Download 6.2 Normal Distribution

6.2 Normal Distribution
Ziad Zahreddine
Importance of
Normal Distribution
1. Describes Many Random Processes or
Continuous Phenomena
2. Can Be Used to Approximate Discrete
Probability Distributions
„
Example: Binomial
3. Basis for Classical Statistical Inference
Ziad Zahreddine
Normal Distribution
1.
‘Bell-Shaped’ &
Symmetrical
2.
Mean, Median,
Mode Are Equal
3.
Random Variable
Has Infinite Range
that cannot be
listed.
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Mean
Median
Mode
Probability
Density Function
The Probability Density Function n (x; μ, σ) is given by
1
− (1 2 )[( x − μ ) σ ] 2
n ( x; μ , σ ) =
e
σ 2π
where
x
μ
σ
π
=
=
=
=
Value of Random Variable (-∞ < x < ∞)
Mean of the random variable x
Population Standard Deviation
3.14159; e = 2.71828
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Effect of Varying Parameters (μ & σ)
The mean and standard deviation affect
the flatness and center of the curve,
but not the basic shape.
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Ziad Zahreddine
6.3 Areas Under the Normal Curve
Probability is
area
under
curve!
d
P (c < X < d ) = ∫ n( x; μ , σ ) dx ?
c
1
P (c < X < d ) =
σ 2π
∫
d
e
− (1 2 )[( x − μ ) σ ] 2
c Ziad Zahreddine
dx
Standardize the
Normal Distribution
Ziad Zahreddine
Standardize the
Normal Distribution
1
− (1 2 )[( x − μ ) σ ] 2
n( x; μ , σ ) =
e
σ 2π
Normal
Distribution
Infinitely Many
Distributions Ziad Zahreddine
Standardize the
Normal Distribution
1
− (1 2 )[( x − μ ) σ ] 2
n( x; μ , σ ) =
e
σ 2π
n( z;0,1) =
Normal
Distribution
Infinitely Many
Distributions Ziad Zahreddine
1 −(1 2 )z 2
e
2π
Standard Normal
Distribution
Only One
Distribution
The Standard
Normal Distribution
Table for Standard Normal Distribution contains probability
for the area between 0 and z.
Partial table below shows components of table
Value of z a
combination of
column and
row
Probability
associated with a
particular z value, in
this case z=.13,
p(0<z<.13) = .0517
Z
.00
.01
.02
.03
.04
.05
.06
.07
.08
.09
.0
.1
.2
.3
.0000
.0398
.0793
.1179
.0040
.0438
.0832
.1217
.0080
.0478
.0871
.1255
.0120
.0517
.0910
.1293
.0160
.0557
.0948
.1331
.0199
.0596
.0987
.1368
.0239
.0636
.1026
.1406
.0279
.0675
.1064
.1443
.0319
.0714
.1103
.1480
.0359
.0753
.1141
.1517
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The Standard
Normal Distribution
What is P(-1.33 < z < 1.33)?
Table gives us area A1
Symmetry about the mean
tell us that A2 = A1
P(-1.33 < z < 1.33) = P(-1.33 < z < 0) +P(0 < z < 1.33)=
A2 + A1 = .4082 + .4082 = .8164
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The Standard
Normal Distribution
What is P(z > 1.64)?
Table gives us area A2
Symmetry about the mean
tell us that A2 + A1 = .5
P(z > 1.64) = A1 = .5 – A2=.5 - .4495 = .0505
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The Standard
Normal Distribution
What is P(z < .67)?
Table gives us area A1
Symmetry about the mean
tell us that A2 = .5
P(z < .67) = A1 + A2 = .2486 + .5 = .7486
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The Standard
Normal Distribution
What is P(|z| > 1.96)?
P(|z| > 1.96) = P(z < -1.96 or z > 1.96)
Table gives us area .5 - A2
=.4750, so A2 = .0250
Symmetry about the mean
tell us that A2 = A1
P(|z| > 1.96) = A1 + A2 = .0250 + .0250 =.05
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Given a standard normal distribution, find the area under the curve
that lies (a) to the right of z = 1.84 and
(b) Between z = -1.97 and z = 0.86.
