REVIEW

LEAST WHOLE NORMAL FUNCTION

"As you read this manuscript, you will see that the least whole normal function, l (τ) = α.τ-√τ-β which I have derived from the normal approximation function or equation is one of the most powerful and useful mathematical function at our time.
It provides us with the necessary tools to analysis information, make predictions and arrive at a decision in virtually every area of human endeavor.
The mathematical of the least whole normal function requires two steps:
    (1)The construction of mathematical models which approximate the physical situations.
    (2)The solutions of the resulting mathematical problems.
To understand God’s purpose, we must study least whole normal model for these are the measure of his purpose. The main purpose of the least whole normal function, like that of any other mathematical function, is to provide a useful model of the real world. All least whole normal models are simplified representations of the reality, ignoring complexities which will have unimportant effects on the final outcome.
The unique feature of least whole normal modeling is that it uses randomness to model those parts of a situation where the details of the process which generate the outcome are unknown, assigning probabilities to possible outcome, rather than predicting definitely which will occur. The least whole normal function constitutes two main categories of models:

a)       Non-product model

b)       Product model

THE FIRST PROOF OF THE LWN FUNCTION

If n is the frequency population of population N and µ is the frequency mean of mean ʘ at population n and N respectively. The time in which µ is occurring is given as τ. The function of the parameters is given as; ɭ(τ)=ατ-√τ-β  where α and β are denoted as quantile coefficient of τ and quantile constant respectively.

The expected mean of the population n is E_s=ʘµτ and its variance is given as V_s=(ʘs²τ+µ²σ²τ)  where s² is variance of the population n, and σ² is variance of population N. At normal approximation;
P(n>-n)=P[n>-(n-ʘµτ)/√(ʘs²τ+µ²σ²τ)]=ϒ%

This implies that;  
-n+ʘµτ=ф^-1(ϒ%)√(ʘs²+µ²σ²)_*√τ  
-n/[ф^-1(ϒ%)√(ʘs²+µ²σ²)+ʘµτ/ф^-1(ϒ%)√(ʘs²+µ²σ²)]= √τ

But let α=ʘµ/ф^-1(ϒ%)√(ʘs²+µ²σ²)

And β=n/[ф^-1(ϒ%)√(ʘs²+µ2σ²)]

Therefore ατ-√τ-β=0,  

  Proving the equation in quadratic form , we have;

√τ =[1+√(1+4αβ)]/2α

The symbols n, N, ʘ, µ, and τ have different meanings, based on the area of application of the Least Whole Normal (LWN) Function.

(a)NON-PRODUCT MODEL
If the normal random variable z which is population available in time or interval or space or quantity τ; and T which is occurrences of each member in the population available in time or interval or space or quantity τ are continuous; then it least whole normal function is given as;

l(τ) = ατ - √τ - β

α=quantile coefficient 

 β=quantil constant

α=(Θμ)/(Φ^-1)(γ%)√(ΘS²+μ²σ²  )

 β= T/(Φ^-1)(γ%)√( ΘS²+μ²σ²  )

To describe the distribution, we have
Z ∽N (Θτ, σ²τ)    
T∽n (μ, S²)

.Where Θ and σ²are population mean and variance respectively. μ and S² are mean of each occurrences  in the population and variance of each occurrences in the population respectively.


LEAST WHOLE TIME OR INTERVAL OR SPACE OR QUANTITY
The parameter τ which represents least whole interval or time or space or quantity is always occur when the function l(τ) is approximately equal to zero. The least whole interval or time or space or quantity is calculated as

√τ= [1+√(1+4αβ)]/2α

THE TOTAL NUMBER OF EACH OCCURRENCES
At least whole time or interval or space or quantity τ, the total number of each occurrence in the entire population is denoted as T and is Calculated as;

T= Θμτ - Φ^-1(γ%)√(ΘS²τ +μ²σ²τ)

AVERAGE NUMBER OF INDIVIDUAL OCCURRENCE
Also at least whole time or interval or space or quantity τ, the average number of each occurrences in the entire population is calculated as
 

μ² =[ (T²+ Φ^-1(γ%)²ΘS²τ)]/[(Θ²τ²- Φ^-1 (γ%)²σ²τ)]

EXAMPLE (I)
The government of Ghana has assigned a specific duty to roof new build houses in Upper East Region. The size of the new built houses has a Poisson random variable with mean 25. The number of roofing sheets that can complete one house has mean 50 and variance 675. The sizes of the houses and the number of roofing sheets are independent. If the government is taken similar projects in all the regions of Ghana, how many roofing sheets are expected to complete 500 houses at probability greater than 95%.

