Class 11th

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.

10. From the given data we can insulate the following.

Take the assumed mean a=120 and h=10

Heigts in cm.	No. of boys fi	Mid-pointsxi		fidi	|xi - x? |	fi |xi - x? |
95-105	9	100	-2	-18	25.3	227.7
105-115	13	110	-1	-13	15.3	198.9
115-125	26	120	0	5.3	137.8
125-135	30	130	1	30	4.7	141
135-145	12	140	2	24	14.7	176.4