4.3 normal probability distributions
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4.3 NORMAL PROBABILITY DISTRIBUTIONS. The Most Important Probability Distribution in Statistics. Normal Distribution. A random variable X with mean m and standard deviation s is normally distributed if its probability density function is given by. Normal Probability Distributions. - PowerPoint PPT PresentationTRANSCRIPT
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4.3 NORMAL PROBABILITY DISTRIBUTIONS
The Most Important Probability Distribution in Statistics
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A random variable X with mean and standard deviation is normally distributed if its probability density function is given by
...71828.2eand...14159.3where
xe2
1)x(f
2x
)2/1(
...71828.2eand...14159.3where
xe2
1)x(f
2x
)2/1(
Normal Distribution
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Normal Probability Distributions
The expected value (also called the mean) can be any number
The standard deviation can be any nonnegative number
There are infinitely many normal distributions
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7: Normal Probability Distributions
4
Parameters μ and σ Normal pdfs have two parameters
μ - expected value (mean “mu”) σ - standard deviation (sigma)
σ controls spreadμ controls location
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The effects of The effects of and and
How does the standard deviation affect the shape of f(x)?
= 2
=3 =4
= 10 = 11 = 12How does the expected value affect the location of f(x)?
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X83 6 9 120
A family of bell-shaped curves that differ only in their means and standard deviations.
µ = the mean of the distribution
= the standard deviation
µ = 3 and = 1
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X3 6 9 120
X3 6 9 120
µ = 3 and = 1
µ = 6 and = 1
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X83 6 9 120
X83 6 9 120
µ = 6 and = 2
µ = 6 and = 1
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EXAMPLE A Normal Random Variable
The following data represent the heights (in inches) of a random sample of 50 two-year old males.
(a) Create a relative frequency distribution with the lower class limit of the first class equal to 31.5 and a class width of 1.
(b) Draw a histogram of the data.
(c ) Do you think that the variable “height of 2-year old males” is normally distributed?
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36.0 36.2 34.8 36.0 34.6 38.4 35.4 36.834.7 33.4 37.4 38.2 31.5 37.7 36.9 34.034.4 35.7 37.9 39.3 34.0 36.9 35.1 37.033.2 36.1 35.2 35.6 33.0 36.8 33.5 35.035.1 35.2 34.4 36.7 36.0 36.0 35.7 35.738.3 33.6 39.8 37.0 37.2 34.8 35.7 38.937.2 39.3
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7: Normal Probability Distributions
14
68-95-99.7 Rule forNormal Distributions
68% of the AUC within ±1σ of μ 95% of the AUC within ±2σ of μ 99.7% of the AUC within ±3σ of μ
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7: Normal Probability Distributions
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Example: 68-95-99.7 Rule Wechsler adult intelligence scores: Normally distributed with μ = 100 and σ = 15; X ~ N(100, 15)
68% of scores within μ ± σ = 100 ± 15 = 85 to 115
95% of scores within μ ± 2σ = 100 ± (2)(15) = 70 to 130
99.7% of scores in μ ± 3σ = 100 ± (3)(15) = 55 to 145
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Property and Notation
Property: normal density curves are symmetric around the population mean , so the population mean = population median = population mode =
Notation: X ~ N( is written to denote that the random variable X has a normal distribution with mean and standard deviation .
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Standardizing
Suppose X~N( Form a new random variable by
subtracting the mean from X and dividing by the standard deviation :
(X This process is called standardizing the
random variable X.
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Standardizing (cont.)
(X is also a normal random variable; we will denote it by Z:
Z = (X has mean 0 and standard deviation
1:E(Z) = = 0; SD(Z) =
The probability distribution of Z is called
the standard normal distribution.
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Standardizing (cont.) If X has mean and stand. dev. , standardizing
a particular value of x tells how many standard deviations x is above or below the mean .