SOLUTION
Θ = 25
σ ² =25   
μ =50
S²=675
τ =500 houses                                               

Applying equation  above, we have
T= Θμτ - Φ^-1(γ%)√(ΘS²τ +μ2.σ²τ)
T= 25*50*500 - Φ^-1(95%)√(25*675*500+502*25*500)
 T= 625000-10363.1776
T=614636.8264≈ 614637.00
Hence, the number of roofing sheets that can complete 500 houses is 614627.
 

EXAMPLE (2)
The number of health insurance claims reported per month at certain health insurance company has mean 110 and variance 750.Individual losses have mean 1101 and variance 4900. The number of claims and amount of individual losses are independent. Estimate the expected losses to be incurred a year at a probability greater than 95%

SOLUTION
Θ  = 110
 σ²=750
 μ=1101
 S²  = 4900
 τ = 12months

Applying equation above, we have;
T= Θμτ - Φ^-1(γ%)√(ΘS²τ +μ²σ²τ)
T= 110*1101*12- Φ-1(95%)√(110*4900*12+11012*750*12)
 T= 1453320-171871.2264
T=1,251,448.774≈ 1,251, 449.00
Hence, the losses that are expected to be incurred a year is 1,251, 449.00

EXAMPLE (3)
The genetic of certain fungal (fungal infections) has average number of chiastmata to be 5.5 per micro interval and variance 14 per micro interval. Average frequency of each chiasmata occurring per micro interval has mean 0.08 and variance 0.23. Estimate the least whole number of micro intervals the chiasmata frequency 0.4 will occur at a probability greater than 99%.

EXAMPLE(4):

Each hour the television in Savanna hall JCR at UDS is switched on, the number of students who go in to watch the television has a distribution with mean 100 and variance 900.

The sachets of pure water purchased by each student had a Poisson distribution with mean 5.  The number of students going in to watch the television each hour it switch’s on and number of sachets  purchased by each student are independent.

(a) Determine the least number of whole hours the television should be left on, so the uncertainty that 10,000 sachets of water will be purchased is greater than 95%.

(b)  If we expect to switch off the television every 12hrs; determine the sachets of pure water that are expected to be purchased in this given whole number of hours.

SOLUTION

(a) Mean of population demand (Θ) = 100

Variance of population demand (σ²) = 900

Production mean of a unit sample (X) = 5

Production variance of a unit sample (S²) = 5

Least whole overall production sample size (n_lw) = 10,000

The quantile coefficient and quantile constant are calculated as;

                        

α =500 /[1.645 x 151.60]  = 2

 β =10,000 / 1.645 x 151.60= 40

The least number of whole hours is calculated as;

 √τ= [1+√(1+4αβ)]/2α

√ T_lw  =  [1 +  (1 + 320)^0.5] /4

√ T_lw  = 4.73

T_lw  = 22.4hrs (round up to 23hrs)

(b) n_lw = 1.645 (2(12) - √12) (500 + 22,500)^0.5

         n_lw = 1.645 x 20.54 x 151.66

          n_lw = 5,124.33 (round up to 5125)

(b)PRODUCT MODEL
For many centuries the problem of physics and its causes had puzzled scientists – it was not until the time Galileo, Newton, Einstein etc. that a real progress in its explanation was made.
Currently, I have realized that more development is needed to solve the scientific situations that we are confronting today.
I have no turning point than to restructure the existing concepts of physics to a new branch of concept called “Least Whole Normal Concept.” The development of the least whole normal concept in physics is mainly depends on a basic theory called “theory of quantitative products” which is discovered by me, whose previous name was Adongo Ayine William and now known as Bill Adongo in the year 2009. The detail of the least whole normal concept in physics will be explained later.

We have seen how the least whole normal function used to solve a long standing technical problems. One of its useful application is the product modeling ( i.e used to fit two or more quantitative products) 

Two Quantitative Product	Three Quantitative Product
Power=force x velocity P=fv E(P)=μf *τv	Electric energy=volt*current*time E=Vit E(E)= Θv*μI* τt
Mass=density*volume M=d*v E(M)= μd *τv	Heat energy=mass*specific heat*temperature E=mct E(E)= Θm*μc* τt

·Radical quantity: the radical quantity is any quantity that has main influence in the quantitative product. Let us consider the quantitative equation for velocity which is given as V = S(1/t): the radical quantity is the time t.

·Subjective quantity: the subjective quantity is any quantity that is determined or has main relationship with the radical quantity. Let us also consider the quantitative equation for velocity which is given as                     V = S(1/t):  the subjective quantity is the distance S.

