Probablity normal

Chapter 6
Continuous Probability Distributions
Uniform Probability Distribution
Normal Probability Distribution
Exponential Probability Distribution
f(x)

x
µ

© 2003 South-Western/Thomson Learning™
Slide
1

Continuous Probability Distributions

A continuous random variable can assume any value
in an 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.

Slide
2


A random variable is uniformly distributed
whenever the probability is proportional to the
interval’s length.
Uniform Probability Density Function

f(x) = 1/(b - a) for a < x < b
=0 elsewhere

where: a = smallest value the variable can assume
b = largest value the variable can assume

Slide
3


Expected Value of x

E(x) = (a + b)/2

Variance of x

Var(x) = (b - a)2/12

where: a = smallest value the variable can assume
b = largest value the variable can assume

Slide
4

Example: Slater's Buffet

Slater customers are charged for the amount of
salad they take. Sampling suggests that the amount
of salad taken is uniformly distributed between 5
ounces and 15 ounces.
The probability density function is
f(x) = 1/10 for 5 < x < 15
=0 elsewhere
where:
x = salad plate filling weight

Slide
5


What is the probability that a customer will
take between 12 and 15 ounces of salad?
f(x)

P(12 < x < 15) = 1/10(3) = .3

1/10

x
5 10 12 15
Salad Weight (oz.)
Slide
6


Expected Value of x
E(x) = (a + b)/2
= (5 + 15)/2
= 10
Variance of x
Var(x) = (b - a)2/12
= (15 – 5)2/12
= 8.33

Slide
7


Graph of the Normal Probability Density Function
f(x)

x
µ

Slide
8


Characteristics of the Normal Probability
Distribution
• The shape of the normal curve is often illustrated
as a bell-shaped curve.
• Two parameters, µ (mean) and σ (standard
deviation), determine the location and shape of
the distribution.
• The highest point on the normal curve is at the
mean, which is also the median and mode.
• The mean can be any numerical value: negative,
zero, or positive.
… continued

Slide
9


Characteristics of the Normal Probability
Distribution
• The normal curve is symmetric.
• The standard deviation determines the width of
the curve: larger values result in wider, flatter
curves.
• 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.

Slide
10


Percent (%) of Values in Some Commonly Used
Intervals
• 68.26% of values of a normal random variable are
within +/- 1 standard deviation of its mean.
within +/- 2 standard deviations of its mean.
within +/- 3 standard deviations of its mean.

Slide
11


Normal Probability Density Function

1 − ( x − µ )2 / 2σ2
f ( x) = e
2 πσ

where:
µ = mean
σ = standard deviation
π = 3.14159
e = 2.71828

Slide
12

Standard Normal Probability Distribution

A random variable that has a normal distribution
with a mean of zero and a standard deviation of one
is said to have a standard normal probability
distribution.
The letter z is commonly used to designate this
normal random variable.
Converting to the Standard Normal Distribution
x−µ
z=
σ
We can think of z as a measure of the number of
standard deviations x is from µ.

Slide
13

Using Excel to Compute
Standard Normal Probabilities
Excel has two functions for computing probabilities
and z values for a standard normal distribution:
• NORMSDIST is used to compute the cumulative
probability given a z value.
• NORMSINV is used to compute the z value given
a cumulative probability.
(The letter S in the above function names reminds us
that they relate to the standard normal probability
distribution.)

Slide
14

Formula Worksheet
A B
1 Probabilities: Standard Normal Distribution
2
3 P (z < 1.00) =NORMSDIST(1)
4 P (0.00 < z < 1.00) =NORMSDIST(1)-NORMSDIST(0)
5 P (0.00 < z < 1.25) =NORMSDIST(1.25)-NORMSDIST(0)
6 P (-1.00 < z < 1.00) =NORMSDIST(1)-NORMSDIST(-1)
7 P (z > 1.58) =1-NORMSDIST(1.58)
8 P (z < -0.50) =NORMSDIST(-0.5)
9

Slide
15

Value Worksheet
A B
1 Probabilities: Standard Normal Distribution
2
3 P (z < 1.00) 0.8413
4 P (0.00 < z < 1.00) 0.3413
5 P (0.00 < z < 1.25) 0.3944
6 P (-1.00 < z < 1.00) 0.6827
7 P (z > 1.58) 0.0571
8 P (z < -0.50) 0.3085
9

Slide
16

Formula Worksheet
A B
1 Finding z Values, Given Probabilities
2
3 z value with .10 in upper tail =NORMSINV(0.9)
4 z value with .025 in upper tail =NORMSINV(0.975)
5 z value with .025 in lower tail =NORMSINV(0.025)
6

Slide
17

Value Worksheet
A B
1 Finding z Values, Given Probabilities
2
3 z value with .10 in upper tail 1.28
4 z value with .025 in upper tail 1.96
5 z value with .025 in lower tail -1.96
6

Slide
18

Example: Pep Zone

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 stockouts while waiting for an order. It has
been determined that leadtime demand 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
stockout, P(x > 20).

