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The Normal
Distribution
Ch. 9, Part b
f(x)

x
Continuous Probability Distributions




A continuous random variable can assume any
real number value within some interval on the
real line or in a collection of intervals.
It is not possible to talk about the probability of
the random variable assuming a particular value.
Instead, we talk about the probability of the
random variable assuming a value within a given
interval.
The probability of the random variable assuming
a value within some given interval from x1 to x2
is defined to be the area under the graph of the
probability density function between x1 and x2.
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Normal Probability Distribution

Graph of the Normal Probability Density Function
f(x)
P  x1  x  x2 
x1

x2
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x
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Normal Probability Distribution
Characteristics of the Normal Probability Distribution
 The shape of the normal curve is often
illustrated as a bell-shaped curve.
 Two parameters
 (mean) determines the location of the
distribution. It can be any numerical value:
negative, zero, or positive.
 (standard deviation) determines the width of
the curve: larger values result in wider, flatter
curves.
 The highest point on the normal curve is at the
mean, which is also the median and mode.
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Normal Probability Distribution
Characteristics of the Normal Probability Distribution
 The normal curve is symmetric.
 The total area under the curve is 1 (.5 to the left of
the mean and .5 to the right).
 Probabilities for the normal random variable are
given by areas under the curve.
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Empirical Rule (68 – 95 – 99.7 Rule)

% of Values in Some Commonly Used Intervals



68.26% of values of a normal random variable are
within +/- 1 standard deviation of its mean.
95.44% of values of a normal random variable are
within +/- 2 standard deviations of its mean.
99.72% of values of a normal random variable are
within +/- 3 standard deviations of its mean.
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Empirical Rule (68 – 95 – 99.7 Rule)
f(x)
P  x2  x  x1   .6826
x6
x4
x2
3
2
1

x1
x3
x5
1
2
3
x
σ
68.26%
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Empirical Rule (68 – 95 – 99.7 Rule)
f(x)
P  x4  x  x3   .9544
x6
x4
x2
3
2
1

x1
x3
x5
1
2
3
x
σ
95.44%
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Empirical Rule (68 – 95 – 99.7 Rule)
f(x)
P  x6  x  x5   .9972
x6
x4
x2
3
2
1

x1
x3
x5
1
2
3
x
σ
99.72%
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Standard Normal Probability Distribution



A normal probability distribution with a mean of zero and
a standard deviation of one.
The letter z is commonly used to designate the standard
normal random variable.
Converting to the Standard Normal Distribution
z
x

