Maths Statistics

$\sum X=200$

$\sum x^2=2125$

For new data

$\sum x=200-25+35=210$

$\overline{x}=10.5$

$\sum x^2=2725$

${\sigma }^2=\frac{2725}{20}-{\left(10.5\right)}^2=26$

$mean=\frac{n\left(n+1\right)/2}{n}$           

  $=\frac{n+1}{2}$            For n = 2k – 1, k   $\in N$

mean = k

mean deviation = avg. of deviations

$=\frac{2\left(\stackrel{0+}{1}+2+...\left(k-1\right)\right)}{2k-1}=\frac{\left(k-1\right)k}{2k-1}$

$=\frac{5\left(2k\right)}{k-1}\Rightarrow k=11\Rightarrow n=2k-1=21$            

               

Z distribution and Chi-Squared are some of the most popular distribution patterns of probability, and it is vital to recognise the variations between them and when to use the distribution pattern. A Z table is of no use when the operation revolves around a smaller sample size. On the other hand, the distribution of a sum of independent regular k squares in standard normal variables is the chi-square distribution of k degrees of freedom. The tests are used for the independence of two variables in an incident table and to assess the observable data for 

Median = $\frac{2k+12}{2}=k+6$

Mean deviation = $\sum \frac{\left|x_i-M\right|}{n}=6$

$\Rightarrow \frac{\left(k+3\right)+\left(k+1\right)+\left(k-1\right)+\left(6-k\right)+\left(6-k\right)+\left(10-k\right)+\left(15-k\right)+\left(18-k\right)}{8}$

$\therefore \frac{58-2k}{8}=6$

k = 5

$Median=\frac{2*5+12}{2}=11$

Arithmetic mean is the measure of central tendency that is most affected by extreme items or outliers in the dataset. The reason behind this is since mean is calculated by summing up all the values and after that, it is divided by the number of values. Extreme values may disproprotionately impact the sum which in turn skews the mean and make it less represenative of dataset as whole.

Median is considered to be the most suitable average for qualitative measurement. It divides an entire frequency distribution into two haves. This is especially useful for ordinal data where values represent categories with meaningful order. However, it is not necessarily a linear scale. The median gives a cental value which is less influenced by extreme values or outliers. This is important while dealing with qualitative data which may not be either symmetrically scaled or evenly distributed.

$\overline{x}=\frac{{\displaystyle \sum _{i=1}^{10}{x}_i}}{10}=15\frac{{\displaystyle \sum _{i=1}^{10}{x_i}^2}}{10}-{\left(\overline{x}\right)}^2=15$

$\Rightarrow \Sigma x_i=150\Sigma {x_i}^2=2400$

Actual mean $\overline{x}=\frac{\Sigma x_i+15-25}{10}=\frac{140}{10}=14$

Actual variance = $\frac{\Sigma {x_i}^2+15^2-25^2}{10}-{\left(14\right)}^2$

$=\frac{2400-400}{10}-196$

$\begin{array}{l}{\sigma }^2=4\\ \sigma =2\end{array}$