The Normal Distribution

Transcription

DEPARTMENT OF QUANTITATIVE METHODS & INFORMATION
SYSTEMS
Introduction to Business Statistics
QM 120
Ch t 6
Chapter
Spring 2008
Dr. Mohammad Zainal
Continuous Probability Distribution
2
When a RV x is discrete, we can assign a positive probability to
each value that x can take and get the probability distribution
for x.
¾
¾
Th
The sum of all probabilities of all values of x is 1.
f ll
b bili i
f ll l
f i 1
¾
Not all experiments result in RVs that are discrete.
Continuous RV, such as heights, weights, lifetime of a
particular product,
product experimental laboratory error…etc.
error etc can
assume infinitely many values corresponding to points on line
interval.
¾
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3
Suppose we have a set of measurements on a continuous RV,
and
d we wantt to
t create
t a relative
l ti
f
frequency
hi t
histogram
t
to
describe their distribution.
¾
For a small number of measurements,
measurements we can use small
number of classes
¾
4
For a larger number of measurements, we must use a larger
number
b off classes
l
and
d reduce
d
th class
the
l
width
idth
¾
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5
As the number of measurements become very large, the class
width
idth become
b
very narrow, the
th relative
l ti frequency
f
hi t
histogram
appears more and more like a smooth curve.
¾
¾
This smooth curve describes the probability distribution of a
continuous random variable.
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6
A continuous RV can take on any of an infinite number of
values
l
on the
th reall line
li
¾
The depth or density of the probability, which varies with x,
maybe described by a mathematical formula f(x), called the
probability density function for the RV x.
¾
The probability distribution f(x); P(a< x <b) is equal to the
shaded area under the curve.
¾
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7
Several important properties of continuous probability
distribution parallel their discrete part.
part
¾
Just as the sum of discrete probabilities is equal to 1, i.e. ΣP(x)
= 1,
1 and
a d the probability
obability that x falls
fall into
i to a certain
e tai interval
i te al can
a
be found by summing the probabilities in that interval,
continuous probability distributions have the following two
characteristics:
¾
¾
Th area under
The
d a continuous
ti
probability
b bilit distribution
di t ib ti is
i equall to
t 1.
1
¾
The probability that x will fall into a particular interval‐ say, from a
to b – is equal to the area under the curve between the two points a
and b
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8
There is a very important difference between the continuous
case and the discrete one which is:
¾
¾
P(x = a) = 0
¾
P(a < x < b) = P(a ≤ x ≤ b)
¾
P( x < a) = P( x ≤ a)
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9
Among many continuous probability distributions, the normal
distribution is the most important and widely used one.
one
¾
A large number of real world phenomena are either exactly or
a
approximately
o i ately normally
o ally distributed.
di t ibuted
¾
The normal distribution or Gaussian distribution is given by a
bell‐shaped curve.
¾
A continuous RV x that has a normal distribution is said to be a
normal RV with a mean μ and a standard deviation σ or simply
x~N(μ,σ).
¾
Note that not all the bell‐shaped curves represent a normal
distribution.
¾
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10
The normal probability distribution, when plotted, gives a bell‐
shaped curve such that
¾
¾
The total area under the curve is 1.0.
¾
The curve is symmetric around the mean.
¾
The two tails of the curve extended indefinitely.
The mean μ and the standard deviation σ are the parameters of
the normal distribution.
¾
Given the values of these two parameters, we can find the area
under the normal curve for any interval.
interval
¾
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There is not just one normal distribution curve but rather a
family
y of normal distribution.
¾
Each different set of values of μ and σ gives different normal
curve.
¾
The value of μ determines the center of the curve on the g
p
horizontal axis and the value of σ gives the spread of the normal curve
¾
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12
Like all other distributions, the normal probability
distribution can be expressed by a mathematical function
¾
f (x ) =
1
σ 2π
e
1 ⎡ x −μ ⎤
− ⎢
2 ⎣ σ ⎥⎦
2
in which the probability that x falls between a and b is the
integral of the above function from a to b, i.e.
b
P (a < x < b ) = ∫
a
1
σ 2π
QM-120, M. Zainal
