Standard Deviation Percentage

 The standard deviation is  most commonly used term in statistics. The relative standard deviations to consider the accuracy of compute the standard deviation of given analytical data. The Standard deviation is the square root of average squared deviation from the mean.

Standard Deviation  :

Standard  deviation is the arithmetic mean of all the deviation of observations taken about their mean

Standard equation When each of the given terms has frequency 1  
Let x₁, x₂, …, x_n be the n given observations and let M be their mean. Then, the variance σ² is given by
σ² =[ (x₁ – M)² + (x₂ – M)² + … + (x_n – M)² ] / n = Σ d²_i /n,
where the deviation from the mean, d_i = (x_i – M)
And, therefore, the standard deviation σ is given by
σ = + √{Σ(x_i – M)²/n} = √Σ d_i²/n, where d_i = (x_i – M)

Formula to Calculate Percentage of Standard Deviation:

When frequencies of the variable are given

In this case, the variance is given by

σ² = (Σ f_i d_i²/Σ f_i) & S.D.= σ = √(Σ f_i d_i²)/n

we proceed in same way  as we have done earlier

but here each d_i²  is multiplied by correponding f_i

and apply the above formula

and we get standard deviation for frequency distribution

Ex :  Find the variance and standard deviation from the following frequency distribution table:

Variable (xi)	2	4	6	8	10	12	14	16
Frequency (fi)	4	4	5	15	8	5	4	5

Sol :  We have

Variable xi	Frequency fi	fi xi	_  di = (xi – M)	di2	fi di2
2	4	8	– 7	49	196
4	4`	16	–     5	25	100
6	5	30	–  3	9	45
8	15	120	–  1	1	15
10	8	80	1	1	8
12	5	60	3	9	45
14	4	56	5	25	100
16	5	80	7	49	245
	Σ fi = 50	Σ fi xi = 450			Σ fi di2 = 754
		... M = 450/50 = 9

.^.. Variance, σ² = Σ f_i d_i²/ Σ f_i = 754/50 = 15.08

And, standard deviation, σ = √15.08 = 3.88

Percentage of standard deviation:

Percentage of standard deviation or relative standard deviation = (standard deviation / mean)  x 100.

Calculating Percentage of Standard Deviation:

Calculate the variance as well as the standard deviation percentage of the given table of the data:

xi	7	10	12	16	18	25	28
fi	2	5	13	10	6	4	1

Solution:

Presenting the data in tabular form, we get

xi	fi	fixi	(xi - mean)	(xi - mean)2	fi(xi - mean)2
7	2	14	-8	64	128
10	5	50	-5	25	125
12	13	156	-3	9	117
16	10	160	1	1	10
18	6	108	3	9	54
25	4	100	10	100	400
28	1	28	13	169	169
	41	616			1003

Here, N= 41 and `sum_(i=1)^7` fixi  = 616.

Therefore, mean = (`sum_(i=1)^7` fixi ) `-:` N = (1/41) x 616   =  15



and                `sum_(i=1)^7`fi(xi - mean)2   = 1003.

Hence, Variance(σ2)  =  (1/ N) `sum_(i=1)^7` fi(xi - mean)2  =  (1/41) x 1003   =  24.46

and

Standard deviation (σ) = `sqrt(24.46)`   =  4.94

Relative standard deviation or standard deviation percentage = (σ / mean) x 100  =  (4.94 / 15 ) x 100     =   32.9%

Division of Decimal Numbers

There are some rules of dividing the decimal numbers. Different methods of division are applied in dividing the decimal numbers.

(1)   Division of a decimal by a whole number:

Rules: (a) Divide as in division of numbers.

(b) When you reach the tenth digit, place the decimal in the quotient.

When the number of digits in the dividend is less and the division is not complete, keep adding zero at every step till the division is complete.

Example:  (1) 100.4 ÷ 25

25)100.4(4.016

100_
40-------------- Put zero

25_
150------------ Put zero

150
X

(2)   1.2 ÷ 25                                    ‌‌‌

25)1.20(0.048

100_
200               Since 12 is not multiply of 25, so 0 is added.

200_
Xx

(2)   Division by decimal numbers by 10, 100, 1000 etc.