We can think of z as a measure of the number of
. standard deviations x is from .
n
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Standard Normal Probability Distribution
P(z)
=0
=1
P = 1.0
0
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z
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Standard Normal Probability Distribution
P(z)
=0
=1
P(z > 0) = .5
P(z < 0) = .5
0
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z
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Standard Normal Probability Distribution
P(z)
=0
=1
P  z  1.00
z
0
1
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Standard Normal Probability Table
z - Table
z
.00
.01
.02
.03
.04
.05
.06
.07
.08
.09
.0 .5000 .5040 .5080 .5120 .5160 .5199 .5239 .5279 .5319 .5359
.1 .5398 .5438 .5478 .5517 .5557 .5596 .5636 .5675 .5714 .5733
.2 .5793 .5832 .5871 .5910 .5948 .5987 .6026 .6064 .6103 .6141
.3 .6179 .6217 .6255 .6293 .6331 .6368 .6406 .6443 .6480 .6517
.4
.6554 .6591 .6628 .6664 .6700 .6736 .6772 .6808 .6844 .6879
.5
.6915 .6950 .6985 .7019 .7054 .7088 .7123 .7157 .7190 .7224
.6
.7257 .7291 .7324 .7357 .7389 .7422 .7454 .7486 .7518 .7549
.7
.7580 .7612 .7642 .7673 .7704 .7734 .7764 .7794 .7823 .7852
.8
.7881 .7910 .7939 .7967 .7995 .8023 .8051 .8078 .8106 .8133
.9
.8159 .8186 .2120 .8238 .8264 .8289 .8315 .8340 .8365 .8389
1.0 .8413 .8438 .8461 .8485 .8508 .8531 .8554 .8577 .8599 .8621
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Standard Normal Probability Table
z - Table
z
.00
.01
.02
.03
.04
.05
.06
.07
.08
.09
.0 .5000 .5040 .5080 .5120 .5160 .5199 .5239 .5279 .5319 .5359
.1 .5398 .5438 .5478 .5517 .5557 .5596 .5636 .5675 .5714 .5733
.2 .5793 .5832 .5871 .5910 .5948 .5987 .6026 .6064 .6103 .6141
.3 .6179 .6217 .6255 .6293 .6331 .6368 .6406 .6443 .6480 .6517
.4
.6554 .6591 .6628 .6664 .6700 .6736 .6772 .6808 .6844 .6879
.5
.6915 .6950 .6985 .7019 .7054 .7088 .7123 .7157 .7190 .7224
.6
.7257 .7291 .7324 .7357 .7389 .7422 .7454 .7486 .7518 .7549
.7
.7580 .7612 .7642 .7673 .7704 .7734 .7764 .7794 .7823 .7852
.8
.7881 .7910 .7939 .7967 .7995 .8023 .8051 .8078 .8106 .8133
.9
.8159 .8186 .2120 .8238 .8264 .8289 .8315 .8340 .8365 .8389
1.0 .8413 .8438 .8461 .8485 .8508 .8531 .8554 .8577 .8599 .8621
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Standard Normal Probability Distribution
f(z)
=0
=1
P  z  1.00  .8413
z
0
1
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Standard Normal Probability Distribution
f(z)
=0
=1
P  z  1.00
z
0
1
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Standard Normal Probability Distribution
f(z)
=0
=1
P  z  1.00  1  P( z  1)
z
0
1
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Standard Normal Probability Distribution
f(z)
=0
=1
1  .8413  .1587
z
0
1
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Using the Standard Normal Probability
Table
P  z  .83
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P  z  .83
n
z
Using the Standard Normal Probability Table (Table 1)
.00
.01
.02
.03
.04
.05
.06
.07
.08
.09
.0 .5000 .5040 .5080 .5120 .5160 .5199 .5239 .5279 .5319 .5359
.1 .5398 .5438 .5478 .5517 .5557 .5596 .5636 .5675 .5714 .5733
.2 .5793 .5832 .5871 .5910 .5948 .5987 .6026 .6064 .6103 .6141
.3 .6179 .6217 .6255 .6293 .6331 .6368 .6406 .6443 .6480 .6517
.4
.6554 .6591 .6628 .6664 .6700 .6736 .6772 .6808 .6844 .6879
.5
.6915 .6950 .6985 .7019 .7054 .7088 .7123 .7157 .7190 .7224
.6
.7257 .7291 .7324 .7357 .7389 .7422 .7454 .7486 .7518 .7549
.7
.7580 .7612 .7642 .7673 .7704 .7734 .7764 .7794 .7823 .7852
.8
.7881 .7910 .7939 .7967 .7995 .8023 .8051 .8078 .8106 .8133
.9
.8159 .8186 .2120 .8238 .8264 .8289 .8315 .8340 .8365 .8389
1.0 .8413 .8438 .8461 .8485 .8508 .8531 .8554 .8577 .8599 .8621
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Using the Standard Normal Probability
Table
P  z  .83  .7967
P  z  .38 
P  z  2.86 
P  0  z  1
P  1  z  1
P  z  1.8
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Example
Annual salaries for sales associates from a
particular store have a mean of $32,500 and a
standard deviation of $2,500.
a. Suppose that the distribution of annual salaries
follows a normal distribution. Calculate and
interpret the z-score for a sales associate who makes
$36,000.
b. Calculate the probability that a randomly selected
sales associate earns $30,000 or less.
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Example
Annual salaries for sales associates from a
particular store have a mean of $32,500 and a
standard deviation of $2,500.
c. Calculate the percentage of sales associates with
salaries between $35,625 and $38,750.
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Example: Pep Zone

Standard Normal Probability Distribution
Pep Zone sells auto parts and supplies including
a popular multi-grade motor oil. When the stock
of this oil drops to 20 gallons, a replenishment
order is placed. The store manager is concerned
that sales are being lost due to stock-outs while
waiting for an order. It has been determined that
lead-time demand (x) is normally distributed
with a mean of 15 gallons and a standard
deviation of 6 gallons. The manager would like
to know the probability of a stock-out, P(x  20).
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Example: Pep Zone
f(x)
 = 15
=6
15
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x
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Example: Pep Zone
f(x)
 = 15
=6
P(x  20)
15 20
0
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x
z
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Example: Pep Zone
f(x)
 = 15
=6
z
x