Slide
19

Example: Pep Zone

The Standard Normal table shows an area of .2967 for
the region between the z = 0 and z = .83 lines below.
The shaded tail area is .5 - .2967 = .2033. The
probability of a stockout is .2033.

z = (x - µ)/σ Area = .2967
= (20 - 15)/6
= .83 Area = .5 - .2967
= .2033

Area = .5
z
0 .83
Slide
20

Example: Pep Zone

Using the Standard Normal Probability Table
z .00 .01 .02 .03 .04 .05 .06 .07 .08 .09
.0 .0000 .0040 .0080 .0120 .0160 .0199 .0239 .0279 .0319 .0359
.1 .0398 .0438 .0478 .0517 .0557 .0596 .0636 .0675 .0714 .0753
.2 .0793 .0832 .0871 .0910 .0948 .0987 .1026 .1064 .1103 .1141
.3 .1179 .1217 .1255 .1293 .1331 .1368 .1406 .1443 .1480 .1517
.4 .1554 .1591 .1628 .1664 .1700 .1736 .1772 .1808 .1844 .1879

.5 .1915 .1950 .1985 .2019 .2054 .2088 .2123 .2157 .2190 .2224
.6 .2257 .2291 .2324 .2357 .2389 .2422 .2454 .2486 .2518 .2549
.7 .2580 .2612 .2642 .2673 .2704 .2734 .2764 .2794 .2823 .2852
.8 .2881 .2910 .2939 .2967 .2995 .3023 .3051 .3078 .3106 .3133
.9 .3159 .3186 .3212 .3238 .3264 .3289 .3315 .3340 .3365 .3389

Slide
21

Example: Pep Zone

If the manager of Pep Zone wants the probability of a
stockout to be no more than .05, what should the
reorder point be?

Area = .05

Area = .5 Area = .45
0 z.05
Let z.05 represent the z value cutting the .05 tail area.

Slide
22

Example: Pep Zone

Using the Standard Normal Probability Table
We now look-up the .4500 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 .4332 .4345 .4357 .4370 .4382 .4394 .4406 .4418 .4429 .4441
1.6 .4452 .4463 .4474 .4484 .4495 .4505 .4515 .4525 .4535 .4545
1.7 .4554 .4564 .4573 .4582 .4591 .4599 .4608 .4616 .4625 .4633
1.8 .4641 .4649 .4656 .4664 .4671 .4678 .4686 .4693 .4699 .4706
1.9 .4713 .4719 .4726 .4732 .4738 .4744 .4750 .4756 .4761 .4767
. . . . . . . . . . .

z.05 = 1.645 is a reasonable estimate.
Slide
23

Example: Pep Zone

The corresponding value of x is given by
x = µ + z.05σ
= 15 + 1.645(6)
= 24.87
A reorder point of 24.87 gallons will place the
probability of a stockout during leadtime at .05.
Perhaps Pep Zone should set the reorder point at 25
gallons to keep the probability under .05.

Slide
24

Using Excel to Compute Normal Probabilities

Excel has two functions for computing cumulative
probabilities and x values for any normal
distribution:
• NORMDIST is used to compute the cumulative
probability given an x value.
• NORMINV is used to compute the x value given a
cumulative probability.

Slide
25


Formula Worksheet for Pep Zone Example
A B
1 Probabilities: Normal Distribution
2
3 P (x > 20) =1-NORMDIST(20,15,6,TRUE)
4
5 Finding x Values, Given Probabilities
6
7 x value with .05 in upper tail =NORMINV(0.95,15,6)
8

Slide
26


Value Worksheet for Pep Zone Example
A B
1 Probabilities: Normal Distribution
2
3 P (x > 20) 0.2023
4
5 Finding x Values, Given Probabilities
6
7 x value with .05 in upper tail 24.87
8

Note: P(x > 20) = .2023 here using Excel, while our
previous manual approach using the z table yielded
.2033 due to our rounding of the z value.

Slide
27


Exponential Probability Density Function
1 − x /µ
f ( x) = e for x > 0, µ > 0
µ

where: µ = mean
e = 2.71828

Slide
28


Cumulative Exponential Distribution Function

P ( x ≤ x0 ) = 1 − e − xo / µ

where:
x0 = some specific value of x

Slide
29

Using Excel to Compute Exponential Probabilities

Excel’s EXPONDIST function can be used to compute
exponential probabilities.
The function has three arguments:
• First – the value of the random variable x
• Second – 1/µ (the inverse of the mean number of
occurrences in an interval)
• Third – “TRUE” or “FALSE” (we will always enter
“TRUE” because we’re seeking a cumulative
probability)

Slide
30


Formula Worksheet

A B
1 Probabilities: Exponential Distribution
2
3 P (x < 18) =EXPONDIST(18,1/15,TRUE)
4 P (6 < x < 18) =EXPONDIST(18,1/15,TRUE)-EXPONDIST(6,1/15,TRUE)
5 P (x > 8) =1-EXPONDIST(8,1/15,TRUE)
6

Slide
31


Value Worksheet

A B
2
3 P (x < 18) 0.6988
4 P (6 < x < 18) 0.3691
5 P (x > 8) 0.5866
6

Slide
32

Example: Al’s Carwash

The time between arrivals of cars at Al’s
Carwash follows an exponential probability
distribution with a mean time between arrivals of 3
minutes. Al would like to know the probability that
the time between two successive arrivals will be 2
minutes or less.

P(x < 2) = 1 - 2.71828-2/3 = 1 - .5134 = .4866

Slide
33

Example: Al’s Carwash

Graph of the Probability Density Function
f(x)

.4
.3 P(x < 2) = area = .4866

.2
.1
x
1 2 3 4 5 6 7 8 9 10
Time Between Successive Arrivals (mins.)

Slide
34


Formula Worksheet for Al’s Carwash Example

A B
2
3 P (x < 2) =EXPONDIST(2,1/3,TRUE)
4

Slide
35


Value Worksheet for Al’s Carwash Example

A B
2
3 P (x < 2) 0.4866
4

Slide
36

Relationship between the Poisson
and Exponential Distributions

(If) the Poisson distribution
provides an appropriate description
of the number of occurrences
per interval

(If) the exponential distribution
provides an appropriate description
of the length of the interval
between occurrences

Slide
37

Probablity normal

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