Solution
(a) The area to right of z = 1.84 is equal to
0.5 minus the area in the table.
That is 0.5 - 0.4671 = 0.0329
1.84
(b) The area between 0 and 0.86 is 0.3051.
By symmetry the area between -1.97 and
0 is the same as the area between 0 and 1.97,
which is 0.4756.
the total area is 0.3051 + 0.4756 = 0.7807.
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-1.97
0.86
Given a standard normal distribution, find the value of k such that
(a) P(Z > k) = 0.3015, and
(b) P(k < Z < -0.18) = 0.4197.
Solution
Why?
(a) k lies to the right of 0.
The total area to the right of zero is 0.5.
0.5 – 0.3015 = 0.1985.
According to the table, the value of k such
that P(0 < Z < k) = 0.1985 is k = 0.52.
(b) The area between -0.18 and 0 is the
0.4197
same as the area between 0 and 0.18
which is 0.0714.
The area between k and 0 becomes:
k
-0.18
0.4197 + 0.0714 = 0.4911.
From the table, the symmetric value of k is 2.37.
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Hence k = -2.37.
0.3015
k
The Normal Distribution
What if values of interest were
not standarized?
We want to know
P (8 < x < 12), with μ = 10 and σ = 1.5
Convert to standard normal using
x = 8 ⇒ z = (8 – 10)/1.5 = -1.33
x = 12 ⇒ z = (12 – 10)/1.5 = 1.33
z=
x−μ
σ
P(8 < x < 12) = P(-1.33 < z < 1.33) = 2(.4082) = .8164
Ziad Zahreddine
The Normal Distribution
We want to know
P (45 < x < 62), with μ = 50 and σ = 10
Convert to standard normal using
x = 45 ⇒ z = (45 – 50)/10 = -0.5
x = 62 ⇒ z = (62 – 50)/10 = 1.2
z=
x−μ
σ
P(45 < x < 62) = P(-0.5 < z < 1.2) =
0.1915 + 0.3849 = 0.5764.
Ziad Zahreddine
The Normal Distribution
Steps for Finding a Probability Corresponding to a
Normal Random Variable
•Sketch the distribution, locate mean, shade area
of interest
•Convert to standard z values using z = x − μ
σ
•Add z values to the sketch
•Use tables to calculate probabilities, making use
of symmetry property where necessary
Ziad Zahreddine
The Normal Distribution
Given that X has a normal distribution
with mean of 27 and standard deviation
of 3, find the probability that X assumes
a value less than 20.
Z value for x = 20 is -2.33
P(x < 20) = P(z < -2.33) = .5 - .4901 = .0099
You could reasonably conclude that this is a rare event
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The Normal Distribution
You can also use the table
in reverse to find a z-value
that corresponds to a
particular probability
What is the value of z that will be exceeded only 10% of the
time?
Look in the body of the table for the value closest to .4, and
read the corresponding z value
Z = 1.28
Ziad Zahreddine
The Normal Distribution
Which values of z enclose the
middle 95% of the standard
normal z values?
Using the symmetry property, z0 must correspond with a
probability of .475
From body of the table, we find that z0 and –z0 are 1.96 and
-1.96 respectively.
Ziad Zahreddine
The Normal Distribution
Given a normally distributed
variable x with mean 100,000 and
standard deviation of 10,000, what
value of x identifies the top 10%
of the distribution?
P(100,000 ≤ x ≤ x0 ) = P⎛⎜ 0 ≤ z ≤ x0 − μ ⎞⎟ = P⎛⎜ 0 ≤ z ≤ x0 − 100,000 ⎞⎟ = .40
σ ⎠
10,000 ⎠
⎝
⎝
The z value corresponding with .40 is 1.28.
