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The Normal Distribution
The Normal Distribution
Dr. Himani Gupta
Continuous Probability Distributions
A continuous random variable is a variable that can assume any value on a continuum (can assume an uncountable number of values) thickness of an item time required to complete a task temperature of a solution height
These can potentially take on any value, depending only on the ability to measure precisely and accurately.
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The Normal DistributionProperties
‘Bell Shaped’ Symmetrical Mean, Median and Mode are equal Location is characterized by the mean, μ Spread is characterized by the standard deviation, σThe random variable has an infinite theoretical range: - to +
Mean = Median = Mode
f(X)
μ
σ
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The Normal DistributionDensity Function
2μ)(X21
e2π1f(X)
The formula for the normal probability density function is
Where e = the mathematical constant approximated by 2.71828π = the mathematical constant approximated by 3.14159μ = the population meanσ = the population standard deviationX = any value of the continuous variable
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The Normal DistributionShape
By varying the parameters μ and σ, we obtain different normal distributions
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The Normal DistributionShape
X
f(X)
μ
σ
Changing μ shifts the distribution left or right.
Changing σ increases or decreases the spread.
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The Standardized Normal Distribution
Any normal distribution (with any mean and standard deviation combination) can be transformed into the standardized normal distribution (Z).
Need to transform X units into Z units. The standardized normal distribution has a
mean of 0 and a standard deviation of 1.
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The Standardized Normal Distribution
σμXZ
Translate from X to the standardized normal (the “Z” distribution) by subtracting the mean of X and dividing by its standard deviation:
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The Standardized Normal Distribution: Density Function
2Z2
e2π1f(Z)
The formula for the standardized normal probability density function is
Where e = the mathematical constant approximated by 2.71828π = the mathematical constant approximated by 3.14159Z = any value of the standardized normal distribution
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The Standardized Normal Distribution: Shape
Z
f(Z)
0
1
Also known as the “Z” distribution Mean is 0 Standard Deviation is 1
Values above the mean have positive Z-values, values below the mean have negative Z-values
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The Standardized Normal Distribution: Example
2.050
100200σμXZ
If X is distributed normally with mean of 100 and standard deviation of 50, the Z value for X = 200 is
This says that X = 200 is two standard deviations(2 increments of 50 units) above the mean of 100.
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The Standardized Normal Distribution: Example
Z100
2.00200 X (μ = 100, σ = 50)
(μ = 0, σ = 1)
Note that the distribution is the same, only the scale has changed. We can express the problem in original units (X) or in standardized units (Z)
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Normal Probabilities
a b
f(X)
(Note that the probability of any individual value is zero)
Probability is measured by the area under the curve
P(a ≤ X ≤ b)
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Normal Probabilities
The total area under the curve is 1.0, and the curve is symmetric, so half is above the mean, half is below.
f(X)
0.50.5
1.0)XP(
0.5)XP(μ 0.5μ)XP(
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Normal Probability Tables
Example: P(Z < 2.00) = .9772
The Standardized Normal table in the textbook (Appendix table E.2) gives the probability less than a desired value for Z (i.e., from negative infinity to Z)
Z0 2.00
.9772
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Normal Probability Tables
The value within the table gives the probability from Z = up to the desired Z value.
.9772
2.0P(Z < 2.00) = .9772
The row shows the value of Z to the first decimal point
The column gives the value of Z to the second decimal point
2.0
...
Z 0.00 0.01 0.02 …
0.0
0.1
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Finding Normal ProbabilityProcedure
Draw the normal curve for the problem in terms of X.
Translate X-values to Z-values.
Use the Standardized Normal Table.
To find P(a < X < b) when X is distributed normally:
Dr. Himani Gupta8.18
Example
Suppose that at another gas station the daily demand for regular gasoline is normally distributed with a mean of 1,000 gallons and a standard deviation of 100 gallons.
The station manager has just opened the station for business and notes that there is exactly 1,100 gallons of regular gasoline in storage.