·Non-subjective quantity: the non-subjective quantity is a chosen quantity in the quantitative product which is not related to the radical quantity under studied. The quantitative equation for force which is given as  F = mvt ^–1:  the non-subjective quantity could be v if m is chosen as subjective quantity or it could be m if v is chosen as subjective quantity.

TWO QUANTITATIVE PRODUCTS

Let us consider E ( Q) = (1_* τ_i )_*µ_i  where 1 is denoted as a unit non-subjective quantity and τ  the radical and µ_i  is the subjective quantity. The non-subjective unit quantity is 1. The quantitative equation E ( Q ) = (I_i * τ )*µ_i    has a distribution. 

T_j  ̴  N( τ , 0) ;     T_j  ̴  n (µ_i , S_i² )   

If the normal random quantity T₁  is total non-subjective unit quantity and  T_i  is total subjective quantity, then the least whole normal function for the above distribution is

ℓ (τ) = α * τ - √ (τ ) – β

Where the constant α  and β are denoted as quantile coefficient of  τ  and quantile constant respectively are calculated as;

α = µ_i / [ϕ ^-1(γ%)(√S_i²)]

β = T_i/ [ϕ ^-1(γ%)(√S_i²)]

The quantity which represents least whole radical quantity is always approximately equal to zero. The least 

 Whole radical quantity is calculated as;

√ ( τ₀ ) = [1+√ (1 + 4αβ)]/ 2α 

THE TOTAL SUBJECTIVE QUANTITY

The quantity T_i   which represents total subjective quantity is calculated as;

T_i = µ_iτ₀ – ϕ ^-1(γ %)√(S_i²τ₀ )

MEAN SUBJECTIVE QUANTITY

The quantity µ_i which represents the mean subjective quantity is calculated as;

µ_i² = [ (1 / τ²)][T_i² + ϕ ^-1( γ%)S²τ] 

APPLYING TWO QUANTITATIVE PRODUCT

DOING WORK:

Doing works are way of transferring energies using forces. The amount of energies transferred. The amount of works done equal to sizes of the forces times the distance move.

Expected work= mean force _*mean distance

i.e E (w) = µ_f τ

Applying the theory of two quantitative products, we have the distribution

T₁ ~ N ( τ, 0)  
T_f  ~ n (µ_f , S_f ²)   

Where µ_f  denoted as mean force and S_f ²   is the variance of the forces.

If the normal random quantity T₁  which is total number of units quantity I which mean is always equal to one and T_f  which is denoted as total number of force, then it least whole normal function is calculated as;

ℓ (τ₀) = α * τ₀  – ( √τ₀ ) – β

 where α and β  are calculated as;

α = µ_f  ^-1/ [ϕ^-1(γ%)(√S_i²)]

β = T_f / [ϕ ^-1(γ%)(√S_i²)] 

THE LEAST WHOLE DISTANCE (RADICAL QUANTITY)

The quantity T_f which represents total subjective distance is always approximately equal to zero. The least whole distance is calculated as;

√τ₀ = [1 + √ (1 + 4αβ) ]/ 2α

TOTAL FORCES (SUBJECTIVE QUANTITY)

The quantity T_f  which represents total subjective quantity or total forces is calculated as;

T_f = µ_f τ – ϕ ^-1[γ%][√(S_i²τ)]

MEAN FORCE (MEAN SUBJECTIVE QUANTITY)

The quantity µ_f  which represent mean force is calculated as;

µ_f ²  = [ (1 / τ₀²)][T_f ² +ϕ ^-1( γ%)S ²τ₀^-1 ] .............(*)

EXAMPLE(1)

Horses pull cars with constant horizontal forces of mean 136N and standard deviation 27N per distance. Calculate the

         i)Total forces at distance 19500m

       ii)Average forces at distance 19500m

(Take γ = 95% )

SOLUTION

                i)Applying equation above, we have

T _f = µ_fτ₀ − ϕ ^-1[γ%][√(S _f ²τ₀) ] 
T _f  = 136 x19500 − ϕ ^-1[95%][√(729 ×19500)]

T_f = 2,652,000 – 1.645x√(14,215,500)

T_f  = 2,645,797.783N

Hence, the total force is 2,645,797.783N

EXAMPLE(2)

Given the formula p=fv, where p is power, f is force and v is velocity: an engineer finds that the mean force of different bodies and standard deviation of a moving bodies are 1400 and 320 respectively per velocity. Calculate

1)       The total force at velocity 150m/s

2)       The average force at velocity 150m/s

3)       Average power at velocity 150m/s

4)       Total power at velocity 150m/s

(Take γ =95)

SOLUTION

(i)Total force at velocity 150m/s is:

T_f= µ _fτ_v − ϕ ^-1[γ%][√(S_f ²τ_v) ]

 T_f=1400_*150-1.645_*√(320²_*150)

T_f=203,552.945N

(ii) µ_i² = [ (1/ τ²)][T_i² + ϕ ^-1( γ%)S²τ]  

    µ_i² =1/150[203,552.943²+1.645_*320²_*150]

EXAMPLE(3)

Given M=d*v, where M is mass, d is density and V is volume. Scientist finds that mean density of different bodies is 800kg/m³ and standard deviation 150kg/m³ per volume. Calculate

i)the total density of the bodies at volume 500m³

ii)the average density of the bodies at volume 500m³

iii)average mass of the bodies at volume 500m³

iv)total masses of the bodies at volume 500m³

i)T_D=394482.5023kg/m³

ii)    µ_D²=789.0119kg/m³

 THREE QUANTITATIVE PRODUCTS

Let us consider E (Q) = (Ä_I*τ )_*µ_i   where Ä_j  is denoted as non-subjective quantity µ_i  is denoted as subjective quantity and τ is the radical quantity.

If the non-subjective quantity Ä_j    is not equal to one, but represents any quantitative value. Then we have the distribution of the quantitative equation E(Q)  which is given as;

T_j ~ N(Ä_j* τ ,  σ ²_*τ ) 
T_j ~ n( µ_i , S_i² )                 

If the normal random quantity T_j  is total non-subjective quantities and T_i  is total subjective quantity, then the least whole normal function is given as

ℓ (τ) = ατ  – ( √τ ) – β  )

the total subject quantity is calculated as:

T= Θμτ - Φ^-1(γ%)√(ΘS²τ +μ²σ²τ)

Average subjective quantity is:

μ² = (T²+ Φ^-1(γ%)²ΘS²τ)/(Θ²τ²- Φ^-1(γ%)²σ²τ)

The radical quantity is calculated as:

√τ= [1+√(1+4αβ)]/2α

EXAMPLE(1)

Energies=volts_*currents_*time: scientist finds that 5.4v with standard deviation 2.10v per second and current 42A with standard deviation 18A per second. Calculate the seconds required for total current 110000A.

(Take γ =95)

SOLUTION

α=5.4*42/[1.645√(5.4*18²+42²*2.10²)] = 1.412398

 β =110000/[1.645√(5.4*18²+42²*2.10²)] = 685.02554

√τ= [1+√(1+4αβ)]/2α
 √τ=22.3797

τ=500.85"


EXAMPLE(2) 

Heat Energies =masses*specific capacities*change in temperature: If the average specific heat capacity is 12.8 and standard deviation is 2.4 per change in temperature and average mass is 2.08 and standard 0.06 per change in temperature.

1) Calculate temperature change if the specific heat capacity is 80

SOLUTION

α=2.08*12.4/[1.645√(2.08*2.4²+12.8²*0.06²)]

α=4.42222

β=80/[1.645√(2.08*2.4²+12.8²*0.06²)

β=13.7165745

√τ=[1+√(1+4*4.42222*13.7165745)]/2*(4.42222)

τ=3.526 ⁰C

APPLICATIONS

CAPACITOR, VOLTAGE AND CHARGE

A capacitor is a passive element designed to store energy in its electric field. Besides resistors, capacitors are the most common electrical components. Capacitors are used extensively in electronics communications, computers, and power systems. For example, they are used in the tuning circuits at radio receivers and as dynamic memory element,

In many practical applications, the plates may be aluminum foil which the dielectric may be air, ceramic, paper, or mica. When a voltage source deposits a positive charge q on one plate and a negative charge   -q on the other, Capacitor is said to store the electric charge. The amount of charge stored, represented by q, is directly proportional to the applied voltage V so that

q=CV

Where C, the constant of proportionality, is known as the capacitance of the capacitor. Finding way of measuring unknown amount of charge stored q sand unknown capacitance of the capacitor C, with only known voltage V is impossible without employing my Least Whole Normal Measurement. From previous publications, I talk of theory of quantitative products where two or more quantitative products in science can be fitted by my Least Whole Normal Function and when applying this theory, the equation above will yield

T= µ_cτ_v- ф^-1(γ%)√(S²τ_v)

E(C)=1/V_p[T+ф^-1(γ%)√(S²V_p)]

E(q)=E(C)V_p

where V_p is the independent voltage for predicting the expected of amount of charge stored E(q) and the expected capacitance of the capacitor E(C).

EXAMPLE

Calculate the capacitance of the capacitor C and amount of Charge stored q if the voltage is 8V based on the previous experiment made on six Capacitance of the capacitors and their known voltage.(take γ=95).