Rules: (a) While dividing a decimal by 10, 100 or 1000 etc, multiples of 10 the decimal shifts to the left by as many places as there are zeros in the divisor.

(b) If the number of places in the integral part is less, then put the required number of zeros to the left of the integral part, then shift the decimal point.



Example: (a) 71.6 ÷ 10



=       716    ÷10
          10

           716      x    1_
=        10           10



=      716
       100

= 7.16 Ans.



(b) 923.07 ÷ 100



92307 ÷ 100

100

=      92307      x     1__
         100              100

=         92307
           10000

=        9.2307 Ans.

Rules of Division of Decimal Numbers Continued

(3)   Division of decimal numbers by multiples of 10, 100, 1000 etc.

Example: 245.1 ÷30

3)245.1(81.7

24___
x  5

3__
21

21_
X                  so, 245.1 ÷ 30 =8.17



245.1

30           = 245.1 =    245.1     x     1_

3x10        3               10

=   81.7 x    1

10       =   8.17 Ans.

(4)   Division of a decimal by a decimal:

Example: 14.7 ÷ 2.1

147 ÷ 21

10    10

=   147   x   10

10         21

= 7 Ans.

Alternative method: (a) count the number of decimal digits in the divisor.

(b) Move the decimal in dividend that many places to right.

(c) Write the divisor without the decimal.

(d) Now divide the numbers as usual.

(5)   Division of a whole number by a decimal:

Rules: (a) Count the number of decimal digits in the divisor.

(b) Add as many zeros to the dividend.

(c) Remove the decimal in the divisor.

(d) Divide as usual.

Example: 42 ÷ 0.7

42.0 ÷ 0.7

= 420 ÷ 7 = 60 Ans.

Exercise of Decimal Number Fractions

Divide: (1) 234.65 by 25

(2) 193.92 by 800

(3) 8.16 by 0.24

(4) 68 by 4.25

(5) 44 by 176

Answer:  (1) 9.386 (2) 0.2424 (3) 34 (4) 16 (5) 0.25

Trig Equation Solving Examples

Trigonometric equations is shortly called as trig equations.  An equation involving trigonometrical function is called a trigonometrical equation.

cosθ =`1/2` , tanθ = 0, cos2θ − 2sinθ =`1/2 `

There are some examples for trigonometrical equations. To solve these equations we find all replacements for the variable θ that make the equations true. A solution of a trigonometrical equation is the value of the unknown angle that satisfies the equation. A trigonometrical equation may have infinite number of solutions. The solution in which the absolute value of the angle is the least is called principal solution. In this article let us study trig equations solving examples.

Trig Equation Solving Examples:

Let us see sample problems for trig equation solving examples.

General solutions of sin θ = 0 ; cosθ = 0 ; tan θ = 0

Find the principal value of the following:

(i) cosx =`sqrt3/2`

Solution: (i) cosx =`sqrt3/2` > 0

∴ x lies in the first or fourth quadrant. Principal value of x must be in[0, π]. Since cosx is positive the principal value is in the first quadrant

cosx =`sqrt3/2` = cos`pi/6` and `pi/6`

∈ [0, π]∴ The principal value of x is `pi/6` .

(ii) cosθ = −`sqrt3/2` < 0

Since cos θ is negative, θ lies in the second or third quadrant. But the  principal value must be in [0, π] i.e.  Within  1st  or  2nd  quadrant. The principal value is in the 2nd quadrant.

cosθ = −`sqrt3/2` = cos (180° − 30°) = cos150°.

The principal value is θ = 150° =`(5pi)/6` .

Trig Equation Solving Examples:

Projection formula

In any triangle ABC  a = b cos C + c cosB

True with usual notations and it is called projection formula.