Solving for x0
x0 = 100,000 +1.28(10,000) = 100,000 +12,800 = 112,800
Ziad Zahreddine
6.4 Applications of the Normal Distribution
Example
A certain type of storage battery lasts, on average, 3 years,
with a standard deviation of 0.5 year.
Assuming that the battery lives are normally distributed, find
the probability that a given battery will last less than 2.3 years.
Solution z = 2.3 − 3 = −1.4
0.5
P( X < 2.3) = P(Z < −1.4) = P(Z > 1.4)
= 0.5 − 0.4192 = 0.0808.
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2.3
3
Example
An electrical firm manufactures light bulbs that have a life,
before burn-out, that is normally distributed with mean equal to
800 hours and a standard deviation of 40 hours.
Find the probability that a bulb burns between 778 and 834
hours.
Solution
834 − 800
778 − 800
z1 =
= −0.55 and z2 =
= 0.85
40
40
P(778 < X < 834) = P(−0.55 < Z < 0.85) = 0.2088 + 0.3023 = 0.5111
σ = 40
778
834
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μ = 800
Example
In an industrial process, the specifications on the diameter of a
ball bearing is 3 ± 0.01 cm.
The implication is that no part falling outside these
specifications will be accepted.
It is known that the diameter of a ball bearing has a normal
distribution with mean 3 and standard deviation 0.005.
On the average, how many manufactured ball bearings will be
scrapped?
0.0228
Solution
z1 =
3.01 − 3
2.99 − 3
= −2 and z2 =
= 2.
0.005
0.005
P(0 < X < 2) = 0.4772
By symmetry
⇒
2.99
0.0228
3
3.01
P( X > 2) = 0.5 − 0.4772 = 0.0228
P( X > 2 or X < −2) = 2 × 0.0228 = 0.0456.
Therefore, on the average, 4.56% of manufactured ball
bearings will be scrapped. Ziad Zahreddine
Example
Gauges are used to reject all components where a certain
dimension is not within the specification 1.50 ± d.
It is known that this measurement is normally distributed with
mean 1.50 and standard deviation 0.2.
Determine the value of d such that the specifications cover
95% of the measurements.
Solution From the table, the Z value
corresponding to the area 0.95/2 = 0.475
is Z = 1.96. In other words
P(0 < Z < 1.96) = 0.475.
0.025
(1.50 + d ) − 1.50
Therefore 1.96 =
0.2
From which we obtain d = (0.2)(1.96) = 0.392.
Ziad Zahreddine
1.5 - d
1.108
0.025
1.50
1.5 + d
1.892
Example
A machine makes electrical resistors having a mean resistance
of 40 Ohms and a standard deviation of 2 Ohms.
Assuming that the resistance follows a normal distribution and
an be measured to any degree of accuracy, what percentage
of resistors will have a resistance exceeding 43 Ohms?
Solution
0.4332
The Z value corresponding to 43 is
Z=
43 − 40
= 1.5
2
0.0668
From the table
P(0 < Z < 1.5) = 0.4332.
40
0
43
1.5
Therefore
P(X > 43) = P(Z > 1.5) = 0.5 − P(0 < Z < 1.5) = 0.5 − 0.4332 = 0.0668.
6.68% of the resistors will have a resistance exceeding 43
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Ohms.
Example
The average grade for an exam is 74, and the standard
deviation is 7.
If 12% of the class are given A’s, and the grades are curved to
follow a normal distribution, what is the lowest possible A and
the highest possible B?
Solution
0.38
0.5 – 0.12 = 0.38.
By examining the table, we take Z to be
0.12
the average of 1.17 and 1.18.
1.17 + 1.18
= 1.175.
Z=
74 82.225
2
0 1.175
Therefore
X − 74
1.175 =
7
Leading to X = (7)(1.175) + 74 = 82.225.
Therefore, the lowest A is 83 and the highest B is 82.
Ziad Zahreddine
Sections 6.1, 6.2, 6.3, 6.4, P. 156
Assignment: 1, 2, 3, 4, 5, 7, 8.
Ziad Zahreddine