LESSON 10: NORMAL DISTRIBUTION
Outline
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Normal distribution
Area under the curve, probability, percentile value
Given z find area
Given percentile value find z
Given x find area
Given percentile value find x
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NORMAL DISTRIBUTION
THE PROBABILITY DENSITY FUNCTION
• If a random variable X with mean  and standard deviation
 is normally distributed, then its probability density function
is given by
 1  1/ 2  x   /  2
f x   
e
  2 
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NORMAL DISTRIBUTION
THE PROBABILITY DENSITY FUNCTION
f (x )
Mean, =50
SD, =10
0.0500
f (x )
0.0400
Area between
the vertical lines
= P (40X 60)
0.0300
Area under
the curve
= 1.00
0.0200
0.0100
0.0000
0
20
40
60
x - VALUES
80
100
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NORMAL DISTRIBUTION
EFFECT OF CHANGING STANDARD DEVIATION
SD,=10
0.0500
SD,=15
f(x)
0.0400
SD,=20
0.0300
0.0200
Mean,= 100
0.0100
0.0000
0
50
100
150
200
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x-values
NORMAL DISTRIBUTION
EFFECT OF CHANGING MEAN
0.0500
f(x)
0.0400
0.0300
0.0200
SD,10
0.0100
0.0000
50
100
150
x -values
200
250
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STANDARD NORMAL DISTRIBUTION
0.4500
Mean=0
SD=1
0.4000
0.3500
f (x )
0.3000
0.2500
0.2000
0.1500
Area under
the curve = 1
0.1000
0.0500
0.0000
-6
-4
-2
0
z -VALUES
2
4
6
6
f(x)
STANDARD NORMAL DISTRIBUTION
f(x)
Area = 1.00
Area
Area
= 0.50 = 0.50
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• If a random
variable X is
normally distributed
with mean  and
standard deviation
, then
z
f(x)
STANDARD NORMAL DISTRIBUTION
RELATIONSHIP BETWEEN x AND z
=50
10
x

or, x    z
-3
-2
-1 z=0
1
2
3
20 30 40 x=50 60 70 80 8
STANDARD NORMAL DISTRIBUTION
TABLE, z-VALUES, AREA AND PROBABILITY
f(x)
Example 1.1: Table D, Appendix A, pp. 536-537 shows the area
under the curve from Z=-∞ to some z value. For example, the
area from Z=-∞ to Z=1.3+.04=1.34 is 0.9099. So,
1.34  0.9099, P   Z  1.34  PZ  1.34  0.9099
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STANDARD NORMAL DISTRIBUTION
TABLE, z-VALUES, AREA AND PROBABILITY
f(x)
Example 1.2: The area shown on the table can be used to get
many other areas. For example, using the fact that the area
under the curve is 1.0, the area from Z=1.34 to Z= is 1.00.9099=0.0901. So, P1.34  Z    PZ  1.34  0.0901
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STANDARD NORMAL DISTRIBUTION
TABLE, z-VALUES, AREA AND PROBABILITY
f(x)
Example 1.3: The area shown on the table can be used to get
area between any two z-values. For example, the area from
z1=-1.25 to z2=1.34 is 0.9099-0.1056=0.8043, where 0.1056 is
the area obtained from Table D for z1=-1.25. So,
P1.25  Z  1.34  0.8043
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STANDARD NORMAL DISTRIBUTION
GIVEN z, FIND PROBABILITY
Example 2: Find the following:
1. PZ  1.63
2. PZ  1.63
3. P 1.05  Z  1.63
4. P1.05  Z  1.63
5. PZ  1.63
6. PZ  1.63
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STANDARD NORMAL DISTRIBUTION
GIVEN z, FIND PROBABILITY : EXCEL
• Excel function NORMSDIST(z) provides the area under
the standard normal distribution curve on the left side of z.
• Example: NORMSDIST(1.34) = 0.9099 = PZ  1.34  1.34
PZ  1.34  ?
f(x)
To get the area on the left of
Z = 1.34, Φ(1.34), use Excel
function NORMSDIST(1.34).
Area =?
