This document discusses the key characteristics and concepts of the normal probability distribution. It outlines six goals related to understanding the normal distribution, its properties, calculating z-values, and using the normal distribution to approximate the binomial probability distribution. The key points covered include defining the mean, standard deviation, and shape of the normal curve; transforming variables to the standard normal distribution; and determining probabilities based on the areas under the normal curve.
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Ranking bertanda Wilcoxon banyak digunakan untuk menguji perbedaan perlakuan yang diberikan kepada objek penelitian dengan mempertimbangkan arah dan magnitude relatif perbedaan dari dua sampel berpsangan.
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Ranking bertanda Wilcoxon banyak digunakan untuk menguji perbedaan perlakuan yang diberikan kepada objek penelitian dengan mempertimbangkan arah dan magnitude relatif perbedaan dari dua sampel berpsangan.
Secara sistematis, dualitas merupakan alat bantu masalah LP, yang secara langasung didefinisikan dari persoalan aslinya atau dari model LP primal. Dalam kebanyakan perlakuan LP, dualitas sangat tergantung pada primal dalam hal tipe kendala, variabel keputusan dan kondisi optimum.
This introduction to probability is brought to you by https://mobilebetting.kiwi. In this presentation you’ll learn how probability affects the bets you place and why it’s a key factor in sports betting.
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Chapter 6: Normal Probability Distribution
6.1: The Standard Normal Distribution
The Normal Distribution:
There are different distributions namely Normal, Skewed, and Binomial etc.
Objectives:
Normal distribution its properties its use in biostatistics
Transformation to standard normal distribution
Calculation of probabilities from standard normal distribution using Z table.
Normal distribution:
- Certain data, when graphed as a histogram (data on the horizontal axis, frequency on the vertical axis), creates a bell-shaped curve known as a normal curve, or normal distribution.
- Two parameters define the normal distribution, the mean (µ) and the standard deviation (σ).
Properties of the Normal Distribution:
Normal distributions are symmetrical with a single central peak at the mean (average) of the data.
The shape of the curve is described as bell-shaped with the graph falling off evenly on either side of the mean.
Fifty percent of the distribution lies to the left of the mean and fifty percent lies to the right of the mean.
-The mean, the median, and the mode fall in the same place. In a normal distribution the mean = the median = the mode.
- The spread of a normal distribution is controlled by the
standard deviation.
In all normal distributions the range ±3σ includes nearly
all cases (99%).
Uni modal:
One mode
Symmetrical:
Left and right halves are mirror images
Bell-shaped:
With maximum height at the mean, median, mode
Continuous:
There is a value of Y for every value of X
Asymptotic:
The farther the curve goes from the mean, the closer it gets to the X axis but it never touches it (or goes to 0).
The total area under a normal distribution curve is equal to 1.00, or 100%.
Using Normal distribution for finding probability:
While finding out the probability of any particular observation we find out the area under the curve which is covered by that particular observation. Which is always 0-1.
Transforming normal distribution to standard normal distribution:
Given the mean and standard deviation of a normal distribution the probability of occurrence can be worked out for any value.
But these would differ from one distribution to another because of differences in the numerical value of the means and standard deviations.
To get out of this problem it is necessary to find a common unit of measurement into which any score could be converted so that one table will do for all normal distributions.
This common unit is the standard normal distribution or Z
score and the table used for this is called Z table.
- A z score always reflects the number of standard deviations above or below the mean a particular score or value is.
where
X is a score from the original normal distribution,
μ is the mean of the original normal distribution, and
σ is the standard deviation of original normal distribution.
Steps for calculating probability using the Z-
score:
-Sketch a bell-shaped curve,
- Shade the area (which represents the probability)
-Use the Z-score formula to calculate Z-value(s)
-Look up Z-values in table
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1. l
Chapter Seven
The Normal Probability DDiissttrriibbuuttiioonn
GOALS
1. List the characteristics of the normal probability
distribution.
2. Define and calculate z values.
3. Determine the probability an observation will lie
between two points using the standard normal
distribution.
4. Determine the probability an observation will be above
or below a given value using the standard normal
distribution.
5. Compare two or more observations that are on
different probability distributions.
6. Use the normal distribution to approximate the
binomial probability distribution.
