Dictionary Rank of a Word | Permutations & Combinations

 PERMUTATIONS & COMBINATIONS Rank of the word or Dictionary order of the English words like COMPUTER, COLLEGE, SUCCESS, SOCCER, RAIN, FATHER, etc. Dictionary Rank of a Word Method of finding the Rank (Dictionary Order) of the word  “R A I N” Given word: R A I N Total letters = 4 Letters in alphabetical order: A, I, N, R No. of words formed starting with A = 3! = 6 No. of words formed starting with I = 3! = 6 No. of words formed starting with N = 3! = 6 After N there is R which is required R ----- Required A ---- Required I ---- Required N ---- Required RAIN ----- 1 word   RANK OF THE WORD “R A I N” A….. = 3! = 6 I……. = 3! = 6 N….. = 3! = 6 R…A…I…N = 1 word 6 6 6 1 TOTAL 19 Rank of “R A I N” is 19 Method of finding the Rank (Dictionary Order) of the word  “F A T H E R” Given word is :  "F A T H E R" In alphabetical order: A, E, F, H, R, T Words beginni

Statistics Formulas & Basic Concepts



Statistics  Formulas & Basic Concepts
Statistic formulas for class 8th, 9th, 10th, Calculation of mean, mode, median, mean deviation
 about mean & median, standard deviation, variance.
https://dinesh51.blogspot.com
Statistics For Class 10

MEAN FOR UNGROUPED DATA
\[Mean=\frac{Sum \; of \; all \; observations}{Number\; of \; observations}\]
\[Mean\left ( \overline{X} \right )=\frac{\sum x_{i}}{n}\]
MEAN FOR A GROUPED DATA
 There are three methods for this
1) DIRECT METHOD
\[Mean\left ( \overline{X} \right )=\frac{\sum f_{i}x_{i}}{\sum f_{i}}\; \; or\; \; \; \; \frac{\sum f_{i}x_{i}}{n}\]
\[Where:\; \; x_{i} = Mid \; values,\; \; x_{i} = \frac{Lower\: limit+Upper\: limit}{2}\]\[\sum f_{i} = sum \; of\; all \; frequencies\]
Table for finding Mean by Direct Method

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\[x_{i}\]

\[f_{i}\]

\[f_{i}x_{i}\]

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\[\sum f_{i}\]

\[\sum f_{i}x_{i}\]

2) ASSUMED MEAN METHOD
\[Mean\left ( \overline{X} \right )=a+\frac{\sum f_{i}d_{i}}{\sum f_{i}}\; \; or\; \; \; \;a+ \frac{\sum f_{i}d_{i}}{n}\]
\[Where:\; \; a=Assumed\; \; Mean,\; \; d_{i}=x_{i}-a\]
3) STEP DEVIATION METHOD
\[Mean\left ( \overline{X} \right )=a+h\times \frac{\sum f_{i}u_{i}}{\sum f_{i}}\; \; or\; \; \; \; a+h\times \frac{\sum f_{i}u_{i}}{n}\]
\[Where:\; \; h=height \; of \; the \; interval,\; \; u_{i}=\frac{x_{i}-a}{h}\]
Table for finding Mean by Step Deviation Method

C-I

\[x_{i}\]

\[f_{i}\]

\[u_{i}=\frac{x_{i}-a}{h}\]

\[f_{i}u_{i}\]

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\[\sum f_{i}\]

 

\[\sum f_{i}u_{i}\]