Exam 1: =80, =10; exam 1 score: 92
Exam 2: =80, =8; exam 2 score: 90
Which score is better?
1 exam on 92 than better is 2 exam on 90
1.258
10
8
8090z
1.210
12
10
8092z
2
1
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X83 6 9 120
µ = 6 and = 2
Z0 1 2 3-1-2-3
.5.5
µ = 0 and = 1
(X-6)/2
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Pdf of a standard normal rv
A normal random variable x has the following pdf:
zez
forandforsubstituteNZ
xexf
z
x
,2
1)(
becomes rv normal standard for the
10)1,0(~
,)(
2
2
2)(21
2
1
21
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Z = standard normal random variable
= 0 and = 1
Z0 1 2 3-1-2-3
.5.5
Standard Normal Distribution
.5.5
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Important Properties of Z
#1. The standard normal curve is symmetric around the mean 0
#2. The total area under the curve is 1;
so (from #1) the area to the left of 0 is 1/2, and the area to the right of 0 is 1/2
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Finding Normal Percentiles by Hand (cont.)
Table Z is the standard Normal table. We have to convert our data to z-scores before using the table.
The figure shows us how to find the area to the left when we have a z-score of 1.80:
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Areas Under the Z Curve: Using the Table
P(0 < Z < 1) = .8413 - .5 = .3413
0 1Z
.1587.3413
.50
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Standard normal probabilities have been calculated and are provided in table Z.
The tabulated probabilities correspondto the area between Z= - and some z0
Z = z0
P(- <Z<z0)
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Example – continued X~N(60, 8)
In this example z0 = 1.25
0.89440.8944
0.89440.89440.89440.8944= 0.8944
60 70 60( 70)
8 8
( 1.25)
XP X P
P z
P(z < 1.25)
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Examples
P(0 z 1.27) =
1.270 z
Area=.3980
.8980-.5=.3980
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P(Z .55) = A1
= 1 - A2
= 1 - .7088 = .2912
0 .55
A2
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Examples
P(-2.24 z 0) =
Area=.4875
.5 - .0125 = .4875
z-2.24 0Area=.0125
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P(z -1.85) = .0322
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Examples (cont.)
A1 A2
0 2.73z
-1.18
P(-1.18 z 2.73) = A - A1
= .9968 - .1190
= .8778
.1190
.9968
A1
A
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P(-1 ≤ Z ≤ 1) = .8413 - .1587 =.6826
vi) P(-1≤ Z ≤ 1)
.8413.1587
.6826
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Is k positive or negative?
Direction of inequality; magnitude of probability
Look up .2514 in body of table; corresponding entry is -.67
6. P(z < k) = .25146. P(z < k) = .2514
.5 .5.2514
-.67
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Examples (cont.) viii)
250 275( 250) ( )
4325
( ) ( .58) 1 .2810 .719043
P X P Z
P Z P Z
.2810
.7190
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Examples (cont.) ix)
225 275 275 375 27543 43 43
) (225 375)
( 1.16 2.33) .9901 .1230 .8671
x
ix P x
P
P z
.9901.1230
.8671
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X~N(275, 43) find k so that P(x<k)=.9846
88.367275)43(16.2
table)normal standard from(16.243275
43275
43275
43275)(9846.
k
k
kzP
kxPkxP
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P( Z < 2.16) = .9846
0 2.16 Z.1587
.4846
Area=.5
.9846
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Example
Regulate blue dye for mixing paint; machine can be set to discharge an average of ml./can of paint.
Amount discharged: N(, .4 ml). If more than 6 ml. discharged into paint can, shade of blue is unacceptable.
Determine the setting so that only 1% of the cans of paint will be unacceptable
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Solution
XX
P X
P x P P zx
= amount of dye discharged into can~ N( , .4); determine so that
(from standard normal table)
= 6 - 2.33(.4) = 5.068
( ) .
. ( )
.
. . .
.
6 01
01 6
2 33
46
46
46
4