Chapter 6 Continuous Random Variables
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Transcript of Chapter 6 Continuous Random Variables
Chapter 6Continuous Random Variables
Nutan S. Mishra
Department of Mathematics and Statistics
University of South Alabama
Probability distributionA 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.
Probability distribution curveA 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) = 0Where a is any specific value of x.
a
P(x=a) = area under the curve =0
Because area is zero
Normal Distribution
If our dataset has bell shape curve then the data is said to have normal distribution
Mean(X) = μ and Var[X] = 2
μ and 2 are called parameters of the normal distribution
f(x)
x
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 .
Normal Distribution• Characteristics
– Bell-shaped curve– - < x < +– μ determines distribution location and is the
highest point on curve– Curve is symmetric about μ – determines distribution spread– Curve has its points of inflection at μ + – μ + 1 covers 68% of the distribution– μ + 2 covers 95% of the distribution– μ + 3 covers 99.7% of the distribution
Normal Distribution
-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
μ + 1 covers 68% μ + 2 covers 95% μ + 3 covers 99.7%
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 -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
to the left of z = -1.85 to the left of z = 2.01
to the right of z = –0.99 to the right of z = –0.99
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
Normal Distribution
• Any normal random variable, X, can be converted to a standard normal random variable:
z = (x – μx)/x
Useful link: (pictures of normal curves borrowed from:
http://www.stat.sc.edu/~lynch/509Spring03/25
Normal DistributionGiven a random Variable X having a normal
distribution with μx = 10 and x = 2, find the probability that X < 8.
-4 -3 -2 -1 0 1 2 3 4
4 6 8 10 12 14 16
z
x
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 2 = 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/normal/tableNormal.html
• http://www.umd.umich.edu/casl/socsci/econ/StudyAids/JavaStat/StandardizeNormalVariable.html
Exercises page 272…6.1In 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).
6.2P(X=a) =06.3P(a<X<b) = P(a ≤ X ≤ b) because equality sign does
not matter since P(X=a) =P(X=b) =0.
Exercise page 272…6.41. 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 µ4. 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…6.51. Standard normal distribution curve is a bell
shaped curve2. The curve represents a normal distribution with
µ=0 and σ=13. The corresponding variable is denoted by Z4. 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…6.7With a constant mean, as the standard deviation
decreases the height of the normal curve decreases and the width increases.
6.8With 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.