2. Distribusi Probabilitas KKoonnttiinnyyuu
Distribusi probabilitas kontinyu: didasarkan
pada variabel acak kontinyu biasanya
diperoleh dengan mengukur.
Dua bentuk distribusi probabilitas kontinyu:
Distribusi probabilitas seragam (uniform)
Distribusi probabilitas normal
3. Normal Distribution
Importance of Normal Distribution
1. Describes many random processes or
continuous phenomena
2. Basis for Statistical Inference
4. Characteristics of a Normal
Probability Distribution
The normal curve is bell-shaped and has a single peak at
the exact center of the distribution.
The arithmetic mean, median, and mode of the
distribution are equal and located at the peak . Thus half
the area under the curve is above the mean and half is
below it.
The normal probability distribution is unimodal=only one
mode.
It is completely described by Mean & Standard Deviation
The normal probability distribution is asymptotic. That is
the curve gets closer and closer to the X -axis but never
actually touches it.
5. Normal Probability distribution
2
2
é ù -ê x
- ú
m
s
( )
Normal Probability
distribution
( ) 1 2
= êë úû
P x e
s p
2
e=2.71828
Do not worry about the formula, you do not need to
calculate using the above formula.
6. The density functions of normal distributions with
zero mean value and different standard deviations
Bell-shaped
Symmetric
Mean=median = mode
Unimodal
Asymptotic
8. Normal Distribution Probability
Probability is the area
under the curve!
f(X) A table may be
constructed to help
us find the probability
c d X
9. Infinite Number of Normal Distribution Tables
Normal distributions differ by
mean & standard deviation.
Each distribution would
require its own table.
X
f(X)
10. Distribusi Probabilitas Normal Standar
Distribusi Normal Standar adalah distribution
normal dengan mean= 0 dan standar
deviasi=1.
Juga disebut z distribution.
A z- value is the distance between a selected
value, designated X , and the population mean
m, divided by the population standard
deviation, s. The formula is:
z =X -m
s
11. The Standard Normal Probability
Distribution
Any normal random variable can be
transformed to a standard normal random
variable
Suppose X ~ N(μ, s 2)
Z=(X-μ)/ s ~ N(0,1)
P(X<k) = P [(X-μ)/ s < (k-μ)/ s ]
12. Standardize the Normal Distribution
Standardized Normal
Distribution
m Z = 0
s z = 1
Z
Normal
Distribution
Z = X -m
m X
s
s
Because we can transform any normal random variable into
standard normal random variable, we need only one table!
13. Transform to Standard Normal Distribution
A numerical example
Any normal random variable can be transformed to a
standard normal random variable
x x-m (x-m)/σ x/σ
0 -2 -1.4142 0
1 -1 -0.7071 0.7071
2 0 0 1.4142
3 1 0.7071 2.1213
4 2 1.4142 2.8284
Mea
n
2 0 0 1.4142
std 1.4142 1.4142 1 1
14. Standardizing Example
Data of weight of children under 5 years old in Kg. Mean
(m)= 5 and s= 10. What is the probability of children
weighted at 5 Kgs to 6.2 Kgs?
s Z = 1
m = 0
.12
Z Z Normal
Distribution
Standardized Normal
Distribution
= - =6.2-5=
s
m = 5 X
s = 10
6.2
0.12
10
Z X m
15. Obtaining the Probability
s Z = 1
m = 0
0.12
Z Z Standardized Normal
Probability Table (Portion)
Z .00 .01
0.0 .0000 .0040 .0080
.0398 .0438
0.2 .0793 .0832 .0871
0.3 .1179 .1217 .1255
0.0478
.02
0.1 .0478
Probabilities
Shaded Area
Exaggerated
16. Example P(3.8 £ X £ 5)
Standardized Normal
Distribution
s Z = 1
m Z = 0 Z
-0.12
Normal
Distribution
0.0478
= - =3.8-5=-
s
m = 5 X
Shaded Area Exaggerated
s = 10
3.8
0.12
10
Z X m
17. Example (2.9 £ X £ 7.1)
Standardized Normal
0
Distribution
s Z = 1
-.21 .21 Z
Normal
Distribution
.1664
.0832 .0832
X
X
-
Shaded Area Exaggerated
5
s = 10
Z
Z
=
2.9 7.1 X
=
-
= -
=
-
=
-
=
m
s
m
s
2 .
