Chapter 6
Continuous Probability Distributions



The Uniform Probability Distribution
The Normal Probability Distribution
Normal Approximation of Binomial Probabilities
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.( It is
zero)
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 .
2
The Uniform Probability Distribution
A random variable is uniformly distributed
whenever the probability is proportional to the
length of the interval.
f(x )
1/(b-a)
x
a
b
3
The Uniform Probability Distribution



Uniform Probability Density Function
f (x) = 1/(b - a) for a < x < b
= 0 elsewhere
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
4
Example: Slater's Buffet
Slater customers are charged for the amount
(weight) of salad they take. Sampling suggests
that the amount of salad taken is uniformly
distributed between 5 ounces and 15 ounces.
 Probability Density Function
f (x ) =
=
1
1

(15  5) 10
0
for 5 < x < 15
elsewhere
where
x = salad plate filling weight
5
Example: Slater's Buffet
What is the probability that a customer will take
between 12 and 15 ounces of salad?
f (x )
P(12 < x < 15) = (15-12)/10 = .3
1/10
x
5
10 12
15
Salad Weight (oz.)
6
Uniform Probability Distribution

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
7
The Normal Probability Distribution

Graph of the Normal Probability Density Function
f (x )

x
8
Normal Probability Distribution

It has been used in a wide variety of applications:
Heights
of people
Scientific
measurements
9
Normal Probability Distribution

It has been used in a wide variety of applications:
Test
scores
Amounts
of rainfall
10
Normal Probability Distribution

Characteristics
The distribution is symmetric; its skewness
measure is zero.
x
11
Normal Probability Distribution

Characteristics
The entire family of normal probability
distributions is defined by its mean  and its
standard deviation s .
Standard Deviation s
Mean 
x
12
Normal Probability Distribution

Characteristics
The highest point on the normal curve is at the
mean, which is also the median and mode.
x
13
Normal Probability Distribution

Characteristics
The mean can be any numerical value: negative,
zero, or positive.
x
-10
0
20
14
Normal Probability Distribution

Characteristics
The standard deviation determines the width of the
curve: larger values result in wider, flatter curves.
s = 15
s = 25
x
15
Normal distributions with the same
mean but different standard deviations:
f(x)
n1
n3
n2

x
16
Normal Probability Distribution

Characteristics
Probabilities for the normal random variable are
given by areas under the curve. The total area
under the curve is 1 (.5 to the left of the mean and
.5 to the right).
.5
.5
x
17
Normal Probability Distribution

Characteristics
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.
18
Normal Probability Distribution

Characteristics
99.72%
95.44%
68.26%
 + 3s
 – 3s
 – 1s   + 1 s
 – 2s
 + 2s
x
19
The Normal Probability Distribution

The Normal Curve
•
•
•
The shape of the normal curve is often
illustrated as a bell-shapedcurve.
The highest point on the normal curve is at
the mean, which is also the median and
mode of the distribution. (The mean can be
any numeric value)
The normal curve is symmetric .
20
The Normal Probability Distribution

The Normal Curve
•
•
•
The standard deviation determines the width or
the spread of the curve: larger values result in
wider, flatter curves.
The total area under the curve is 1.
Probabilities for the normal random variable are
given by areas under the curve. The total area
under the curve to the left of the mean is .50 and
the total area under the curve to the right of the
mean is also .50.
21
The Normal Probability Distribution

Normal Probability Density Function
1
 ( x   ) 2 / 2s 2
f (x) 
e
2 s
where
 = mean
s = standard deviation
 = 3.14159
e = 2.71828
22
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
standard normal random variable.
Converting to the Standard Normal Distribution
z

x
s
Given the random variable x has a normal distribution
with mean  and standard deviation s.
We can think of z as a measure of the number of
standard deviations x is from .
23
The Standard Normal Probability Distribution - the
probability distribution associated with any normal
random variable that has  = 0 and s = 1.
There are tables that give the results of the
integration
b
P(a  x  b)  f(x)dx
a
b

