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Copyright 2012 Pearson Education, Inc. publishing as Prentice Hall Chap 6-1Chap 6-1
Mirza Amin ul Haq
The Normal Distribution andSampling Distribution
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Copyright 2012 Pearson Education, Inc. publishing as Prentice Hall Chap 6-2Chap 6-2
Continuous Probability Distributions
A continuous random variableis a variable thatcan assume any value on a continuum (canassume an uncountable number of values)
thickness of an item time required to complete a task
temperature of a solution
height, in inches
These can potentially take on any valuedepending only on the ability to precisely andaccurately measure
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Copyright 2012 Pearson Education, Inc. publishing as Prentice Hall Chap 6-3Chap 6-3
The Normal Distribution
Bell Shaped
Symmetrical
Mean, Median and Mode
are EqualLocation is determined by themean,
Spread is determined by thestandard deviation,
The random variable has aninfinite theoretical range:+ to
Mean= Median= Mode
X
f(X)
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Copyright 2012 Pearson Education, Inc. publishing as Prentice Hall Chap 6-4Chap 6-4
By varying the parameters and , we obtaindifferent normal distributions
Many Normal Distributions
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Copyright 2012 Pearson Education, Inc. publishing as Prentice Hall Chap 6-5Chap 6-5
The Normal DistributionShape
X
f(X)
Changing shifts thedistribution left or right.
Changing increasesor decreases thespread.
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Copyright 2012 Pearson Education, Inc. publishing as Prentice Hall Chap 6-6Chap 6-6
The Standardized Normal
Anynormal distribution (with any mean andstandard deviation combination) can be
transformed into the standardized normaldistribution (Z)
Need to transform X units into Z units
The standardized normal distribution (Z) has amean of 0 and a standard deviation of 1
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7/41Copyright 2012 Pearson Education, Inc. publishing as Prentice Hall Chap 6-7Chap 6-7
Translation to the StandardizedNormal Distribution
Translate from X to the standardized normal(the Z distribution) by subtracting the mean
of X and dividing by its standard deviation:
XZ
The Z distribution always has mean = 0 and
standard deviation = 1
8/10/2019 Normal Distribution and Sampling Theory
8/41Copyright 2012 Pearson Education, Inc. publishing as Prentice Hall Chap 6-8Chap 6-8
The StandardizedNormal Distribution
Also known as the Z distribution
Mean is 0
Standard Deviation is 1
Z
f(Z)
0
1
Values above the mean have positiveZ-values,values below the mean have negativeZ-values
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9/41Copyright 2012 Pearson Education, Inc. publishing as Prentice Hall Chap 6-9Chap 6-9
Example
If X is distributed normally with mean of $100and standard deviation of $50, the Z valuefor X = $200 is
This says that X = $200 is two standarddeviations (2 increments of $50 units) abovethe mean of $100.
2.0$50
100$$200
XZ
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10/41Copyright 2012 Pearson Education, Inc. publishing as Prentice Hall Chap 6-10Chap 6-10
Comparing X and Z units
Z
$100
2.00
$200 $X
Note that the shape of the distribution is the same,only the scale has changed. We can express theproblem in the original units (X in dollars) or in
standardized units (Z)
(= $100, = $50)
(= 0, = 1)
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Finding Normal Probabilities
a b X
f(X) P a X b( )
Probability is measured by the areaunder the curve
P a X b( )
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f(X)
X
Probability asArea Under the Curve
0.50.5
The total area under the curve is 1.0, and the curve issymmetric, so half is above the mean, half is below
1.0)XP(
0.5)XP( 0.5)XP(
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The Standardized Normal Table
The Cumulative Standardized Normal tablegives the probability less thana desired valueof Z (i.e., from negative infinity to Z)
Z0 2.00
0.9772Example:
P(Z < 2.00) = 0.9772
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14/41Copyright 2012 Pearson Education, Inc. publishing as Prentice Hall Chap 6-14Chap 6-14
The Standardized Normal Table
The value within thetable gives theprobability from Z =up to the desired Z-value
.9772
2.0P(Z < 2.00) =0.9772
The row showsthe value of Zto the first
decimal point
The columngives the value ofZ to the second decimal point
2.0
.
.
.
(continued)
Z 0.00 0.01 0.02
0.0
0.1
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15/41Copyright 2012 Pearson Education, Inc. publishing as Prentice Hall Chap 6-15Chap 6-15
General Procedure forFinding Normal Probabilities
Draw the normal curve for the problem interms of X
Translate X-values to Z-values
Use the Standardized Normal Table
To find P(a < X < b) when X isdistributed normally:
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Finding Normal Probabilities
Let X represent the time it takes (in seconds)to download an image file from the internet.
Suppose X is normal with a mean of 18.0
seconds and a standard deviation of 5.0seconds. Find P(X < 18.6)
18.6
X18.0
8/10/2019 Normal Distribution and Sampling Theory
17/41Copyright 2012 Pearson Education, Inc. publishing as Prentice Hall Chap 6-17Chap 6-17
Let X represent the time it takes, in seconds to download an image filefrom the internet.