MODE
 (FOR UNGROUPED DATA)
Most frequent observation is called mode, in other words, the observation whose frequency is maximum is called mode.
MODE :- (FOR GROUPED DATA)
First of all find the modal class
MODAL CLASS:- Class corresponding to the highest frequency is called the modal class. 
\[MODE=l+\left ( \frac{f_{1}-f_{0}}{2f_{1}-f_{0}-f_{2}} \right )\times h\]
                     Where:-\[l=Lower \: limit \: of\: the\: modal\: \: class\]
\[f_{1}=Frequency\: \: of\: the\: modal\: class\]
\[f_{0}=Frequency\: \: of\: the\:class\: preceding\: the\: modal\: class\]
\[f_{2}=Frequency\: \: of\: the\:class\: succeeding \: the\: modal\: class\]
MEDIAN
FOR UNGROUPED DATA
ALGORITHM:-  a) First write the given observation in ascending order.
b)Find the number of terms let it is n.
\[If\; n\; is\; odd \; then\; \; Median= \left ( \frac{n+1}{2} \right )^{th}term\]
\[If\; n\; is\; even \; then,\; \; Median= \frac{1}{2}\left [ \left ( \frac{n}{2} \right )^{th}term\; +\left ( \frac{n}{2}+1 \right )^{th}term \right ]\]
MEDIAN FOR GROUPED DATA  
ALGORITHM:-     1) Find the cumulative frequency of the observations.
 \[2)\; Find\; \; \sum f_{i}=N,\; then\; find\; \; \frac{N}{2}\]
3) Then find the Median class and then Median.
\[Median \; Class= Class \; corresponding \; to \; the\; c.f\geq \frac{N}{2}\; is\; called\: \; Median\; Class\]\[Median=l+\left ( \frac{\frac{N}{2}-c.f}{f} \right )\times h\]
\[Where\: \: l=Lower \: limit \: of\: Median\: Class\]
\[c.f=Cumulative\: frequency \: of\: the\: class\: preceding \: the\: Median\: Class\]
\[f=Frequency\; of\: \: the\: Median\; \; Class\]
\[h=Height \: of\: the\: Median\: Class\]
Table for finding the Median

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\[f_{i}\]

cf

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\[\sum f_{i}\] 

 

EMPIRICAL FORMULA 
It is the formula which shows the relationship between mean, mode and median
Mode = 3 Median - 2 Mean
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Statistics for Class 10+1

Measure of central tendency:
Representative value of the given data is called the central tendency. Meam, Mode, Median are the three measures of central tendency. A measure of central tendency gives us a rough idea where the  data points are centered.

Dispersion
Measures of central tendency are not sufficient to give complete information about a given data. Variability is another factor which is required to be studied under statistics. Like measures of central tendency we want to have a single number to describe variability. This single number is called Dispersion.

Measures of Dispersion:

There are following measures of Dispersion:

i) Range,   ii) Quartile Deviation,  iii) Mean Deviation   iv) Standard Deviation

In this chapter we will study all measures except Quartile Deviation.

Range :  Maximum Value - Minimum value

Mean Deviation:\[Mean \: of \: Deviations=\frac{Sum\: of\: deviations}{Number\: of\: Deviations}\]In this chapter we discuss two types of Mean Deviation.

i) Mean Deviation about mean   ii) Mean Deviation about Median

MEAN DEVIATION ABOUT MEAN

FOR UNGROUPED DATA
\[Mean \; Deviation \; About(\overline{X})=\frac{\sum \left | x_{i}-\overline{x} \right |}{n}\]
FOR GROUPED DATA
\[Mean \; Deviation \; About(\overline{X})=\frac{\sum f_{i}\left | x_{i}-\overline{x} \right |}{n}\]

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\[x_{i}\]

\[f_{i}\]

\[\left |x_{i}-\bar{x} \right |\]

\[f_{i}\left |x_{i}-\bar{x} \right |\]

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 \[\sum f_{i}\]

 

 \[\sum f_{i}\left |x_{i}-\bar{x} \right |\]


MEAN DEVIATION ABOUT MEDIAN

FOR UNGROUPED DATA
\[M.D.(M)=\frac{\sum \left | x_{i} -M\right |}{N}, \:\: \: Where\, M=Median\]
FOR GROUPED FREQUENCY
\[M.D.(M)=\frac{\sum f_{i}\left | x_{i} -M\right |}{N}, \:\: \: Where\, N=Sum \; of\; all\; frequencies.\]

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\[x_{i}\]

\[f_{i}\]

\[\left |x_{i}-M \right |\]

\[f_{i}\left |x_{i}-M \right |\]

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\[\sum f_{i}\]

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Note: For calculating mean class interval may or may not be continuous. But for calculating median class intervals should be continuous.
Limitations
While calculating Mean deviation about mean and about median we take absolute values and ignore the negative sign. So this calculation is not become very scientific. Also in many cases it gives unsatisfactory results.
This imply that we need another measure of Dispersion. Standard Deviation is such a measure of central tendency.
Variance and Standard Deviation.
While calculating mean deviation about mean or median, the absolute values of the deviations were taken otherwise deviations may cancel among themselves.

Another way to overcome this difficulty which arose due to the sign of deviations, is to take square of all the deviations. Mean of squares of deviations about mean is called Variance.