a
1
e
2 s
  x   2
2s 2
dx
for the standard normal random variable (Table 1,
Appendix B).
24
The Standard Normal Distribution (Appendix B, Table 1)
z
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.7
0.8
0.9
1.0
1.1
1.2
1.3
1.4
1.5
1.6
1.7
1.8
1.9
2.0
2.1
2.2
2.3
2.4
2.5
2.6
2.7
2.8
2.9
3.0
0.00
0.5000
0.5398
0.5793
0.6179
0.6554
0.6915
0.7257
0.7580
0.7881
0.8159
0.8413
0.8643
0.8849
0.9032
0.9192
0.9332
0.9452
0.9554
0.9641
0.9713
0.9772
0.9821
0.9861
0.9893
0.9918
0.9938
0.9953
0.9965
0.9974
0.9981
0.9987
0.01
0.5040
0.5438
0.5832
0.6217
0.6591
0.6950
0.7291
0.7611
0.7910
0.8186
0.8438
0.8665
0.8869
0.9049
0.9207
0.9345
0.9463
0.9564
0.9649
0.9719
0.9778
0.9826
0.9864
0.9896
0.9920
0.9940
0.9955
0.9966
0.9975
0.9982
0.9987
0.02
0.5080
0.5478
0.5871
0.6255
0.6628
0.6985
0.7324
0.7642
0.7939
0.8212
0.8461
0.8686
0.8888
0.9066
0.9222
0.9357
0.9474
0.9573
0.9656
0.9726
0.9783
0.9830
0.9868
0.9898
0.9922
0.9941
0.9956
0.9967
0.9976
0.9982
0.9987
0.03
0.5120
0.5517
0.5910
0.6293
0.6664
0.7019
0.7357
0.7673
0.7967
0.8238
0.8485
0.8708
0.8907
0.9082
0.9236
0.9370
0.9484
0.9582
0.9664
0.9732
0.9788
0.9834
0.9871
0.9901
0.9925
0.9943
0.9957
0.9968
0.9977
0.9983
0.9988
0.04
0.5160
0.5557
0.5948
0.6331
0.6700
0.7054
0.7389
0.7704
0.7995
0.8264
0.8508
0.8729
0.8925
0.9099
0.9251
0.9382
0.9495
0.9591
0.9671
0.9738
0.9793
0.9838
0.9875
0.9904
0.9927
0.9945
0.9959
0.9969
0.9977
0.9984
0.9988
0.05
0.5199
0.5596
0.5987
0.6368
0.6736
0.7088
0.7422
0.7734
0.8023
0.8289
0.8531
0.8749
0.8944
0.9115
0.9265
0.9394
0.9505
0.9599
0.9678
0.9744
0.9798
0.9842
0.9878
0.9906
0.9929
0.9946
0.9960
0.9970
0.9978
0.9984
0.9989
0.06
0.5239
0.5636
0.6026
0.6406
0.6772
0.7123
0.7454
0.7764
0.8051
0.8315
0.8554
0.8770
0.8962
0.9131
0.9279
0.9406
0.9515
0.9608
0.9686
0.9750
0.9803
0.9846
0.9881
0.9909
0.9931
0.9948
0.9961
0.9971
0.9979
0.9985
0.9989
0.07
0.5279
0.5675
0.6064
0.6443
0.6808
0.7157
0.7486
0.7794
0.8078
0.8340
0.8577
0.8790
0.8980
0.9147
0.9292
0.9418
0.9525
0.9616
0.9693
0.9756
0.9808
0.9850
0.9884
0.9911
0.9932
0.9949
0.9962
0.9972
0.9979
0.9985
0.9989
0.08
0.5319
0.5714
0.6103
0.6480
0.6844
0.7190
0.7517
0.7823
0.8106
0.8365
0.8599
0.8810
0.8997
0.9162
0.9306
0.9429
0.9535
0.9625
0.9699
0.9761
0.9812
0.9854
0.9887
0.9913
0.9934
0.9951
0.9963
0.9973
0.9980
0.9986
0.9990
0.09
0.5359
0.5753
0.6141
0.6517
0.6879
0.7224
0.7549
0.7852
0.8133
0.8389
0.8621
0.8830
0.9015
0.9177
0.9319
0.9441
0.9545
0.9633
0.9706
0.9767
0.9817
0.9857
0.9890
0.9916
0.9936
0.9952
0.9964
0.9974
0.9981
0.9986
0.9990
25