9 5
.
21
10 7 .
1 5
.
21
10
18. Example (2.9 £ X £ 7.1)
Standardized Normal
0
Distribution
s Z = 1
-.21 .21 Z
Normal
Distribution
.1664
.0832 .0832
X
X
-
Shaded Area Exaggerated
5
s = 10
Z
Z
=
2.9 7.1 X
=
-
= -
=
-
=
-
=
m
s
m
s
2 .
9 5
.
21
10 7 .
1 5
.
21
10
19. Example P(X ³ 8)
s Z = 1
.5000 .3821
m Z = 0
.30
Z Normal
Distribution
Standardized Normal
Distribution
.1179
Shaded Area Exaggerated
Z
X
=
-
=
-
=
m
s
8 5
10
.30
s = 10
m = 5 8
X
20. Example P (7.1 £ X £ 8)
Standardized Normal
m z = 0
Distribution
s Z = 1
.1179 .0347
.21 .30 Z
Normal
Distribution
.0832
Z X
Shaded Area Exaggerated
Z
X
=
-
=
-
=
=
-
=
-
=
m
s
m
s
71 .
5
.
21
10 8 5
.
30
10
s = 10
m = 5
7.1 8 X
21. Normal Distribution Thinking Challenge
You work in Quality Control for GE. Light
bulb life has a normal distribution with μ=
2000 hours & s = 200 hours. What’s the
probability that a bulb will last
between 2000 & 2400 hours?
less than 1470 hours?
22. Solution P (2000 £ X £ 2400)
Standardized Normal
Distribution
s Z = 1
m Z Z = 0
2.0
Normal
Distribution
.4772
Z
X
=
-
=
-
=
m
s
2400 2000
200
2.0
s = 200
m = 2000 2400
X
23. Solution P (X £ 1470)
Standardized Normal
Distribution
s Z = 1
.5000
m Z= 0 Z
-2.65
Normal
Distribution
.0040 .4960
Z
X
=
-
=
-
= -
m
s
1470 2000
200
2.65
m = 2000 X
s = 200
1470
24. Finding Z Values for Known
Probabilities
Standardized Normal
Probability Table (Portion)
.1217 .01
Z .00 .02
0.0 .0000 .0040 .0080
0.1 .0398 .0438 .0478
0.2 .0793 .0832 .0871
.1179 .1255
s Z = 1
m Z = 0 .31
Z
0.3 .1217
What Is Z Given
P(Z) = 0.1217?
Shaded Area
Exaggerated
25. Finding X Values for Known
Probabilities
Normal Distribution Standardized Normal Distribution
s Z = 1
m = 5 X .31
m Z = 0 Z
s = 10
?
.1217 .1217
X =m +Z×s =5+(0.31)×10=8.1
Shaded Area Exaggerated
26. EXAMPLE 1
The bi-monthly starting salaries of recent MBA
graduates follows the normal distribution with
a mean of $2,000 and a standard deviation of
$200. What is the z- value for a salary of
$2,200?
z = X -
m
$2,200 $2,000 1.00
$200
s
= - =
27. EXAMPLE 1 continued
What is the z-value of $1,700 ?
z = X -
m
s
=$1,700 -$2,200 =-
1.50
$200
A z-v alue of 1 indicates that the value of
$2,200 is one standard deviation above
the mean of $2,000.
A z-v alue of –1.50 indicates that $1,700 is
1.5 standard deviation below the mean of
$2000.
28. Areas Under the Normal Curve
About 68 percent of the area under the normal
curve is within one standard deviation of the
mean. m ± s
P(m - s < X < m + s) = 0.6826
About 95 percent is within two standard
deviations of the mean. m ± 2 s
P(m - 2 s < X < m + 2 s) = 0.9544
Practically all is within three standard
deviations of the mean. m ± 3 s
P(m - 3 s < X < m + 3 s) = 0.9974
29. Areas Under the Normal Curve
Between:
± 1 s - 68.26%
± 2 s - 95.44%
± 3 s - 99.74%
μ
μ-2σ μ+2σ
μ-3σ μ-1σ μ+1σ
μ+3σ
30. EXAMPLE 2
The daily water usage per person in
Surabaya is normally distributed with a
mean of 20 gallons and a standard
deviation of 5 gallons. About 68
percent of those living in Surabaya will
use how many gallons of water?
