}(n)}
\displaystyle =\sqrt{\frac{1}{n}\sum\limits_{i=1}^{n}x_i^2-2\overline{X}^{\,2}+\overline{X}^{\,2}}
\displaystyle =\sqrt{\frac{1}{n}\sum\limits_{i=1}^{n}x_i^2-\overline{X}^{\,2}}
\displaystyle \therefore \sigma=\sqrt{\frac{1}{n}\sum\limits_{i=1}^{n}x_i^2-\overline{X}^{\,2}}
\displaystyle \therefore \text{The two formulae for the standard deviation are equivalent.}
\displaystyle \\

 

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