The next delivery is scheduled later today at the close of business. The manager would like to know the probability that he will have enough regular gasoline to satisfy today’s demands.
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The demand is normally distributed with mean µ = 1,000 and standard deviation σ = 100. We want to find the probability
P(X < 1,100)Graphically we want to calculate:
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P(X < 1,100) = = P(Z < 1.00)
100000,1100,1XP
The first step is to standardize X. However, if we perform any operations on X we must perform the same operations on 1,100. Thus,
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probability that we seek is
P(X < 1,100) = P( Z < 1.00) = .8413
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Example : A placement company has conducted a written test to recruit people in a software company. Assume that the test marks are normally distributed with mean 120 and standard deviation 50.
Calculate the following:
(a) Probability of randomly obtaining scores greater than 200 in the test.(b) Probability of randomly obtaining a score that is 180 or less.(c) Probability of randomly obtaining a score less than 80.(d) Probability of randomly selecting a score between 70 to 170 for the exam.(e) Probability of randomly obtaining a score between 80 to 110.
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APPLICATIONS IN FINANCE: Measuring Risk
Computed variance and standard deviation as numerical measure of risk An important application in finance where the emphasis was placed on reducing the variance of the returns on a portfolio. Demonstrate, why risk is measured by the variance and deviation.
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Example
Consider an investment whose return is normally distributed with a mean of 10% and a standard deviation of 5%. a. Determine the probability of losing money.b. Find the probability of losing money when the standard deviation is equal to 10%.
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a The investment loses money when the return is negative. Thus we wish to determine
=P(X < 0)
=P(X < 0) = = P(Z < – 2.00)
5100XP
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From Table
P(Z < − 2.00) = .0228
Therefore the probability of losing money is .0228
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b. If we increase the standard deviation to 10% the probability of suffering a loss becomes
P(X < 0) =
= P(Z < –1.00) = .1587
10100XP
Conclusion:
As increasing the standard deviation increases the probability of losing money. It should be noted that increasing the standard deviation will also increase the probability that the return will exceed some relatively large amount.
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Application in operation management: Determining the reorder point
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Application in Project management
PERT/ CPM
The expected value and variance of the completion time of the project are based on the expected value and variances of the completion times of the activities on the critical path
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z .07 .08 .09. . . . .. . . . .. . . . .1.1 . . . 0.3790 0.3810 0.38301.2 . . . 0.3980 0.3997 0.40151.3 . . . 0.4147 0.4162 0.4177. . . . . . . . . . . . . . .
The area within k of the mean is the same for all normal random variables. To find a probability associated with any interval of values for any normal random variable, all that is needed is to express the interval in terms of numbers of standard deviations from the mean. That is the purpose of the standard normal transformation. If X~N(50,102),
That is, P(X >70) can be found easily because 70 is 2 standard deviations above the mean of X: 70 = + 2. P(X > 70) is equivalent to P(Z > 2), an area under the standard normal distribution.
The area within k of the mean is the same for all normal random variables. To find a probability associated with any interval of values for any normal random variable, all that is needed is to express the interval in terms of numbers of standard deviations from the mean. That is the purpose of the standard normal transformation. If X~N(50,102),
That is, P(X >70) can be found easily because 70 is 2 standard deviations above the mean of X: 70 = + 2. P(X > 70) is equivalent to P(Z > 2), an area under the standard normal distribution.
Example 4-12 X~N(124,122)P(X > x) = 0.10 and P(Z > 1.28) 0.10x = + z = 124 + (1.28)(12) = 139.36
Example 4-12 X~N(124,122)P(X > x) = 0.10 and P(Z > 1.28) 0.10x = + z = 124 + (1.28)(12) = 139.36
18013080
0.04
0.03
0.02
0.01
0.00
X
f(x)
Normal Distribution: = 124, = 12
The Inverse Transformation
0.01
139.36
P X P x P Z P Z( ) ( )
70 70 70 50
102
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Problem
The US environment Protection Agency publishes figures on solid waste in the United states. One year, the average number of waste generated per person per day was 3.58 pounds. Suppose the daily amount of waste is normally distributed, with s.d. 1.04 pounds. Of the daily amounts of waste generated per person, 67.72% would be greater than what amount?
