Chapter 6
Continuous Probability Distributions

Normal Probability Distribution
f(x)

x
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.
Normal Probability Distribution



The normal probability distribution is the most
important distribution for describing a continuous
random variable.
It has been used in a wide variety of applications:
• Heights and weights of people
• Test scores
• Scientific measurements
• Amounts of rainfall
It is widely used in statistical inference
Normal Probability Distribution

Normal Probability Density Function
1
 ( x   ) 2 / 2 2
f ( x) 
e
2 
where:
 = mean
 = standard deviation
 = 3.14159
e = 2.71828
Normal Probability Distribution

Graph of the Normal Probability Density Function
f(x)

x
Normal Probability Distribution

Characteristics of the Normal Probability
Distribution
• The distribution is symmetric, and is often
illustrated as a bell-shaped curve.
• Two parameters,  (mean) and  (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
Normal Probability Distribution

Characteristics of the Normal Probability
Distribution
• The standard deviation determines the width of
the curve: larger values result in wider, flatter
curves.
 = 10
 = 50
Normal Probability Distribution

Characteristics of the Normal Probability Distribution
• 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.
The standard normal curve is symmetric about its
mean (μ=0) and has a total area of one under it
and above the z axis.
point of inflection
0
P( z  0)  .5
Z
Find the area under the standard normal
distribution for z values between z = 0 and
z = 2.53. [Or P(0<z<2.53)=?]
.4943
0
2.53
Z
Find P(z>-1.14).
-1.14
0
P(z>-1.14)=.3729+.5000
=.8729
.3729
-1.14
.5000
0
Find P(-1.28<z<1.83).
-1.28
0
1.83
Z
P(-1.28<z<1.83)=.3997+.4664
=.8661
.4664
.3997
-1.28
1.83
Find P(-2.54<z<-.42).
-2.54
-.42
P(-2.54<z<-.42) =.4945-.1628
=.3317
Dark blue shaded area is
.1628=P(0<z<.42).
Total shaded area is
.4945=P(0<z<2.54).
-2.54
-.42
Find c so that P(0<z<c) = .4772.
Find c so that the
shaded area is .4772.
0
z=c
Z
P(0<z<2.0)=.4772
0
z=2.0
Z
Normal Probability Distribution

Characteristics of the Normal Probability
Distribution
• 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.
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
z

x

We can think of z as a measure of the number of
standard deviations x is from .
What if we have a normal population with mean of 50 and a
standard deviation of 5.
That is
  50
 5
50
X
If we have a normal population with mean of 50 and a standard
deviation of 5. We can standardize the scores by finding their
corresponding z-scores.
That is
z ( x) 
x

x  50

.
5
50
0
X
Z
If we have a normal population with mean of 50 and a
standard deviation of 5. What is the probability that a
randomly selected element of the population will have a
value larger than 58?
  50
 5
50
We must first find a z-score for x = 58.
z
x

58  50

.
5
58
X
x
58  50
z

.

5
8

P( x  58)  P  z  
5

 P( z  1.60)
 .5000  .4452
P(0<z<1.6)=.4452
 .0548
0
1.6
P(z>1.6)=.0548
Z
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 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).

Example: Pep Zone

Standard Normal Probability Distribution
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.
Area = .2967
z = (x - )/
= (20 - 15)/6
= .83
Area = .5 - .2967
= .2033
Area = .5
0
.83
z
Example: Pep Zone

z
Using the Standard Normal Probability Table
.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
Example: Pep Zone

Standard Normal Probability Distribution
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
z.05
0
Let z.05 represent the z value cutting the .05 tail area.
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.
.
.
Example: Pep Zone

Standard Normal Probability Distribution
The corresponding value of x is given by
x =  + z.05
= 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.
Applications Where the Area Under a Normal Curve is
Provided
Let X be a normal random variable describing life of
a certain brand of light bulbs with mean of 500 hours
and standard deviation of 65 hours. Find the time
that at least 75% of the light bulbs will last longer
than.
First find a z so that 75% of the area under the normal
curve lies to the right of z.
.2500
z = -.67
.5000
Now convert this z score to hours of life for the light
bulbs.
z
x

x  500
 .67 
65
x  500  .67(65)
 456.45
75% of the light bulbs should last longer than
456.45 hours