Suppose X is normal with a mean of 18.0 seconds and a standarddeviation of 5.0 seconds. Find P(X < 18.6)
Z0.120X18.618
= 18
= 5
= 0
= 1
(continued)
Finding Normal Probabilities
0.125.0
8.0118.6
X
Z
P(X < 18.6) P(Z < 0.12)
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18/41Copyright 2012 Pearson Education, Inc. publishing as Prentice Hall Chap 6-18Chap 6-18
Z
0.12
Z .00 .01
0.0 .5000 .5040 .5080
.5398 .5438
0.2 .5793 .5832 .5871
0.3 .6179 .6217 .6255
Solution: Finding P(Z < 0.12)
0.5478.02
0.1 .5478
Standardized Normal ProbabilityTable (Portion)
0.00
= P(Z < 0.12)
P(X < 18.6)
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19/41Copyright 2012 Pearson Education, Inc. publishing as Prentice Hall Chap 6-19Chap 6-19
Finding NormalUpper Tail Probabilities
Suppose X is normal with mean 18.0and standard deviation 5.0.
Now Find P(X > 18.6)
X
18.6
18.0
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20/41Copyright 2012 Pearson Education, Inc. publishing as Prentice Hall Chap 6-20Chap 6-20
Now Find P(X > 18.6)
(continued)
Z
0.12
0Z
0.12
0.5478
0
1.000 1.0 - 0.5478= 0.4522
P(X > 18.6) = P(Z > 0.12) = 1.0 - P(Z 0.12)
= 1.0 - 0.5478 = 0.4522
Finding NormalUpper Tail Probabilities
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21/41Copyright 2012 Pearson Education, Inc. publishing as Prentice Hall Chap 6-21Chap 6-21
Finding a Normal ProbabilityBetween Two Values
Suppose X is normal with mean 18.0 andstandard deviation 5.0. Find P(18 < X < 18.6)
P(18 < X < 18.6)
= P(0 < Z < 0.12)
Z0.120X18.618
05
8118
XZ
0.125
8118.6
XZ
Calculate Z-values:
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Z
0.12
Solution: Finding P(0 < Z < 0.12)
0.0478
0.00
= P(0 < Z < 0.12)
P(18 < X < 18.6)
= P(Z < 0.12)P(Z 0)
= 0.5478 - 0.5000 = 0.0478
0.5000
Z .00 .01
0.0 .5000 .5040 .5080
.5398 .5438
0.2 .5793 .5832 .5871
0.3 .6179 .6217 .6255
.02
0.1 .5478
Standardized Normal ProbabilityTable (Portion)
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23/41Copyright 2012 Pearson Education, Inc. publishing as Prentice Hall Chap 6-23Chap 6-23
Suppose X is normal with mean 18.0and standard deviation 5.0.
Now Find P(17.4 < X < 18)
X
17.418.0
Probabilities in the Lower Tail
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Copyright 2012 Pearson Education, Inc. publishing as Prentice Hall Chap 6-24Chap 6-24
Probabilities in the Lower Tail
Now Find P(17.4 < X < 18)
X17.4 18.0
P(17.4 < X < 18)
= P(-0.12 < Z < 0)
= P(Z < 0)P(Z -0.12)
= 0.5000 - 0.4522 = 0.0478
(continued)
0.0478
0.4522
Z-0.12 0
The Normal distribution issymmetric, so this probabilityis the same as P(0 < Z < 0.12)
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Exercise
Given a normal distribution with = 50 and =10, find the probability that X assumes a valuebetween 45 and 62.
Given a normal distribution with = 300 and = 50, find the probability that X assumes avalue greater than 362.
A certain type of battery lasts on average 3years with a SD of 0.5 years. Assuming that thebattery lives are normally distributed, find theprobability that a given battery will last less than
2.3 yearsCopyright 2012 Pearson Education, Inc. publishing as Prentice Hall Chap 6-25
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Exercise
An electrical firm manufactures light bulbs thathave a length of life that is normally distributedwith mean 800 hours and a standard deviation
of 40 hours. Find the probability that a bulbburns between 778 and 834 hours.
Copyright 2012 Pearson Education, Inc. publishing as Prentice Hall Chap 6-26
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Copyright 2012 Pearson Education, Inc. publishing as Prentice Hall Chap 6-27Chap 6-27
Steps to find the X value for a knownprobability:
1. Find the Z-value for the known probability2. Convert to X units using the formula:
Given a Normal ProbabilityFind the X Value
ZX
Fi di th X l f
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Copyright 2012 Pearson Education, Inc. publishing as Prentice Hall Chap 6-28Chap 6-28
Finding the X value for aKnown Probability
Example:
Let X represent the time it takes (in seconds) todownload an image file from the internet.
Suppose X is normal with mean 18.0 and standarddeviation 5.0
Find X such that 20% of download times are less thanX.
