Binomial Applet
http://stattrek.com/Tables/Binomial.aspx
Chapter 6
Continuous Probability Distributions
Discrete vs. Continuous Random
Variables
1. For discrete random variables we find the
probability the variable will take on a specific
value. For continuous random variable we
find the probability the variable will fall in
some range.
2. For discrete random variables we find the
probability associated with some value using
a probability function. For a continuous
random variable we find the area under a
curve between two points.
Probability Density Curves
1. The area under the curve is equal to 1
2. The area under the curve between two values of
the variable measures the probability the value will
fall between those values
3. A probability density curve is defined by a
probability density function
Uniform Probability Distribution
The probability of the variable falling into some
interval is the same for all equally-sized intervals.
Uniform Probability Density
Function
d c
f (c  x  d ) 
ba
for a  x  b, given a  c  b and a  d  b
for x  a and x  b f ( x)  0
Mean and Variance of a Uniform
Distribution
ab
E ( x) 
2
2

b  a
Var ( x) 
12
Uniform Distribution, Example
Assume that a delivery truck takes between 200
minutes and 250 minutes to complete its delivery
route. Assume that the amount of time taken is
uniformly distributed.
What is the probability the truck will take between 210
and 230 minutes?
Uniform Distribution, Example
What is the probability the truck will take between 210
and 230 minutes?
(230 – 210)/(250 – 200) = 20/50 = 0.4
Uniform Distribution, Example
What is the probability the truck will take more than
240 minutes?
Uniform Distribution, Example
What is the probability the truck will take more than
240 minutes?
(250 – 240)/(250 – 200) = 10/50 = 0.2
Uniform Distribution, Example
What is the probability the truck will take less than 220
minutes?
Uniform Distribution, Example
What is the probability the truck will take less than 220
minutes?
(220 – 200)/(250 – 200) = 20/50 = 0.4
Uniform Distribution, Example
What is the mean and variance of this distribution?
E(x) = (250 + 200)/2 = 225
Var(x) = (250 – 200)2/12 = 2500/12 = 208.33
Importance of the Normal
Distribution
•Many variables have a distribution similar to that of
the normal distribution
•The means of samples are normally distributed (given
samples of 30 or more)
Characteristics of the Normal
Distribution
• Symmetrical and bell-shaped
• The tails of the distribution asymptotically converge
on the x axis
• The location of the distribution on the x axis is
determined by the mean of the distribution
• The shape of the distribution is determined by the
variance
• The total area under the curve is equal to 1, the areas
to either side of the mean equal 0.5
• Moving out a given number of standard deviations
will always capture the same share of the distribution
Normal Distributions
Wikipedia
Normal Probability Distribution
99.72%
95.44%
68.26%
m – 3s
m – 2s
m – 1s
m
m + 1s
m + 2s
m + 3s
x
Standard Normal Distribution
m=0
s1
To convert a normal
distribution to
standard normal:
z = (x - m)/s
Characteristics of the Normal
Distribution
http://www-stat.stanford.edu/~naras/jsm/NormalDensity/NormalDensity.html
Normal Probability Density
Function
1
f ( x) 
e
s 2
  x  m 2
2s 2
Standard Normal
Table
Standard Normal Table
Standard Normal Table
Normal Distribution, Exercises
Find: P(z < 2)
P(z < 2) = .9772
Normal Distribution, Exercises
Find: P(z < -2)
P(z < -2) = .0228
Normal Distribution, Exercises
Find: P(0 < z < 2)
P(0 < z < 2) = .9772 - .5 = .4772
Or
P(0 < z < 2) = .5 - .0228 = .4772
Normal Distribution, Exercises
Find: P(-1 < z < 0)
P(-1 < z < 0) = P(0 < z < 1) = .8413 - .5 = .3413
Or
P(-1 < z < 0) = .5 - .1587 = .3413
Normal Distribution, Exercises
Find: P(z > -1)
P(z > -1) = P(z < 1) = .8413
Or
P(z > -1) = 1 - .1587 = .8413
Normal Distribution, Exercises
Find: P(z > 2)
P(z > 2) = 1 - .9772 = .0228
Or
P(z > 2) = P(z < -2) = .0228
Normal Distribution, Exercises
Find: P(1 < z < 1.5)
P(1 < z < 1.5) = .9332 - .8413 = 0.0919
Or
P(1 < z < 1.5) = P(-1.5 < z < -1) = .1587 - .0668 = .0919
Normal Distribution, Exercises
Find: P(-1.5 < z < -0.5)
P(-1.5 < z < -0.5) = P(0.5 > z > 1.5) = .9332 - .6915 = .2417
Or
P(-1.5 < z < -0.5) = .3085 -.0668 = .2417
Normal Distribution, Exercises
Find: P(-1 < z < 2)
P(-1 < z < 2) = .9772 - .1587 = 0.8185
Standard Normal Probability
Distribution, Example
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
Pep
replenishment order is placed.
