Answer $11$

$$\begin{array}{rl}\text{Mean}& =\frac{7+4+x+10}{4}\\ \\ 8& =\frac{21+x}{4}\\ \\ 21+x& =32\\ x& =32-21\\ x& =\boxed{{\displaystyle 11}}\end{array}$$

Answer $mode$

Answer $149cm$

$$\begin{array}{rl}\text{Ascending order}& =145,146,148,150,152,153\\ \text{No of data}& =6(even)\\ \text{Median}& =\text{Mean of two middle most data}\\ \\ & =\frac{148+150}{2}\\ \\ & =\frac{298}{2}\\ \\ \text{Median}& =149cm\end{array}$$

Answer $\text{between 52 kg and 53 kg}$

$$\begin{array}{rl}\text{Mean weight of 5 students}& =51kg\\ \text{Sum of weight of 5 students}& =51\times 5\\ & =255kg\\ \\ \text{Mean weight of 4 students}& =54kg\\ \text{Sum of weight of 4 students}& =54\times 4\\ & =216kg\\ \\ \text{Total weight of 9 students}& =255+216\\ & =471kg\\ \\ \text{Mean}& =\frac{471}{9}\\ \\ \text{Mean weight of 9 students}& =52.3kg\\ \end{array}$$

Answer $10$

$$\begin{array}{rl}\text{Mean}& =\frac{1+3+5+7+9+11+13+15+17+19}{10}\\ \\ & =\frac{100}{10}\\ \\ \text{Mean}& =10\end{array}$$

$$\begin{array}{rl}\text{Ascending order}& =1.1,1.2,1.3,1.3,1.3,1.6,1.6,1.8\\ \text{Mode}& =\boxed{{\displaystyle 1.3}}\end{array}$$

Answer $Mode=\boxed{{\displaystyle 1.3}}$

Answer $Mean,Median,Mode$

$$\begin{array}{rl}& \text{Prime numbers between 70 and 10}\\ & =71,73,79,83,89,97\\ \\ \text{Mean}& =\frac{71+73+79+83+89+97}{6}\\ \\ & =\frac{492}{6}\\ \\ \text{Mean}& =82\\ \end{array}$$

Answer $Mean=\boxed{{\displaystyle 82}}$

$$\begin{array}{rl}\text{Ascending order}& =29,36,48,62,71,84\\ \text{Number of observations}& =6(even)\\ \text{Median}& =\text{Mean of two middle most data}\\ & =\frac{48+62}{2}\\ \\ & =\frac{110}{2}\\ \\ \text{Median}& =55\end{array}$$

Answer $Median=\boxed{{\displaystyle 55}}$

Answer $NewMean=\boxed{{\displaystyle 10}}$

Solution

$$\begin{array}{rl}& \boxed{{\displaystyle \text{Mean}=\frac{\text{Sum of all observations}}{\text{Number of observations}}}}\\ \\ & =\frac{35+32+35+42+35+32+34}{7}\\ \\ & =\frac{245}{7}\\ \\ & \text{Mean}=\boxed{{\displaystyle 35}}\\ \end{array}$$

$$\begin{array}{rl}& \text{Ascending order}=32,32,34,35,35,35,42\\ \\ & \text{Number of observations}=7(\text{odd})\\ \\ \text{Median}& ={\left(\frac{n+1}{2}\right)}^{th}\text{ observation}\\ \\ & ={\left(\frac{7+1}{2}\right)}^{th}\text{ observation}\\ \\ & =4^{th}\text{ observation}\\ \\ \text{Median}& =\boxed{{\displaystyle 35}}\\ \end{array}$$

$$\begin{array}{rl}& \text{Mode}=\text{Most frequently occurring observation}\\ \\ & \text{Observations: }32,32,34,\boxed{{\displaystyle 35,35,35}},42\\ \\ & \text{Mode}=\boxed{{\displaystyle 35}}\\ \end{array}$$

Answer $Mean=\boxed{{\displaystyle 35}},Median=\boxed{{\displaystyle 35}},Mode=\boxed{{\displaystyle 35}}$

Solution

$$\begin{array}{rl}\text{Mean of 40 observations}& =160\\ \text{Sum of 40 observations}& =160\times 40\\ & =6400\\ \\ \text{Correct sum of observations}& =6400+165-125\\ & =6400+40\\ & =6440\\ \\ \text{Correct mean}& =\frac{6440}{40}\\ \\ & =161\end{array}$$

Answer Correct Mean $=\boxed{{\displaystyle 161}}$

Solution

$$\begin{array}{rl}\text{No. of observations}& =10\\ \text{Mean of 10 numbers}& =20\\ \text{Sum of 10 numbers}& =20\times 10\\ & =200\\ \\ \text{Subtracting 5 from each number}& =5\times (\text{No. of observations})\\ & =5\times 10\\ & =50\\ \\ \text{New Sum}& =200-50\\ & =150\\ \\ \text{New mean}& =\frac{150}{10}\\ \\ & =15\\ \end{array}$$

Answer New Mean $=\boxed{{\displaystyle 15}}$

Solution

$$\begin{array}{rl}& \text{Ascending order:}\\ & 33,35,41,46,55,58,64,77,87,90,92\\ & \text{Number of observations:}11(\text{odd})\\ \text{Median}& =\text{Middle most observation}\\ \text{Median}& =58\\ \\ & \text{Ascending order after replacing the data:}\\ & 33,35,43,46,55,58,64,77,87,90,99\\ \text{Median}& =\text{Middle most observation}\\ \text{New Median}& =58\\ \end{array}$$

Answer New Median $=\boxed{{\displaystyle 58}}$

Solution

$$\begin{array}{rl}& \text{Convert mixed fractions to decimals}\\ & =2.25,3.5,4.1,3.5,2.5,0.5\\ \text{Ascending order}& =0.5,2.25,2.5,\boxed{{\displaystyle 3.5,3.5}},4.1\\ \text{Mode}& =\text{Most frequently occurring observation}\\ \text{Mode}& =\boxed{{\displaystyle 3.5}}\end{array}$$

Answer Mode $=\boxed{{\displaystyle 3.5}}$

Answer