Capacitance of the capacitor(C)	Voltage(V)
60mF=60x10-3	5
50mf=50x10-3	6
40mF=40x10-3	7.5
30mF=30x10-3	10
20mF=20x10-3	15
10mF=10x10-3	30

SOLUTION

MEAN AND STANDARD DEVIATION

C	V	CV	Deviation of c from mean(d)	d2	Vd2
60mF=60x10-3	5	0.3	0.03551	0.001261	0.006305
50mf=50x10-3	6	0.3	0.02551	0.000651	0.003905
40mF=40x10-3	7.5	0.3	0.01551	0.000241	0.001804
30mF=30x10-3	10	0.3	0.00551	0.000030	0.003036
20mF=20x10-3	15	0.3	0.00449	0.000020	0.000302
10mF=10x10-3	30	0.3	-0.01449	0.000210	0.006299

Mean(µ_c)=∑CV/∑V=1.8/73.5=0.02449

Standard Deviation(S)=√(∑Vd2/∑V=√(0.021651/73.5=0.017147

From the data above we have;

T= µ_cτ_v- ф^-1(γ%)√(S²τ_v)

But τ_v=∑V/N=73.5/6=12.25

T=0.02449x12.25- ф^-1(95%)√(0.017147²x12.25)

T=0.2012761475

E(C)=µ_cV_p=1/V_p[T+ф^-1(γ%)√(S²V_p)]

E(C)= 1/8[0.2012761475+ф^-1(95%)√(0.017147²x8)]

E(C)=0.035=35x10^-3=35mF

q=E(C)V_p=35x10^-3x8=0.3

Hence, the capacitance of the capacitor and charge stored are 35mF and 0.3 respectively if the voltage is 8V.

HEART AS A PUMP

In previous attempt to explain the practical important of my Least Whole Normal Function, a medical student asked me concerning the practical application of my Least Whole Normal Function in medical field.

I am writing this to help those who are willing to think and know the practical important of my Least Whole Normal Function and not to jokers. More again, my function is a UNIVERSAL function that applies many field of knowledge. For example, knowing the heart as a pump and its intrinsic regulation is an important issue in medical sciences. Nowhere can we predict the volume of blood pumped and the blood flowing from a heart if only we measured the heartbeat of a patient at a given time (=heart rate). The Least Whole Normal Function is the possibility of predicting the amount of blood flowing from a heart (=Cardiac Output) and the volume of blood pumped (=Stroke Volume), if only we measured the heart rate based on each patient.

The sound associated with the values of the heart closing can be heard by using stethoscope. The stethoscope is the main instrument in medical field where a patient heartbeat can be measured.

The key importance of projecting the cardiac output and the stroke volume is to examine heart diseases like hypertension, arteriosclerosis, rheumatic fever which often leads to rheumatic heart diseases later in life. My Least Whole Normal Function is not only important for Cardiologists but also for other specialists.

Below are an experiment conducted by physician and we can use this data to project the stroke volume and the cardiac output if the heart rate is 75 per minute.

Time	Volume(x)	Heart rate(f)
0.8	50	45
1.6	100	89
2.4	150	137

MEAN AND STANDARD DEVIATION

TIME	VOLUME(x)	HEART RATE(f)	fx	DEVIATION OF THE X FROM MEAN(d)	d2	fd2
0.8	50	45	2250	-66.974	4485.517	201848.265
1.6	100	89	8900	-16.974	288.117	25462.413
2.4	150	137	120550	33.026	1090.717	149428.229
		∑f=271	∑fx=31700			∑fd2=376738.907

Mean(µ)=∑fx/∑f= 31700/271=116.974

Standard deviation(s)=√∑fd² /∑f=√(376738.907/271)=37.285

CALCULATING CARDIAC OUTPUT AND STROKE VOLUME

from the data above;

T=µf_m -φ^-1(γ%)√(S²f_m )

but f_m =∑f/n=271/3=90.3333

T=116.974*90.3333- φ^-1(95%)√(37.285²*90.3333)

T=9983.707258

the average Stroke volume at given heart rate 75 is:

Stroke Volume=1/f[T+ φ^-1(γ%)√(S²f)]

Stroke Volume=1/75[9983.707258 +  φ^-1(95%)√(37.285²*75)]

Stroke Volume=140.1983169

The Cardiac Output is calculated as:

Cardiac Output=Stroke Volume*Heart Rate

Cardiac Output=140.1983169*75=10514.87376

Hence, the Cardiac Output and the Stroke Volume of the patient at given heart rate75, are 10514.87376

and 140.1983169 respectively.