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:
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 expected mean of the population n is Es=ʘµτ and its variance is given as Vs=(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,
√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.
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 Es=ʘµτ and its variance is given as Vs=(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;
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
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)
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
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;
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
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
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.
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%
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
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
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%.
SOLUTION
The least number of whole hours is calculated as;
(b) nlw = 1.645 (2(12) - √12) (500 + 22,500)0.5
nlw = 1.645 x 20.54 x 151.66
nlw = 5,124.33 (round up to 5125)
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 (nlw)
= 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
√ Tlw = [1
+ (1 + 320)0.5] /4
√ Tlw = 4.73
Tlw = 22.4hrs (round up to 23hrs)
(b) nlw = 1.645 (2(12) - √12) (500 + 22,500)0.5
(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.
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* τi )*µi where 1 is denoted as a unit
non-subjective quantity and τ the
radical and µi is the
subjective quantity. The non-subjective unit quantity is 1. The quantitative
equation E ( Q ) = (Ii * τ )*µi has a distribution.
Tj ̴ N( τ
, 0) ; Tj
̴ n (µi , Si2 )
If the normal random quantity T1 is total non-subjective unit quantity and Ti 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 = µi / [ϕ
-1(γ%)(√Si2)]
β = Ti/ [ϕ -1(γ%)(√Si2)]
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 Ti which
represents total subjective quantity is calculated as;
Ti = µiτ0
– ϕ -1(γ %)√(Si2τ0 )
MEAN SUBJECTIVE QUANTITY
The quantity µi
which represents the mean subjective quantity is calculated as;
µi2 = [ (1 / τ2)][Ti2 + 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) = µf τ
Applying
the theory of two quantitative products, we have the distribution
T1 ~ N ( τ, 0)
Tf ~ n (µf , Sf 2)
Where
µf denoted as mean force and Sf 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 Tf 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 = µf -1/ [ϕ-1(γ%)(√Si2)]
β = Tf / [ϕ -1(γ%)(√Si2)]
THE LEAST WHOLE DISTANCE (RADICAL QUANTITY)
The quantity Tf
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 Tf which represents total subjective quantity or
total forces is calculated as;
Tf = µf τ – ϕ -1[γ%][√(Si2τ)]
MEAN FORCE (MEAN SUBJECTIVE QUANTITY)
The quantity µf which represent mean force is calculated as;
µf 2
= [ (1 / τ02)][Tf 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
= µfτ0 − ϕ
-1[γ%][√(S f 2τ0) ]
T f = 136 x19500 − ϕ -1[95%][√(729 ×19500)]
T f = 136 x19500 − ϕ -1[95%][√(729 ×19500)]
Tf = 2,652,000 – 1.645x√(14,215,500)
Tf = 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:
Tf= µ fτv − ϕ -1[γ%][√(Sf 2τv) ]
Tf=1400*150-1.645*√(3202*150)
Tf=203,552.945N
(ii) µi2 = [ (1/ τ2)][Ti2
+ ϕ -1( γ%)S2τ]
µi2
=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)TD=394482.5023kg/m3
ii) µD2=789.0119kg/m3
THREE QUANTITATIVE PRODUCTS
Let us consider E (Q) = (ÄI*τ )*µi where Äj is denoted as non-subjective quantity µi is denoted as subjective quantity and τ is the radical quantity.
If the non-subjective quantity Äj 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;
Tj ~ N(Äj* τ , σ 2*τ )
Tj ~ n( µi , Si2 )
If the normal random quantity Tj is total non-subjective quantities and Ti 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=22.3797
τ=500.85"
EXAMPLE(2)
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=
µcτv- ф-1(γ%)√(S2τv)
E(C)=1/Vp[T+ф-1(γ%)√(S2Vp)]
E(q)=E(C)Vp
where Vp 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(µc)=∑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=
µcτv- ф-1(γ%)√(S2τv)
But
τv=∑V/N=73.5/6=12.25
T=0.02449x12.25-
ф-1(95%)√(0.0171472x12.25)
T=0.2012761475
E(C)=µcVp=1/Vp[T+ф-1(γ%)√(S2Vp)]
E(C)=
1/8[0.2012761475+ф-1(95%)√(0.0171472x8)]
E(C)=0.035=35x10-3=35mF
q=E(C)Vp=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=µfm -φ-1(γ%)√(S2fm
)
but fm =∑f/n=271/3=90.3333
but fm =∑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.