Proof:

In triangle ABC, draw AD perpendicular to BC.From the right angled triangles ABD and ADC,

cosB =BD

AB ⇒ BD = AB × cosB

cosC =DC

AC ⇒ DC = AC × cosC

But BC = BD + DC = AB cosB + AC cosC

a = c cosB + b cosC

or a = b cosC + c cosB


Solve : sin2x + sin6x + sin4x = 0

Solution:

sin2x + sin6x + sin4x = 0 or (sin6x + sin2x) + sin4x = 0 or 2sin4x. cos2x + sin4x = 0

sin4x (2 cos2x + 1) = 0

when sin4x = 0 ⇒ 4x = nπ or x =`(npi)/4` ; n ∈ Z

When 2 cos2x + 1 = 0 ⇒ cos 2x =− `1/2`

= − cos`pi/3` = cos (π −π3)= cos`(2pi/3)`

∴ 2x = 2nπ ±`(2pi)/3` or x = n π ±`pi/3`

Hence x =`(npi)/4` or x = nπ ±`pi/3`

; n ∈ Z

Please visit this website and Know more on Derivative of Cosine .

The Number -‘One followed by Six zeros’

The number system started with the natural number system .Now we have huge numbers
being represented. They can be either positive or negative. They can be fractions. They can
be rational or irrational. The natural numbers start with 1 and extend up to infinity. One
million is also natural number. One million is nothing but 10 lakhs. It is the number of zeros
that follow ‘1’ that is important. The number of zeros decides the value of the number.
In ten lakhs, we have six zeros. In ten we have one zero. In hundred two zeros, thousand
three zeros, ten thousand four zeros, lakh five zeros and finally in ten lakhs six zeros. So,
the number of zeros is very important. Now we will try to write 1 million in numbers to
understand this. It comes after the number 999999 in the natural number system. It is also
the number which comes before the number 1000001 in the natural number system. So,
we now understand how to write 1 million after coming to know its position in the natural
number system. Expressing  million in numbers is quite easy and counting up to a million can
be a difficult task.

In the modern era we use the term a million rather than saying ten lakhs or thousand
thousands to represent the same number. Using this term has become very easy. So, we
need to know how to write 1 million in numbers as this term is very commonly used. The
word million is derived is one of the most beautiful languages in the world, Greek. This
shows how to write one million in numbers and can be clearer if we understand it to be just
a number on the natural number system number line.

Just studying a million is not enough we need to study its multiples too. If we do that
we can be clearer with the concepts. So, how to write millions in numbers if there are
numbers greater than a million. We now come to the concept of a billion. In a billion there
are nine zeros. One is followed by nine zeros to form a billion. This shows that there are
1000 millions in a billion. A billion is also known as a milliard. The natural number system
further increases to trillion, quadrillion and so on. A trillion consists of a one followed by 12
zeros.These are very large numbers.

Derivative Math Problems

In math, differentiation is the process of finding the derivative which means measuring how a function changes with respect to its input. The derivative of y with respect to x is given by`(dy)/(dx)` . The reverse process of derivative is antiderivative. Following is the example and practice problems of math derivative which helps you for learning derivative in math.

Example Problems of Math Derivative:

Learn math derivative with Example problem 1:

Find the derivative of the function y = 7x8 + 2x6 + 5x5

Solution:

Step 1: Given function

y = 7x8 + 2x6 + 5x5

Step 2: Differentiate the given function y = 7x8 + 2x6 + 5x5 with respect to ' x ', to get `(dy)/dx`

`(dy)/dx` = 56x7 + 12x5 + 25x4

Learn math derivative with example problem 2:

Find the derivative of the function y = sin (7x4 + 1)

Solution:

Step 1: Given function

y = sin (7x4 + 1)

Step 2: Differentiate the given function y = sin (7x4 + 1) with respect to ' x ', to get `(dy)/dx`

`(dy)/dx` = [cos(7x4 + 1)] (28x3)

= 28x3 cos (7x4 + 1)

Learn math derivative with example problem 3:

Find the derivative of the function y = ecos(9x - 14)

Solution:

Step 1: Given function

y = ecos (9x - 14)

Step 2: Differentiate the given function y =  ecos (9x - 14) with respect to ' x ', to get `(dy)/dx`

`(dy)/dx` = ecos (9x - 14) (- 9sin (9x - 14))

= - 9 ecos (9x - 14) sin (9x - 14)

Practice Problems of Math Derivative:

A few practice problems are given below with solutions which gives you idea how to differentiate the function.

