Maths

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,

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

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

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

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

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

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

13. The given data can be tabulated as.

we have,

mean,  $\overline{x}=\frac{{\displaystyle \sum _{i=1}^n{x}_i}}{n}=\frac{6+7+10+12+13+4+8+12}{8}=\frac{72}{8}=9.$

So, variance,  $\begin{array}{c}{a}^2=\frac{1}{8}{\displaystyle \sum _{i=1}^n{\left(x_i-\overline{x}\right)}^2}\\ \end{array}$

$=\frac{1}{8}*74=9.25$

12. The given data is made'continuous by subtracting 0.5 from the lower limit and adding 0.5 to the upper limit of each class. So, we cam tabulate as.

11. From the given data we can tabulate the following.