A small part of the Standard Normal
Probability Distribution Table
Example: for a standard normal random variable z,
what is the probability that z is less than 0.43?
z
0.0
0.1
0.2
0.3
0.4
0.5
0.6
0.00
0.5000
0.5398
0.5793
0.6179
0.6554
0.6915
0.7257
0.01
0.5040
0.5438
0.5832
0.6217
0.6591
0.6950
0.7291
0.02
0.5080
0.5478
0.5871
0.6255
0.6628
0.6985
0.7324
0.03
0.5120
0.5517
0.5910
0.6293
0.6664
0.7019
0.7357
0.04
0.5160
0.5557
0.5948
0.6331
0.6700
0.7054
0.7389
26
Example: for a standard normal random variable z,
what is the probability that z is less than 2.0?
And what is the probability that z is greater than
2.0?
f(z)
0
2.0
z
27
Again, looking at a small part of the Standard Normal
Probability Distribution Table, we find the probability
that a standard normal random variable z is less than
2.00?
z
:
:
1.6
1.7
1.8
1.9
2.0
2.1
2.2
0.00
:
:
0.9452
0.9554
0.9641
0.9713
0.9772
0.9821
0.9861
0.01
:
:
0.9463
0.9564
0.9649
0.9719
0.9778
0.9826
0.9864
0.02
:
:
0.9474
0.9573
0.9656
0.9726
0.9783
0.9830
0.9868
0.03
:
:
0.9484
0.9582
0.9664
0.9732
0.9788
0.9834
0.9871
0.04
:
:
0.9495
0.9591
0.9671
0.9738
0.9793
0.9838
0.9875
28
What is the probability that z is at least 2.0?
Area (Probability)=
1-0.9772 = 0.0228
{
f(z)
0
2.0
z
29
Example: for a standard normal random variable z,
what is the probability that z is between 0 and 2.0?
}
Area (Probability) = 0.4772
f(z)
0
2.0
z
30
What is the probability that z is between -1.5 and 2.0?
Area (P<2) = 0.9772
Area (P<-1.5) = 0.0668
-1.5
2.0
P(-1.5<X<2.0)= P(X<2.0)-P(X<-1.5)
=0.9772-0.0668=0.9107
31
What is the probability that z is between -1.5 and -2.0?
f(z)
-2.0 -1.5
0
z
P(-2.0<X<-1.5)= P(X<-1.5)-P(X<-2)
=0.0668-0.0228=0.0440
32
What is the probability that z is exactly 1.5?
Probability of any continuous random variable
takes exactly some number is always ZERO.
f(z)
0
1.5
z
33
z-Transform - mathematical means by which any
•
•
normal random variable with a mean  and
standard deviation s can be converted into a
standard normal random variable .
to make the mean equal to 0, we simply subtract 
from each observation in the population
to then make the standard deviation equal to 1, we
divide the results in the first step by s
The resulting transform is given by
x 
z
s
34
Example: for a normal random variable x with a
mean of 5 and a standard deviation of 3, what is
the probability that x is between 5.0 and 7.0?
f(x)
}
Area
(Probability)
 =5
0
7.0
x
z
35
Using the z-transform, we can restate the problem
in the following manner:
 5.0-5.0
P(5.0  x  7.0) P 