1) Find the derivative of the function y = 16x2 + 11x + 5

2) Find the derivative of the function y = - 5tan (7x2 + 2)

3) Find the derivative of the function y = `6/5` e sin (5x - 9)

Solutions:

1) 32x + 11

2) - 70x sec2 (7x2 + 2)

3) 6e sin (5x - 9) cos (5x - 9)

Solving Equations with Exponents

Introduction:

Exponent equations are the equations in which variable appear as an exponent.

To solve these equations rules and laws of exponents are used. Exponent equations are of two types
(1) Exponent equations in which bases are same
(2) Exponent equations in which bases are different.

Steps to Solve Equations with Exponent

Solving Exponential Equations of the same base

1) Ignore the bases, and simply set the exponents equal to each other
2) Solve for the variable

 When the bases of the terms are different

1) Ignore the exponents; rewrite both of the bases as powers of same number.
For example if there are 2 and 4 in the bases, then convert base 4, in to base 2
by writing it again as (2)^2
2) once the bases are same , ignore them
3) Equalize the exponents
4) Solve for variable

Simple Problems of Equations with Exponents

Solve for variable Answer 1. 3m  =  35 Since the bases are the same, set the exponents equal to one another: m = 5 2. 5t   = 125 125can be expressed as a power of 5: 5t  = 53 t = 3 3.  493y=343 49 and 343 can be expressed as a power of 7: [(7)2]3y = 73 76y = 73 6y = 3 y = 1/2

More Problems of Equations with Exponents

Solve for x. Answer 1.  52x+1  =  53x-2 Since the bases are the same, set the exponents equal to one another: 2x + 1 = 3x - 2 3 = x 2.  32x-1  = 27x 27 can be expressed as a power of 3: 32x-1  = 33x 2x - 1 = 3x -1 = x 3.   43x-8  = 162x 16 can be expressed as a power of 4: 43x-8= [(4)2]2x 3x - 8 = 4x   -8 = x

Units of Meters

In this article discuss about standard units (international system) of meters. The meter (or meter), symbol m, is the base unit of length in the International System of Units (SI).The meters is the length of path travelled by light in vacuum during a time interval of 1/299792458 of a seconds.

The Basic Units of Length is Meters (m):

Linear measures     Short form of writing

Millimeter                         mm

Centimeter                         cm

Decimeter                          dm

Meter                                m

Decameter                         dam

Hectometer                         hm

Kilometer                           km

Some examples of units values:

1 mm = 0.001 m

1 cm = 0.01 m

1 dc = 0.1 m

1 km = 1000 m

Units of Meters - Examples:

Units of meters - Example 1:

Convert 378.6 cm into meter

Solution:

Here the conversion is from centimeter to meter. i.e. from lower unit to higher unit

100cm = 1m.

Hence 378.6cm = 378.6/100 m

= 3.786 m (shifting the decimal two digits to the left)

Units of meters - Example 2:

Convert 40.1735 km into meter.

Solution:

Here the conversion is from higher to lower. Hence we have to shift the decimal point to the right.

1km = 1000 m

40.1735 km = 40.1735 × 1000 m

= 40173.5 m

Units of meters - Example 3:

Convert 6m into millimeter.

Solution:

Note: Here we are converting the unit meter into millimeter. i.e. from higher unit to lower unit. So the operation should be multiplication.

1 m = 1000 mm

Hence 6 m = 6 × 1000 mm

= 6000 mm

Units of meters - Example 4:

Convert 778.6 cm into meter

Solution:

Here the conversion is from centimeter to meter. i.e. from lower unit to higher unit

100cm = 1m.

Hence 778.6cm = 778.6/100 m

= 7.786 m (shifting the decimal two digits to the left)

Units of meters - Example 5:

Convert 90.1736 km into meter.

Solution:

Here the conversion is from higher to lower. Hence we have to shift the decimal point to the right.

1km = 1000 m

90.1736 km = 90.1736 × 1000 m

= 90173.6 m

Units of meters - Example 6:

Convert 9m into millimeter.

Solution:

Note: Here we are converting the unit meter into millimeter. i.e. from higher unit to lower unit. So the operation should be multiplication.

1 m = 1000 mm

Hence 9 m = 9 × 1000 mm

= 9000 mm.