VARIANCE
VARIANCE:- Mean of the squares of the deviations from the mean is called the variance.
VARIANCE FOR UNGROUPED DATA
\[Variance(\sigma ^{2})=\frac{\sum \left ( x_{i}-\overline{x} \right )^{2}}{N}, Where \: N \; is\; the\; sum\; of\; all\; frequencies\]
VARIANCE FOR GROUPED DATA
\[Variance(\sigma ^{2})=\frac{\sum f_{i}\left ( x_{i}-\overline{x} \right )^{2}}{N}, Where \: N = \sum f_{i}\]
Important Result
\[Derivation\: of\: \frac{1}{N}\sum \left ( x_{i}-\bar{x} \right )^{2}=\frac{\sum (x_{i})^{2}}{N}-\left (\bar{x} \right )^{2}\]
\[\frac{1}{N}\sum \left ( x_{i}-\bar{x} \right )^{2} =\frac{\sum (x_{i})^{2}}{N}+ \frac{\sum (\bar{x})^{2}}{N}-\frac{\sum 2x_{i}\bar{x}}{N}\]
\[\frac{1}{N}\sum \left ( x_{i}-\bar{x} \right )^{2} =\frac{\sum (x_{i})^{2}}{N}+ \frac{(\bar{x})^{2}N}{N}-2\bar{x}\left (\frac{\sum x_{i}}{N} \right )\]
\[\frac{1}{N}\sum \left ( x_{i}-\bar{x} \right )^{2} =\frac{\sum (x_{i})^{2}}{N}+ (\bar{x})^{2}-2\bar{x}\bar{x}\]
\[\frac{1}{N}\sum \left ( x_{i}-\bar{x} \right )^{2} =\frac{\sum (x_{i})^{2}}{N}+ (\bar{x})^{2}-2(\bar{x})^{2}\]
\[\frac{1}{N}\sum \left ( x_{i}-\bar{x} \right )^{2} =\frac{\sum (x_{i})^{2}}{N}-(\bar{x})^{2}\]
NCERT Exercise 15.2
Q 2:  Find mean and variance of first n natural numbers?
Ans:  Let first n natural numbers are  1, 2, 3, 4, ................, N
\[Mean(\bar{X})=\frac{1+2+3....n}{n}=\frac{n(n+1)}{2n}=\frac{n+1}{2}\]
\[Variance(\sigma^{2} )=\frac{1}{n}\sum (x_{i}-\bar{x})^{2}\]\[=\frac{\sum \left (x_{i} \right )^{2}}{n}-(\bar{x})^{2}=\frac{n(n+1)(2n+1)}{6n}-\left ( \frac{n+1}{2} \right )^{2}\]\[=\left (\frac{n+1}{2} \right )\left [ \frac{2n+1}{3}-\frac{n+1}{2} \right ]=\left ( \frac{n+1}{2} \right )\left ( \frac{n-1}{6} \right )\]\[=\frac{n^{2}-1}{12}\]
SHORT-CUT METHOD OF FINDING THE VARIANCE
\[Variance(\sigma ^{2})=\frac{h^{2}}{N}\left [ \sum f_{i}y_{i}^{2}-\frac{\left ( \sum f_{i}y_{i} \right )^{2}}{N} \right ]\]
\[Variance(\sigma ^{2})=\frac{h^{2}}{N}\left [ \sum y_{i}^{2}-\frac{\left ( \sum y_{i} \right )^{2}}{N} \right ], if\; f_{i}\; is\; not \; \: given\]\[Where \; \; y_{i}=\frac{x_{i}-a}{h}\]

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\[x_{i}\]

\[f_{i}\]

\[y_{i}=\frac{x_{i}-a}{h}\]

\[f_{i}y_{i}\]

\[f_{i}y_{i}^{2}\]

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\[\sum f_{i}\]

 

\[\sum f_{i}y_{i}\]

\[\sum f_{i}y_{i}^{2}\]

STANDARD DEVIATION 
\[Standard Deviation (S.D.) = \sqrt{Variance},\; It\; is \; \; denoted \; \; by\; \; "\sigma "\]
COEFFICIENT OF VARIATION
\[C.V=\frac{Standard \; \; Deviation}{Mean}\times 100\; \; or\; \; C.V=\frac{\sigma }{\overline{x}}\times 100\]
If C.V is more then observations are more variable, unstable and less consistence and vice-versa.
STATISTICS-CBSE Mathematics

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