Download Chapter 6 Continuous Random Variables

Chapter 6
Continuous Random Variables
Nutan S. Mishra
Department of Mathematics and Statistics
University of South Alabama
Probability distribution
A survey finds the following probability distribution for the age
of a rented car.
Age (Years)
Probability
0 to 1
0.2
1 to 2
0.28
2to 3
0.2
3 to 4
0.15
4 to 5
0.1
5 to 6
0.05
6 to 7
0.02
Histogram
Probability curve
Probability distribution
Suppose now that we want to calculate the probability that a
rented car is between 0 and 4 years old. Referring to the
table,
P(0 ≤ X ≤ 4) = 0.20 + 0.28 + 0.20 + 0.15 = 0.83.
Referring to the following figure, notice that we can obtain the
same result by adding the areas of the corresponding bars,
since each bar has a width of 1 unit.
P(0 ≤ X ≤ 4) = 0.20 + 0.28 + 0.20 + 0.15 = 0.83.
Now what happens if we want to find P(2 ≤ X ≤ 3.5)?
Probability is the area under the curve
Probability distribution curve
A probability distribution of a continuous
variable can be represented by its
probability curve. Total area underneath
the curve is assumed to be 1.00
f(x)
x
Probability distribution curve
Probability that x takes values between two
numbers a and b is given by the area
underneath the curve between the two
points
P(a<x<b) = shaded area
a
b
Probability distribution curve
For a continuous random variable
P(X = a) = 0
Where a is any specific value of x.
P(x=a) = area under the
curve =0
Because area is zero
a
Normal Distribution
f(x)
x
If our dataset has bell shape curve then the data is said to
have normal distribution
Mean(X) = μ and Var[X] = s2
μ and s2 are called parameters of the normal distribution
Normal Distribution
• Many phenomena in nature, industry and
research follow this bell-shaped
distribution.
– Physical measurements
– Rainfall studies
– Measurement error
• There are an infinite number of normal
distributions, each with a specified μ and
s.
Normal Distribution
• Characteristics
– Bell-shaped curve
– - < x < +
– μ determines distribution location and is the
highest point on curve
– Curve is symmetric about μ
– s determines distribution spread
– Curve has its points of inflection at μ + s
– μ + 1s covers 68% of the distribution
– μ + 2s covers 95% of the distribution
– μ + 3s covers 99.7% of the distribution
Normal Distribution
s
s
s
s
μ
-4
-3
-2
-1
0
1
2
3
4
Normal Distribution
-4
-3
-2
-1
0
1
2
μ + 1s covers 68%
3
4
-4
-3
-2
-1
0
1
2
μ + 2s covers 95%
3
4
-4
-3
-2
-1
0
1
2
3
μ + 3s covers 99.7%
4
Normal Distribution
n(x; μ = 0, s = 1)
n(x; μ = 5, s = 1)
f(x)
-4
-3
-2
-1
0
1
2
x
3
4
5
6
7
8
Normal Distribution
n(x; μ = 0, s = 0.5)
f(x)
n(x; μ = 0, s = 1)
-4
-3
-2
-1
0
x
1
2
3
4
Normal Distribution
n(x; μ = 5, s = .5)
n(x; μ = 0, s = 1)
f(x)
-4
-3
-2
-1
0
1
2
x
3
4
5
6
7
8
Standard Normal Distribution
The distribution of a normal random
variable with mean 0 and variance 1 is
called a standard normal distribution.
-4
-3
-2
-1
0
1
2
3
4
Standard Normal Distribution
• The letter Z is traditionally used to
represent a standard normal random
variable.
• z is used to represent a specific value of Z.
• The standard normal distribution has been
tabulated in the textbook.
Standard Normal Distribution
Given a standard normal distribution, find
the area under the curve
(a) to the left of z = -1.85
(b) to the left of z = 2.01
(c) to the right of z = –0.99
(d) to right of z = 1.50
(e) between z = -1.66 and z = 0.58
-4
-3
-2
-1
0
1
2
3
4
-4
-3
to the left of z = -1.85
-4
-3
-2
-1
0
1
-2
-1
0
1
2
3
4
to the left of z = 2.01
2
to the right of z = –0.99
3
4
-4
-3
-2
-1
0
1
2
to the right of z = –0.99
3
4
Standard Normal Distribution
Given a standard normal distribution, find
the value of k such that
(a) P(Z < k) = .1271
(b) P(Z < k) = .9495
(c) P(Z > k) = .8186
(d) P(Z > k) = .0073
(e) P( 0.90 < Z < k) = .1806
(f) P( k < Z < 1.02) = .1464
-4
-3
-2
-1
0
1
2
3
4
-4
-3
-2
-1
0
1
2
3
4
-4
-3
-2
-1
0
1
2
3
4
-4
-3
-2
-1
0
1
2
3
4
Normal Distribution
• Any normal random variable, X, can be
converted to a standard normal random variable:
z = (x – μx)/sx
Useful link: (pictures of normal curves borrowed from:
http://www.stat.sc.edu/~lynch/509Spring03/25
Normal Distribution
Given a random Variable X having a normal
distribution with μx = 10 and sx = 2, find the
probability that X < 8.
z
-4
x
-3
-2
-1
0
1
2
3
4
6
8
10
12
14
16
4
Relationship between the Normal
and Binomial Distributions
• The normal distribution is often a good
approximation to a discrete distribution when the
discrete distribution takes on a symmetric bell
shape.
• Some distributions converge to the normal as
their parameters approach certain limits.
• Theorem 6.2: If X is a binomial random variable
with mean μ = np and variance s2 = npq, then
the limiting form of the distribution of Z = (X –
np)/(npq).5 as n  , is the standard normal
distribution, n(z;0,1).
Useful links
• http://psych.colorado.edu/~mcclella/java/n
ormal/tableNormal.html
• http://www.umd.umich.edu/casl/socsci/eco
n/StudyAids/JavaStat/StandardizeNormal
Variable.html
Exercises page 272…
6.1
In case of discrete distribution we define P(X=a) , we can
compute this probability.
In case of continuous variable we can not assign P(X=a)
because P(X=a) =0. Instead we talk about P(a <X <b).
6.2
P(X=a) =0
6.3
P(a<X<b) = P(a ≤ X ≤ b) because equality sign does
not matter since P(X=a) =P(X=b) =0.
Exercise page 272…
6.4
1. It has bell shaped curve.
2. To find the different probabilities we must know
the values of µ and σ .They are called
parameters of the normal distribution.
3. The curve is symmetric about the mean µ
4. Smaller is the value of σ, steeper is the curve.
Larger value of σ causes a flat curve.
5. Area under the curve measures the probability.
6. 99.73% of the X-values fall between µ±3σ
limits.
Exercise page 272…
6.5
1. Standard normal distribution curve is a bell
shaped curve
2. The curve represents a normal distribution with
µ=0 and σ=1
3. The corresponding variable is denoted by Z
4. It is symmetric about 0.
5. Any other normal distribution can be
transformed to standard normal.
6. The area under the standard normal curve are
tabulated.
Exercise page 272…
6.7
With a constant mean, as the standard deviation
decreases the height of the normal curve
decreases and the width increases.
6.8
With a constant standard deviation, change in the
value of mean does not cause any change in the
shape of the normal curve.
Standard deviation is sometimes called shape parameter and mean is called
location parameter.
Exercise page 272…