About 68% of the daily water usage will
lie between 15 and 25 gallons.
31. EXAMPLE 3
What is the probability that a person from
Surabaya selected at random will use
between 20 and 24 gallons per day?
= - =20-20 =
0.00
5
z X m
s
= - =24-20 =
0.80
5
z X m
s
P(20<X<24)
=P[(20-20)/5 < (X-20)/5 < (24-20)/5 ]
=P[ 0<Z<0.8 ]
32. The Normal Approximation to the
Binomial
The normal distribution (a continuous
distribution) yields a good approximation of
the binomial distribution (a discrete
distribution) for large values of n.
The normal probability distribution is
generally a good approximation to the
binomial probability distribution when n p
and n(1- p ) are both greater than 5.
33. The Normal Approximation continued
Recall for the binomial experiment:
There are only two mutually exclusive
outcomes (success or failure) on each
trial.
A binomial distribution results from
counting the number of successes.
Each trial is independent.
The probability is fixed from trial to trial,
and the number of trials n is also fixed.
35. Continuity Correction Factor
The value 0.5 subtracted or added, depending on
the problem, to a selected value when a binomial
probability distribution (a discrete probability
distribution) is being approximated by a continuous
probability distribution (the normal distribution).
Four cases may arise:
For the P at least X occur, use the area above (X – 0,5)
For the P that more than X occur, use the area above (X +
0,5)
For the P that X or fewer occur, use the area below (X +
0,5)
For the P that fewer than X occur, use the area below (X –
0,5)
36. Continuity Correction Factor
Because the normal distribution can take all real
numbers (is continuous) but the binomial distribution
can only take integer values (is discrete), a normal
approximation to the binomial should identify the
binomial event "8" with the normal interval "(7.5, 8.5)"
(and similarly for other integer values). The figure
below shows that for P(X > 7) we want the magenta
region which starts at 7.5.
37. n=20 and p=0.25 P(X ≥ 8)=?
First step: determine mean (m) & standard deviation (s)
m = np = (20)(0.25) = 5
s = np (1-p ) = (5)(1-0.25) = 3.57 =1.94
Second step: determine the z-value
Third step: check the area below normal curve
38. 1. Without correction factor of 0.5:
= - = - =
8 5 1.55
1.94
z X m
s
check in table-Z: P(X ≥ 8)=0.5-0.4394=0.0606
2. With correction factor of 0.5: P(X ≥ 8) P(X ≥ 8-0.5=7.5)
= - = - =
7.5 5 1.29
1.94
z X m
s
check in table-Z: P(X ≥ 7.5)=0.5-0.4015=0.0985
The exact solution from binomial distribution function is
0.1019.
The continuity correct factor is important for the
accuracy of the normal approximation of binomial.
The approximation is quite good.
39. EXAMPLE 5
A recent study by a marketing research firm showed
that 15% of American households owned a video
camera. For a sample of 200 homes, how many of the
homes would you expect to have video cameras?
What is the mean?
m =np =(.15)(200) =30
What is the variance?
s 2 = np(1-p ) =(30)(1-.15) = 25.5
What is the standard deviation?
s= 25.5 =5.0498
40. EXAMPLE 5 continued
What is the probability that less than 40
homes in the sample have video cameras?
We use the correction factor, so X is 39.5.
The value of z is 1.88.
1.88
= - = 39.5 -30.0 =
5.0498
z X m
s
41. Example 5 continued
From Appendix D the area between 0 and 1.88
on the z scale is .4699.
So the area to the left of 1.88 is
.5000 + .4699 = .9699.
The likelihood that less than 40 of the 200
homes have a video camera is about 97%.
42. EXAMPLE 5
P(z< 1.88)
=.5000+.4699
=.9699
z =1.88
0 1 2 3 4
z
43. Soal
KPS MP melakukan polling dengan
menyebarkan 60 kuisioner. Probabilitas
kembalinya kuisioner adalah 80%. Hitung
probabilitas :
50 kuisioner kembali
Antara 45-55 kuisioner kembali
Kurang dari 55 kuisioner kembali
20 atau lebih kuisioner kembali
44. Chapter Seven
The Normal Probability DDiissttrriibbuuttiioonn
- END -