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Given Normal Probability, Find the X Value
Let X represent the time it takes (in seconds) to download an image file from the internet.
Suppose X is normal with mean 8.0 and standard deviation 5.0 Find X such that 20% of download times are less than X.
X? 8.0
.2000
Z? 0
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Given Normal Probability, Find the X Value First, find the Z value corresponds to the
known probability using the table.
.2296
.2005
.1736
.04
.2266.2327….-0.7
.1977.2033….-0.8
.1711.1762….-0.9
.05.03….Z
X? 8.0
.2000
Z-0.84 0
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Given Normal Probability, Find the X Value Second, convert the Z value to X units using
the following formula.
80.30.5)84.0(0.8
ZσμX
So 20% of the download times from the distribution with mean 8.0 and standard deviation 5.0 are less than 3.80 seconds.
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Normal Approximation of the Binomial Distribution The normal distribution can be used to
approximate binomial probabilities Procedure
Convert binomial parameters to normal parameters Does the interval lie between 0 and n? If so,
continue; otherwise, do not use the normal approximation.
Correct for continuity Solve the normal distribution problem
σμ
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Conversion equations
Conversion example:
Normal Approximation of Binomial: Parameter Conversion
n p
n p q
Given that X has a binomial distribution , find and P X n p
n p
n p q
( | . ).( )(. )
( )(. )(. ) .
25 60 3060 30 18
60 30 70 3 55
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Normal Approximation of Binomial: Interval Check
3 18 3 355 18 10653 7353 2865
( . ) ...
0 10 20 30 40 50 60n
70
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Normal Approximation of Binomial: Correcting for Continuity
Values Being
DeterminedCorrection
XXXXXX
+.50-.50-.50+.05
-.50 and +.50+.50 and -.50
The binomial probability, and
is approximated by the normal probabilitP(X 24.5| and
P X n p( | . )
. ).
25 60 30
18 3 55
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0
0.02
0.04
0.06
0.08
0.10
0.12
6 8 10 12 14 16 18 20 22 24 26 28 30
Normal Approximation of Binomial: Graphs
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Normal Approximation of Binomial: Computations
252627282930313233
Total
0.01670.00960.00520.00260.00120.00050.00020.00010.00000.0361
X P(X)
The normal approximation,P(X 24.5| and
18 35524 5 18
355183
5 0 1835 46640336
. )..
( . ). .. ..
P Z
P ZP Z
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Assessing Normality
It is important to evaluate how well the data set is approximated by a normal distribution.
Normally distributed data should approximate the theoretical 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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Assessing Normality
Construct charts or graphs For small- or moderate-sized data sets, do stem-and-
leaf display and box-and-whisker plot look symmetric?
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 σ? Is the range approximately 6 σ?
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Assessing Normality
Observe the distribution of the data set Do approximately 2/3 of the observations lie within
mean ± 1 standard deviation? Do approximately 80% of the observations lie within
mean ± 1.28 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 with positive slope?
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The Normal Probability Plot
Normal probability plot (steps): Arrange data into ordered array Find corresponding standardized normal quantile (Z)
values Plot the pairs of points with observed data values (X) on
the vertical axis and the standardized normal quantile (Z) values on the horizontal axis
Evaluate the plot for evidence of linearity
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The Normal Probability Plot
30
60
90
-2 -1 0 1 2 Z
X
A normal probability plot for data from a normal distribution will be approximately linear:
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The Normal Probability PlotLeft-Skewed Right-Skewed
Rectangular
30
60
90
-2 -1 0 1 2 Z
X
30
60
90
-2 -1 0 1 2 Z
X
30
60
90
-2 -1 0 1 2 Z
XNonlinear plots indicate a deviation from normality