X? 18.0
0.2000
Z? 0
(continued)
Fi d th Z l f
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Copyright 2012 Pearson Education, Inc. publishing as Prentice Hall Chap 6-29Chap 6-29
Find the Z-value for20% in the Lower Tail
20% area in the lowertail is consistent with aZ-value of -0.84Z .03
-0.9 .1762 .1736
.2033
-0.7 .2327 .2296
.04
-0.8 .2005
Standardized Normal ProbabilityTable (Portion)
.05
.1711
.1977
.2266
X? 18.0
0.2000
Z-0.84 0
1. Find the Z-value for the known probability
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Copyright 2012 Pearson Education, Inc. publishing as Prentice Hall Chap 6-30Chap 6-30
2. Convert to X units using the formula:
Finding the X value
8.13
0.5)84.0(0.18
ZX
So 20% of the values from a distributionwith mean 18.0 and standard deviation5.0 are less than 13.80
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Exercise
Given a normal distribution with = 40 and =6, find X that has
38% of the area below it
5% of the area above it
On a examination the average grade was 74and SD was 7. If 12% of the class are given A
and the grades are curved to follow normaldistribution, what is the lowest possible A andhighest possible B?
Copyright 2012 Pearson Education, Inc. publishing as Prentice Hall Chap 6-31
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Copyright 2012 Pearson Education, Inc. publishing as Prentice Hall Chap 6-32
Using Excel With The NormalDistribution
Chap 6-32
Finding Normal Probabilities
Finding X Given A Probability
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Copyright 2012 Pearson Education, Inc. publishing as Prentice Hall Chap 6-33Chap 6-33
Evaluating Normality
Not all continuous distributions are normal
It is important to evaluate how well the data set isapproximated by a normal distribution.
Normally distributed data should approximate thetheoretical normal distribution:
The normal distribution is bell shaped (symmetrical)where the mean is equal to the median.
The empirical rule applies to the normal distribution. The interquartile range of a normal distribution is 1.33
standard deviations.
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Copyright 2012 Pearson Education, Inc. publishing as Prentice Hall Chap 6-34Chap 6-34
Evaluating Normality
Comparing data characteristics to theoreticalproperties
Construct charts or graphs For small- or moderate-sized data sets, construct a stem-and-leaf
display or a boxplot to check for symmetry
For large data sets, does the histogram or polygon appear bell-shaped?
Compute descriptive summary measures Do the mean, median and mode have similar values?
Is the interquartile range approximately 1.33 ?
Is the range approximately 6 ?
(continued)
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Copyright 2012 Pearson Education, Inc. publishing as Prentice Hall Chap 6-35Chap 6-35
Evaluating Normality
Comparing data characteristics to theoreticalproperties
Observe the distributionof the data set Do approximately 2/3 of the observations lie within mean 1
standard deviation?
Do approximately 80% of the observations lie within mean1.28 standard deviations?
Do approximately 95% of the observations lie within mean 2
standard deviations?
Evaluate normal probability plot Is the normal probability plot approximately linear (i.e. a straight
line) with positive slope?
(continued)
Using A Normal Distribution To
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Copyright 2012 Pearson Education, Inc. publishing as Prentice Hall Chap 6-36
Using A Normal Distribution ToApproximate A Binomial Probability
A binomial distribution is a discrete distributionwhich can only take on the values of 0, 1, 2, . . ,n.
When n gets large the calculations associatedwith the binomial distribution become tedious.
In these situations can use a normal distribution
with the same mean and standard deviation asthe binomial to approximate the binomialprobability
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Copyright 2012 Pearson Education, Inc. publishing as Prentice Hall Chap 6-37
For a binomial random variable X, P(X = c) is nonzerofor c = 0, 1, 2, . . . n.
For a normal random variable W, P(W = c) for any value
c is zero. So to approximate a binomial probability using the
normal distribution have to use a continuity adjustment.
If X is binomial and W is normal we approximate P(X=c)
by P(c0.5 < W < c + 0.5) where W has the samemean and standard deviation as X.
Adding and subtracting the 0.5 is the continuityadjustment
The Need For A Continuity Adjustment
Wh C Th N l
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Copyright 2012 Pearson Education, Inc. publishing as Prentice Hall Chap 6-38
When Can The NormalApproximation Be Used
The normal approximation can be used as long as:
np 5 and
n(1p) 5
Recall
Mean of a binomial is = np
Standard deviation of a binomial is =SQRT(np(1p))
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Copyright 2012 Pearson Education, Inc. publishing as Prentice Hall Chap 6-39
An Example
You select a random sample of n = 1600 tiresfrom a production process with a defect rate of8%. You want to calculate the probability that
150 or fewer tires will be defective.
Here = 1600*0.08 = 128 and =SQRT(1600*0.08*0.92) = 10.85.
Let X be a normal random variable with thismean and standard deviation then the desiredprobability is P(X < 150.5)
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Copyright 2012 Pearson Education, Inc. publishing as Prentice Hall Chap 6-40
Example (Cont)
This is P(Z
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Exercise
The probability that a patient recovers from arare bold disease is 0.6. If 100 people areknown to have contracted this disease, what is
the probability that less than one-half survive?
A Quiz has 200 questions, each with 4 possibleanswers, of which only 1 is the correct answer.
What is the probability that sheer guessworkyields from 25 to 30 correct answers for 80 ofthe 200 problems about which the student hasno knowledge?