Zone
5w-20
Motor Oil
Standard Normal Probability
Distribution, Example
The store manager is concerned that sales are
being lost due to stockouts while waiting for an
order.
It has been determined that demand
during replenishment lead-time is
Pep
Zone
normally distributed with a mean of
5w-20
Motor Oil
15 gallons and a standard deviation of
6 gallons.
The manager would like to know the
probability of a stockout, P(x > 20).
Standard Normal Probability
Distribution, Example
1. Solve for the z score:
Z = (x – m)/s
= (20 – 15)/6
= .83
2. Look up value in standard normal table
Standard Normal Probability
Distribution, Example
Pep
Zone
5w-20
Motor Oil
z
.00
.01
.02
.03
.04
.05
.06
.07
.08
.09
.
.
.
.
.
.
.
.
.
.
.
.5
.6915 .6950 .6985 .7019 .7054 .7088 .7123 .7157 .7190 .7224
.6
.7257 .7291 .7324 .7357 .7389 .7422 .7454 .7486 .7517 .7549
.7
.7580 .7611 .7642 .7673 .7704 .7734 .7764 .7794 .7823 .7852
.8
.7881 .7910 .7939 .7967 .7995 .8023 .8051 .8078 .8106 .8133
.9
.8159 .8186 .8212 .8238 .8264 .8289 .8315 .8340 .8365 .8389
.
.
.
.
.
.
.
P(z < .83)
.
.
.
.
Standard Normal Probability
Distribution, Example
3. Compute the area under the curve to the
right of 0.83
P(z > .83) = 1 – P(z < .83)
= 1- .7967
= .2033
Standard Normal Probability
Distribution, Example
If the manager of Pep Zone wants the probability
of a stockout to be no more than .05, what should
the reorder point be?
Pep
Zone
5w-20
Motor Oil
Standard Normal Probability
Distribution, Example

Solving for the Reorder Point
Area = .9500
Area = .0500
z
0
z.05
Pep
Zone
5w-20
Motor Oil
Standard Normal Probability
Distribution, Example
Pep
Zone
5w-20
Motor Oil
Step 1: Find the z-value that cuts off an area of .05
in the right tail of the standard normal distribution.
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
look up.9761
the .9767
1.9 .9713 .9719 .9726 .9732 .9738 .9744 .9750We.9756
complement of the
.
.
.
.
.
.
.
. tail area
. (1 - .05
. = .95)
.
Standard Normal Probability
Distribution, Example
Step 2: Convert z.05 to the corresponding
value of x.
x = m+z.05s
= 15 + 1.645(6)
= 24.87 or 25
A reorder point of 25 gallons will place the
probability of a stockout during lead-time at
(slightly less than) .05.
Pep
Zone
5w-20
Motor Oil
Standard Normal Probability
Distribution, Example
Pep
Zone
5w-20
Motor Oil
By raising the reorder point from 20 gallons to
25 gallons on hand, the probability of a stockout
decreases from about .20 to .05.
This is a significant decrease in the chance that Pep
Zone will be out of stock and unable to meet a
customer’s desire to make a purchase.
Practice Homework
P. 241, #12, 20
P. 246, #30