3.0

x-  7.0-5.0 
s
3.0 
= P(0.0  z  0.67)
then use the standard normal probability table to
find the ultimate answer:
P(0.0  z  0.67) = 0.2486
36
which graphically looks like this:
f(x)
}
Area (Probability)
= 0.2486
7.0
x
0.0 0.67
z
 =5
37
Computing Normal Probabilities
• Draw the normal curve and determine the area
(probability) to calculate.
• Convert to standard normal using z-transform
x
z
s
• Look up the Standard Normal Probability table
• Note facts
a. total area under curve is 1 (total probability)
b. curve is symmetric, the area to the left (right)
of mean = 0.50 always.
38
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).
39
Example: Pep Zone
Step1: Draw Curve
  15 , s  6
Area = 0.7967
Step 2: z-transform
Area = 0.2033
x = 20
z = (x -  )/s
= (20-15)/6
= 0.83
0
15
.83
20
Step 3: Standard Normal table
P( z < 0.83) = 0.7967
The desired area is 1 – 0.7967 = 0.2033
Answer: The probability of a stockout is 0.2033.
z
x
40
Example: Pep Zone

z
Using the Standard Normal Probability Table
.00
.01
.02
.03
.0
.5120
.1
.5517
.2
.5910
.3
.6293
.4
.6664
.5
.7019
.6
.7357
.7
.7673
.04
.05
.06
.07
.08
.09
.8
.7881 .7910 .7939 .7967 .7995 .8023 .8051 .8078 .8106 .8133
.9
.3159 .3186 .3212 .3238 .3264 .3289 .3315 .3340 .3365 .3389
41
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
z.05
0
Let z.05 represent the z value cutting the tail area of .05.
It is called the cut-off point.
42
Example: Pep Zone

z
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.05 = 1.645 is a reasonable estimate.
.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
43
Example: Pep Zone
The corresponding value of x is given by
x =  + z.05s
= 15 + 1.645*(6)
= 24.87
24.87 gallons will place the probability of
A reorder point of
a stockout during leadtime at .05. Perhaps Pep Zone should set
the reorder point at 25 gallons to keep the probability
under .05.
Or you could do in this way:
z.05 = (x -  )/s
1.645= (x - 15 )/6
x=24.87
44
Example: Cost of Attending
Major League Baseball Games

Team Marketing Report, a sports-business
newsletter, reported the mean cost for a family of
four to attend a major league baseball game is
$95.80 (The Wall Street Journal, April 5, 1994).
Assume that a normal probability distribution
applies and that the standard deviation is $10.00.
a. What is the probability a family of four will
spend between $85 and $100 at a major league
baseball game?
45
Example: Baseball
Step1: Draw Curve
  95.8, s  10
Area=0.6628
Step 2: z-transform
Area=0.1401
z1 = (x1 -  )/s
= (85-95.8)/10
= -1.08
z2 = (x2 -  )/s
=(100-95.8)/10
85
95.8 100
= 0.42
-1.08
0
0.42
x
z
Step 3: Standard Normal table
The desired area is 0.6628-0.1401=0.5227
46

a. What is the probability a family of four will spend
between $85 and $100 at a major league baseball
game?
P(85  x  100)= P(
85-95.8 x   100-95.8


)
10
s
10
=P(-1.08  z  0.42)
=P(z<0.42)-P(z<-1.08)
=0.6628-0.1401=0.5227
47
b. 95% of families of four spend at least 79.35 when
attending a major league baseball game.
Body of table area =0.95 occurs at z= -1.645
x is 1.645 std. Dev. below the mean
x =  + zs
=95.8+(-1.645)(10)
= 79.35
Or z = (x -  )/s
-1.645= (x-95.8)/10
Area =
.05
_____
Area=.95
?
=79.3
5
?
=-1.645
x
z
0
48
c. The top 2% of families of four spend more than $116.3
when attending a major league baseball game.
Body of table area =0.98 occurs at z=
X is 2.05 std. dev. above the mean
x =  + zs
2.05
=95.8+(2.05)(10)
= 116.30
Area = 2%
Area=.5
Area =.48 ____
95.8
?
=$116.3
0
?
=2.05
x
z
49
Why is the normal probability distribution
considered so important?
•
many random variables are naturally normally
distributed
•
many distributions, such as the Poisson and the
binomial, can be approximated by the normal
distribution
•
the distribution of many statistics, such as the
sample mean and the sample proportion, are
approximately normally distributed if the sample
is sufficiently large (Central Limit Theorem)
50