Class 11th

22. The given data is converted into continuous frequency duration by subtracting and adding 0.5 from lower and upper limit respectively. Lit the assumed mean be A=42.5 and h=4

21. Let the assumed mean be A=92.5 and h=5

20.

We have,  $N={\displaystyle \sum _{i=1}^{n}fi}=50$ .

So, mean,  $\overline{x}=\frac{1}{N}*{\displaystyle \sum _{i=1}^{n}fi}xi=\frac{1350}{5}=27.$

$\begin{array}{c}=\frac{1}{50}*6600\\ \end{array}$

= 132.

19. Let the assumed mean be A=105 and class width, h=30. The given data can be tabulated as

= 2280 - 4

= 2276.

18. Let the assumed mean be A=64 and it the width, h=1.

17. The given data can be tabulated as follow

16. The given data can be tabulated as follow.

15. We have, first 10 multiples of 3=3,6,9,12,15,18,21,24,27,30.

So,  $\overline{x}=\frac{3+6+9+12+15+18+21+24+27+30}{10}=\frac{165}{10}=16.5$

We can now tabulate the given data as following.

Therefore, variance,  $a^2=\frac{1}{n}{\displaystyle \sum _{i=1}^n{\left(x_i-\overline{x}\right)}^2}$

$\begin{array}{c}=\frac{1}{10}*742.5\\ \end{array}$

= 74.25.

14. We know that,

Sum of first'n ' natural no $=\frac{n(n+1)}{2}$

So, mean, $\begin{array}{c}\overline{x}=\frac{\text{\hspace{0.17em}sumoffirst}\left(n\right)\text{\hspace{0.17em}naturalno}.=}{}\frac{n(n\text{\hspace{0.17em}+1})/2}{n}\\ no\; of\; observations\end{array}$

$=\frac{n+1}{2}$

So, Variance, $a^2=\frac{1}{n}{\displaystyle \sum _{i=1}^n{\left(x_i-\overline{x}\right)}^2}=\frac{1}{n}{\displaystyle \sum _{i=1}^n{\left(x_i-\left(\frac{n+1}{2}\right)\right)}^2}$

$\begin{array}{l}{a}^2=\frac{1}{n}\left[{\displaystyle \sum _{i=1}^n{x}_i^2}-{\displaystyle \sum _{i=1}^{n}2}{x}_i\left(\frac{n+1}{2}\right)+{\displaystyle \sum _{i=1}^n{\left(\frac{n+1}{2}\right)}^2}\right]\\ \_\_\_\_\_\_\_(1)\end{array}$

So, $\begin{array}{c}{\displaystyle \sum _{i=1}^n{x}_i^2}={(1)}^2+{(2)}^2+{(3)}^2+\dots +{(n)}^2\\ =\frac{n(n+1)(2n+1)}{6}\_\_\_\_\_\left(2\right).\end{array}$

And ${\displaystyle \sum _{i=1}^n{\left(\frac{n+1}{2}\right)}^2}=\frac{{(n+1)}^2}{4}{\displaystyle \sum _{i=1}^{n}1}.=\frac{n{(n+1)}^2}{4\cdot }\_\_\_\_\left(4\right).$

Putting (2), (3) and (4) in (1) we get,

$\begin{array}{c}{a}^2=\frac{1}{n}\left[\frac{n(n+1)(2n+1)}{6}-\frac{n{(n+1)}^2}{2}+\frac{n{(n+1)}^2}{4}\right]\\ \end{array}$

$\begin{array}{c}=\frac{(n+1)(2n+1)}{6}-\frac{{(n+1)}^2}{2}+\frac{{(n+1)}^2}{4}\\ \end{array}$

$\begin{array}{c}=\frac{(n+1)(2n+1)}{6}-\frac{{(n+1)}^2}{4}.\\ \end{array}$

$\begin{array}{c}=(n+1)\left[\frac{2n+1}{6}-\frac{(n+1)}{4}\right]\\ \end{array}$

$=(n+1)\left[\frac{4n+2-3n-3}{12}\right]$

$=\frac{(n+1)(n-1)}{12}=\frac{n^2-1}{12}.$