e
1 ⎡ x −μ ⎤
− ⎢
2 ⎣ σ ⎥⎦
2
dx
13
However we will not use the formula to find the probability.
instead, we will use normal probability table.
¾
QM-120, M. Zainal
The Standard Normal Distribution
14
The standard normal distribution is a special case of the
normal distribution where the μ is zero and the σ is 1.
1
¾
The RV that possesses the standard normal distribution is
d
denoted
d by
b z and
d it
i is
i called
ll d z values
l
or z scores.
¾
μ=00
σ=1
-4
-2
0
z
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2
4
15
Since μ is zero and the σ is 1 for the standard normal, a specific value of z gives the distance between the mean and the point
value of z gives the distance between the mean and the point represented by z in terms of the standard deviation.
¾
The z values to the right side of the mean are positive and Th
l
h i h id f h
ii
d
those on the left are negative BUT the area under the curve is always positive.
l
iti
¾
For a value of z = 2, we are 2 standard deviations from the mean (to the right).
¾
Similarly, for z Similarly,
for z = ‐2,
2, we are 2 standard deviations from the we are 2 standard deviations from the
mean (to the left)
¾
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16
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17
The standard normal table lists the area under the standard
normal curve between z = 0 to the values of z to 3.09.
3 09
¾
To read the standard normal table, we always start at z = 0,
which
hi h represents the
h mean.
¾
Always remember that the normal curve is symmetric, that is
the area from 0 to any positive z value is equal to the area from 0
to that value in the negative side.
¾
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Example: Find the area under the standard normal curve
between z = 0 to z = 1.95
1 95
Example: Find the area under the standard normal curve
between z = ‐2.17
betwee
. to z = 0
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19
Example: Find the following areas under the standard normal curve.
curve
a) Area to the right of z = 2.32
b) A ea to t e e t o
b) Area to the left of z = ‐1.54
.5
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20
Example: Find the following probabilities for the standard normal curve
normal curve.
a) P(1.19 < z < 2.12)
b) P(‐1.56
b) P(
1.56 < z < 2.31)
< z < 2.31)
c) P(z > ‐.75)
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21
¾
Remember: When two points are on the same side of the mean, first find
the
h areas between
b
the
h mean and
d each
h off the
h two points.
i
Th
Then,
subtract the smaller area from the larger area
When two points are on different side of the mean, first find
the areas between the mean and each of the two points. Then,
add these two area
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¾
Empirical rule
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Example: Find the following probabilities for the standard normal curve
normal curve.
a) P(0 < z < 5.65)
b) P( z < ‐
b)
(
5.3)
QM-120, M. Zainal
Standardizing a Normal Distribution
24
As it was shown in the previous section, Table VII of Appendix
C can be used only to find the areas under the standard normal
curve.
¾
However, in
H
i real‐world
l
ld applications,
li i
most off continuous
i
RV
RVs
that are normally distributed come with mean and standard
d i ti different
deviation
diff
t from
f
0 and
d 1,
1 respectively.
ti l
¾
What shall we do? Is there any way to bring μ to zero and σ to
1.
¾
Yes, it can be done by subtracting μ from x and dividing the
result by σ (standardizing)
¾
z =
x −μ
σ
25
Example: For the following data [3.2, 1.9, 2.7, 4.3, 3.5, 2.8, 3.9,
4.4, 1.5, 1.8], standardize each value in the sample
x
Transformation
z
3.2
[3.2 ‐ 3]/1
0.2
1.9
/
[1.9 ‐ 3]/1
‐1.1
2.7
[2.7 ‐ 3]/1
‐0.3
4.3
[[4.3 ‐ 3]/1
]
1.3
3.5
[3.5 ‐ 3]/1
0.5
2.8
[2.8 ‐ 3]/1
‐0.2
39
3.9
[3 9 3]/1
[3.9 ‐
0.9
4.4
[4.4 ‐ 3]/1
1.4
1.5
[1.5 ‐ 3]/1
‐1.5
1.8
[1.8 ‐ 3]/1
Mean
3
‐1.2
0
St deviation
St deviation
1
1
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26
Example: Let x be a continuous RV that has a normal
distribution with a mean 50 and a standard deviation of 10.
10
Convert the following x values to z values
a) x = 55
b) x = 35
c)) x = 50
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27
To find the area between two values of x for a normal
distribution
¾
¾Convert
both values of x to their respective z values
¾Find the area under the standard normal curve between those two
values
Example:
E
l Let
L t x be
b a continuous
ti
RV that
th t is
i normally
ll distributed
di t ib t d
with a mean of 25 and a standard deviation of 4. Find the area
a)) between
b t
