Download Chapter6a--Area as a measure of probability

Continuous Probability
Distributions
•
•
•
•
•
f(x)
Uniform Probability Distribution
Area as a measure of Probability
The Normal Curve
The Standard Normal Distribution
Computing Probabilities for a Standard Normal
Distribution
X
Uniform Probability Distribution
NY
Chicago
•Consider the random variable x representing the flight time of
an airplane traveling from Chicago to NY.
•Under normal conditions, flight time is between 120 and 140
minutes.
•Because flight time can be any value between 120 and 140
minutes, x is a continuous variable.
Uniform Probability Distribution
With every one-minute
interval being equally
likely, the random
variable x is said to have
a uniform probability
distribution
1

f (x )   20
0
for 120  x  140
elsewhere
0 elsewhere
Uniform Probability Distribution
 1

f ( x)   b  a
0
for a  x  b
elsewhere
For the flight-time
random variable,
a = 120 and b = 140
Uniform Probability Density
Function for Flight time
f (x )
The shaded area
indicates the probability
the flight will arrive in
the interval between
120 and 140 minutes
1
20
120
125
130
135
140
x
Basic Geometry
Remember when we
multiply a line
segment times a line
segment, we get an
area
Probability as an Area
Question: What is the probability that arrival time
will be between 120 and 130 minutes—that is:
P (120  x  130)
f (x )
1
20
P(120  x  130)  Area  1 / 20(10)  10 / 20  .50
10
120
125
130
135
140
x
When  is measured by area,  ( x)  0
Notice that in the continuous
case we do not talk of a
random variable assuming a
specific value. Rather, we talk
of the probability that a random
variable will assume a value
within a given interval.
E(x) and Var(x) for the Uniform
Continuous Distribution
E ( x) 
ab
2
Applying these formulas to
the example of flight times
of Chicago to NY, we have:
(b  a) 2
Var ( X ) 
12
E ( x) 
(120  140)
 130
2
(140  120) 2
Var ( X ) 
 33.3
12
Thus
  33.33  5.77 minutes
Normal Probability Distribution
The normal distribution is by
far the most important
distribution for continuous
random variables. It is widely
used for making statistical
inferences in both the natural
and social sciences.
Normal Probability Distribution

It has been used in a wide variety of applications:
Heights
of people
Scientific
measurements
Normal Probability Distribution

It has been used in a wide variety of applications:
Test
scores
Amounts
of rainfall
The Normal Distribution
1
( x   ) 2 / 2 2
f ( x) 
e
 2
Where:
μ is the mean
σ is the standard deviation
 = 3.1459
e = 2.71828
Normal Probability Distribution

Characteristics
The distribution is symmetric, and is bell-shaped.
x
Normal Probability Distribution

Characteristics
The entire family of normal probability
distributions is defined by its mean  and its
standard deviation  .
Standard Deviation 
Mean 
x
Normal Probability Distribution

Characteristics
The highest point on the normal curve is at the
mean, which is also the median and mode.
x
Normal Probability Distribution

Characteristics
The mean can be any numerical value: negative,
zero, or positive.
x
-10
0
20
Normal Probability Distribution

Characteristics
The standard deviation determines the width of the
curve: larger values result in wider, flatter curves.
 = 15
 = 25
x
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
The Standard Normal Distribution
The Standard Normal Distribution is a
normal distribution with the special
properties that is mean is zero and its
standard deviation is one.
 0
 1
Standard Normal Probability Distribution
The letter z is used to designate the standard
normal random variable.
1
z
0
Cumulative Probability
Probability that z ≤ 1 is the area under the curve
to the left of 1.
P ( z  1)
0
1
z
What is P(z ≤ 1)?
To find out, use the Cumulative Probabilities
Table for the Standard Normal Distribution
Z
.00
.01
.02
●
●
●
.9
.8159
.8186
.8212
1.0
.8413
.8438
.8461
1.1
.8643
.8665
.8686
1.2
.8849
.8869
.8888
●
●
P ( z  1)
Exercise 1
a) What is P(z ≤2.46)?
Answer:
b) What is P(z ≥2.46)?
a) .9931
b) 1-.9931=.0069
2.46
z
Exercise 2
a) What is P(z ≤-1.29)?
Answer:
b) What is P(z ≥-1.29)?
a) 1-.9015=.0985
b) .9015
Red-shaded area is
equal to greenshaded area
-1.29
Note that:
P( z  1.29)  1  P( z  1.29)
1.29
z
Note that, because of the symmetry, the area to the left of -1.29 is
the same as the area to the right of 1.29
Exercise 3
What is P(.00 ≤ z ≤1.00)?
P(.00 ≤ z ≤1.00)=.3413
0
1
P(.00  z  1)  P( z  1)  P( z  0)
 .8413  .5000  .3413
z
Exercise 4
What is P(-1.67 ≥ z ≥ 1.00)?
P(-1.67 ≤ z ≤1.00)=.7938
Thus P(-1.67 ≥ z ≥ 1.00)
=1 - .7938 = .2062
-1.67
0
1
P(1.67  z  1)  P( z  1)  [1  P( z  1.67)]
 .8413  (1  .9525)  .7938
z