20  15
z
 .83
6
15 20
0
.83
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x
z
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Example: Pep Zone

Standard Normal Probability Distribution
The Standard Normal table shows an area of .7967 for
the region below z = .83. The shaded tail area is
1.00 - .7967 = .2033. The probability of a stock-out is
.2033.
Area = .7967
Area = 1.00 - .7967
= .2033
0
.83
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z
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Example: Pep Zone

Standard Normal Probability Distribution
If the manager of Pep Zone wants the probability
of a stock-out to be no
more than .05, what
should the reorder
point be?
Area = .05
Area = .95
0
z.05
Let z.05 represent the z value cutting the .05 tail area.
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Example: Pep Zone

Using the Standard Normal Probability Table
We now look-up the .9500 area in the Standard
Normal Probability table to find the corresponding
z.05 value.
z
.00
.01
.02
.03
.04
.05
.06
.07
.08
.09
.
1.5 .9332 .9345 .9357 .9370 .9382 .9394 .9406 .9418 .9429 .9441
1.6 .9452 .9463 .9474 .9484 .9495 .9505 .9515 .9525 .9535 .9545
1.7 .9554 .9564 .9573 .9582 .9591 .9599 .9608 .9616 .9625 .9633
1.8 .9641 .9649 .9656 .9664 .9671 .9678 .9686 .9693 .9699 .9706
1.9 .9713 .9719 .9726 .9732 .9738 .9744 .9750 .9756 .9761 .9767
.
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Example: Pep Zone

Using the Standard Normal Probability Table
We now look-up the .9500 area in the Standard
Normal Probability table to find the corresponding
z.05 value.
z
.00
.01
.02
.03
.04
.05
.06
.07
.08
.09
.
1.5 .9332 .9345 .9357 .9370 .9382 .9394 .9406 .9418 .9429 .9441
1.6 .9452 .9463 .9474 .9484 .9495 .9505 .9515 .9525 .9535 .9545
1.7 .9554 .9564 .9573 .9582 .9591 .9599 .9608 .9616 .9625 .9633
1.8 .9641 .9649 .9656 .9664 .9671 .9678 .9686 .9693 .9699 .9706
1.9 .9713 .9719 .9726 .9732 .9738 .9744 .9750 .9756 .9761 .9767
.
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Example: Pep Zone

Using the Standard Normal Probability Table
We now look-up the .9500 area in the Standard
Normal Probability table to find the corresponding
z.05 value.
z
.00
.01
.02
.03
.04
.05
.06
.07
.08
.09
.
1.5 .9332 .9345 .9357 .9370 .9382 .9394 .9406 .9418 .9429 .9441
1.6 .9452 .9463 .9474 .9484 .9495 .9505 .9515 .9525 .9535 .9545
1.7 .9554 .9564 .9573 .9582 .9591 .9599 .9608 .9616 .9625 .9633
1.8 .9641 .9649 .9656 .9664 .9671 .9678 .9686 .9693 .9699 .9706
1.9 .9713 .9719 .9726 .9732 .9738 .9744 .9750 .9756 .9761 .9767
.
z.05 = 1.645 is a reasonable estimate.
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Example: Pep Zone

Standard Normal Probability Distribution
If the manager of Pep Zone wants the probability
of a stock-out to be no
more than .05, what
should the reorder
point be?
Area = .05
Area = .95
0
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1.645
z
34
Example: Pep Zone

Standard Normal Probability Distribution
The corresponding value of x is given by
z 
x

x  15
1.645 
6
x  24.87

A reorder point of 24.87 gallons will place the
probability of a stock-out during lead-time at .05.
Perhaps Pep Zone should set the reorder point at
25 gallons to keep the probability under .05.
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Now You Try

Annual salaries for sales associates from a particular store
have a mean of $32,500 and a standard deviation of $2,500.
How much does a sales associate make if no more than 5% of
the sales associates make more than he or she?
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Now You Try

a)
b)
A certain type of light bulb has an average life of 500
hours with a standard deviation of 100 hours. The
length of life of the bulb can be closely
approximated by a normal curve. An amusement
park buys and installs 10,000 such bulbs. Find the
total number that can be expected to last …
between 650 and 780 hours.
more than 300 hours.
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End of Chapter 9, Part b
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