x = 25 and
d x = 32
b) between
bet ee x = 18 and
a d x = 34
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5 Find
the following probabilities
a) P(x > 55)
b) P(x
( < 49)
9)
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8 Find
the probability P(30 < x < 39)
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12 Find
the following probabilities
a) P(70 <x < 135)
b) P(x
( < 27))
QM-120, M. Zainal
Application of the Normal Distribution
31
Example: According to the U.S. Bureau of Census, the mean
income of all U.S.
U S families was $43,237
$43 237 in 1991.
1991 Assume that the
1991 incomes of all U.S. families have a normal distribution with
a mean of $43,237
$43 237 and a standard deviation of $10,500.
$10 500 Find the
probability that the 1991 income of a randomly selected U.S.
family was between $30,000 and $50,000.
QM-120, M. Zainal
32
Example: A racing car is one of the many toys manufactured by
Mack Corporation.
Corporation The assembly time for this toy follows a
normal distribution with a mean of 55 minutes and a standard
deviation of 4 minutes.
minutes The company closes at 5 P.M.
P M every day.
day
If one worker starts assembling a racing car at 4 P.M., what is
the probability that she will finish this job before the company
closes for the day?
QM-120, M. Zainal
33
Example: Hupper Corporation produces many types of soft
drinks,, including
g Orange
g Cola. The filling
g machines are
adjusted to pour 12 ounces of soda in each 12‐ounce can of
Orange Cola. However, the actual amount of soda poured into
each can is not exactly 12 ounces; it varies from can to can. It is
found that the net amount of soda in such a can has a normal
distribution with a mean of 12 ounces and a standard
deviation of .015 ounces.
(a) What is the probability that a randomly selected can of
Orange Cola contains 11.97 to 11.99 ounces of soda?
(b) What percentage of the Orange Cola cans contain 12.02 to
12 07 ounces of soda?
12.07
QM-120, M. Zainal
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Example: The life span of a calculator manufactured by lntal
Corporation has a normal distribution with a mean of 54
months and a standard deviation of 8 months. The company
guarantees that any calculator that starts malfunctioning within
36 months of the purchase will be replaced by a new one. About
what percentage of such calculators made by this company are
expected to be replaced?
QM-120, M. Zainal
Determining the z and x values
35
We reverse the procedure of finding the area under the normal
curve for a specific value of z or x to finding a specific value of z
or x for a known area under the normal curve.
¾
Example: Find a point z such that the area under the standard
normal curve between 0 and z is .4251 and the value of z is
positive
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QM-120, M. Zainal
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Example: Find the value of z such that the area under the
standard normal curve in the right tail is .0050.
0050
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Example: Find the value of z such that the area under the
standard normal curve in the left tail is .050.
050
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39
Finding an x value for a normal distribution:
To find an x value when an area under a normal distribution
curve is given, we do the following
1. Find the z value corresponding to that x value from the
standard normal curve.
2. Transform the z value to x by substituting the values of μ, σ,
and z in the following formula
x = μ +σz
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40
Example: Recall the calculators example, it is known that the life
of a calculator manufactured by Intal Corporation has a normal
distribution with a mean of 54 months and a standard deviation
of 8 months.
months What should the warranty period be to replace a
malfunctioning calculator if the company does not want to
replace more than 1 % of all the calculators sold
QM-120, M. Zainal
41
Example: Most business schools require that every applicant for
admission to a graduate degree program take the GMAT.
GMAT
Suppose the GMAT scores of all students have a normal
distribution with a mean of550 and a standard deviation of 90.
90
Matt Sanger is planning to take this test soon. What should his
score be on this test so that only 10% of all the examinees score
higher than he does?
QM-120, M. Zainal
42
Example: Find the value of z such that 0.95 of the area is within
±z standard deviations of the mean
QM-120, M. Zainal

The Normal Distribution

Transcription

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