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z=1.34
STANDARD NORMAL DISTRIBUTION
AREA AND PERCENTILE
• A percentile is the value at or below which the stated
percentage of units lie. Therefore, percentile corresponds
to an area under the curve. For example, if GMAT scores
are normally distributed with the 78th percentile 600, then
78% scores are less than 600 and P X  600  0.78
Then, the area on the left of X = 600 is 0.78.
f(x)
If the 78th percentile is
600, then the area on
the left of X=600 is 0.78.
Area = 0.78
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X=600
STANDARD NORMAL DISTRIBUTION
GIVEN AREA OR PERCENTILE, FIND z
2. 30th
f(x)
Example 3: If return on investment of a fund has a mean 0
and standard deviation 1, find the returns that corresponds
to following percentiles:
1. 96th
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STANDARD NORMAL DISTRIBUTION
GIVEN AREA OR PERCENTILE, FIND z : EXCEL
• Excel function NORMSINV(p) provides the value of z
corresponding to the 100pth percentile. For example,
NORMSINV(0.33)=-0.44. So, for the standard normal
distribution, the 33rd percentile is -0.44.
f(x)
To get the z value for which
area on the left, Φ(z) = 0.33,
use Excel function
NORMSINV(0.33).
Area
=0.33
z=?
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NORMAL DISTRIBUTION
GIVEN x, FIND PROBABILITY
2. Find area from the Table
f(x)
Example 4.1: A retailer has observed that the monthly
demand of an item is normally distributed with a mean of
650 and standard deviation of 50 units. What is the
probability that the demand of the item in the next month
will not exceed 700 units?
P X  700  ?
1. Compute z
3. Find probability
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NORMAL DISTRIBUTION
GIVEN x, FIND PROBABILITY
2. Find area from the Table
f(x)
Example 4.2: A retailer has observed that the monthly
demand of an item is normally distributed with a mean of
650 and standard deviation of 50 units. What is the
probability that the demand of the item in the next month
will exceed 600 units?
P X  600  ?
1. Compute z
3. Find probability
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NORMAL DISTRIBUTION
GIVEN x, FIND PROBABILITY
2. Find areas from the Table
f(x)
Example 4.3: A retailer has observed that the monthly
demand of an item is normally distributed with a mean of
650 and standard deviation of 50 units. What is the
probability that the demand of the item in the next month
will be between 600 and 700 units?
P600  X  700  ?
1. Compute z1 and z2
3. Find probability
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NORMAL DISTRIBUTION
GIVEN x, FIND PROBABILITY : EXCEL
• Excel function NORMDIST(x,μ,,TRUE) provides the area
under the standard normal distribution curve on the left
side of x. For example, NORMDIST(700,650,50,TRUE) =
0.8413
P X  700  ?
μ=600
σ=50
f(x)
To get the area on the left of
X = 700, when μ=600, σ=50,
use Excel function
NORMSDIST(700,650,50,TRUE)
.
Area=?
X = 700
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NORMAL DISTRIBUTION
GIVEN AREA OR PERCENTILE, FIND x
2. Find x
f(x)
Example 5: A retailer has observed that the monthly demand
of an item is normally distributed with a mean of 650 and
standard deviation of 50 units. If the retailer wants to meet
demand with probability 0.90, how many units should be
ordered for the next month? Assume that there is no units
in the inventory.
1. Find z from the table
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STANDARD NORMAL DISTRIBUTION
GIVEN AREA OR PERCENTILE, FIND x : EXCEL
• Excel function NORMINV(p,μ,) provides 100pth percentile
when mean is μ and standard deviation . So, the function
also gives that value of X for which the area on the left
side of X is p. For example, NORMINV(0.90,650,50) = 714.
So, the 90th percentile is 714.
μ=600
σ=50
f(x)
To get the x value for which
area on the left = 0.90, use
Excel function
NORMSINV(0.90,650,50) if
mean is 650 and standard
deviation 50.
Area
= 0.90
X=?
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READING AND EXERCISES
Lesson 10
Reading:
Section 7-4, pp. 216-225
Exercises:
7-35, 7-36, 7-37
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