Maths Statistics

Since a frequency polygon represents the data distribution, by interpreting the areas under curve, it is possible to infer probabilities for some range and interval. Let us take a look at it:

• Understanding the Data: You must be able to comprehend the data represented by frequency polygon. In most cas

The following are different types of frequency polygons:

• Simple Frequency Polygon: This is a standard form that connects the midpoints of tops of the bars in a histogram with straight lines.
• Relative Frequency Polygon: This type of frequency polygon uses relative frequencies (proportions or percentage

The following points highlight the importance of frequency polygons:

• Frequency polygons are useful for comparing distributions of multiple datasets on the same graph. It becomes easy to visually compare shapes and trends of different datasets by overlaying multiple frequency polygons.
• These use lines

The Interquartile range is also a measure of statistical dispersion that indicates the range within which middle 50% of dataset remains. It is the difference between the third and first quartile of the given dataset. Also known as IQR, it represents the length of the box which illustrates the spread

Q3 denotes the third Quartile which is the statistical measure to represent value below which 75% of the data falls in dataset when arranged in ascending order. It is the median of all the observations of the upper half of the data set i.e. 25% of the data. It is commonly used in conjunction with Q1

$\left(\begin{array}{cc}\hfill 1+i\hfill & \hfill 1\hfill \\ \hfill -i\hfill & \hfill 0\hfill \end{array}\right)$

$A^2=\left(\begin{array}{cc}\hfill 1+i\hfill & \hfill 1\hfill \\ \hfill -i\hfill & \hfill 0\hfill \end{array}\right)\left(\begin{array}{cc}\hfill 1+i\hfill & \hfill 1\hfill \\ \hfill -i\hfill & \hfill 0\hfill \end{array}\right)=\left(\begin{array}{cc}\hfill i\hfill & \hfill 1+i\hfill \\ \hfill 1-i\hfill & \hfill -i\hfill \end{array}\right)$

$A^4=\left(\begin{array}{cc}\hfill i\hfill & \hfill 1+i\hfill \\ \hfill 1-i\hfill & \hfill -i\hfill \end{array}\right)\left(\begin{array}{cc}\hfill i\hfill & \hfill 1+i\hfill \\ \hfill 1-i\hfill & \hfill -i\hfill \end{array}\right)=\left(\begin{array}{cc}\hfill 1\hfill & \hfill 0\hfill \\ \hfill 0\hfill & \hfill 1\hfill \end{array}\right)$

$\therefore A^5=A$

A⁹ = A

$\therefore forn=1,5,9,.....,97$

total possible values of n = 25

$Mean=\frac{3+12+7+a+\left(43-a\right)}{5}=13$

Variance = $\frac{3^2+12^2+7^2+a^2+{\left(43-a\right)}^2}{5}-{\left(13\right)}^2$

$\Rightarrow \frac{2a^2-a+1}{5}\to Naturalnumber$

Let 2a² – a + 1 = 5x

D = 1 – 4 (2) (1 – 5n)

= 40n – 7, which is not $4\lambda or4\lambda +1from.$

As each square form is $4\lambda or4\lambda +1$

$l=\frac{48}{{\pi }^4}{\displaystyle {\int }_0^{\pi }\left[{\left(\frac{\pi }{2}-x\right)}^3-\frac{3{\pi }^2}{4}\left(\frac{\pi }{2}-x\right)+\frac{{\pi }^3}{4}\right]\frac{sinxdx}{1+co{s}^{2}x}}$

Using

${\displaystyle {\int }_a^{b}f\left(x\right)dx={\displaystyle {\int }_a^{b}f\left(a+b-x\right)dx}}$

we get

$l=\frac{48}{{\pi }^4}{\displaystyle {\int }_0^{\pi }\left[-{\left(\frac{\pi }{2}-x\right)}^3+\frac{3{\pi }^2}{4}\left(\frac{\pi }{2}-x\right)+\frac{{\pi }^3}{4}\right]\frac{sinxdx}{1+co{s}^{2}x}}$

Adding these two equations, we get

$\Rightarrow l=\frac{12}{\pi }{\left[-ta{n}^{-1}\left(cosx\right)\right]}_0^{\pi }=\frac{12}{\pi }.\frac{\pi }{2}=6$