chapter 16 continuous random variables
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Chapter 16 Continuous Random Variables. Chapter 16: Continuous Random Variables Chapter Objectives. To introduce continuous random variables and discuss density functions. To discuss the normal distribution, standard units, and the table of areas under the standard normal curve. - PowerPoint PPT PresentationTRANSCRIPT
INTRODUCTORY MATHEMATICAL INTRODUCTORY MATHEMATICAL ANALYSISANALYSISFor Business, Economics, and the Life and Social Sciences
2011 Pearson Education, Inc.
Chapter 16 Chapter 16 Continuous Random VariablesContinuous Random Variables
2011 Pearson Education, Inc.
• To introduce continuous random variables and discuss density functions.
• To discuss the normal distribution, standard units, and the table of areas under the standard normal curve.
• To show the technique of estimating the binomial distribution by using normal distribution.
Chapter 16: Continuous Random Variables
Chapter ObjectivesChapter Objectives
2011 Pearson Education, Inc.
Continuous Random Variables
The Normal Distribution
The Normal Approximation to the Binomial Distribution
16.1)
16.2)
16.3)
Chapter 16: Continuous Random Variables
Chapter OutlineChapter Outline
2011 Pearson Education, Inc.
Chapter 16: Continuous Random Variables
16.1 Continuous Random Variables16.1 Continuous Random Variables
• If X is a continuous random variable, the (probability) density function for X has the following properties:
b
a
dxxfbXa. P
dxxf.
x. f
3
1 2
01
2011 Pearson Education, Inc.
Chapter 16: Continuous Random Variables
16.1 Continuous Random Variables
Example 1 – Uniform Density FunctionThe uniform density function over [a, b] for the random variable X is given by
Find P(2 < X < 3).Solution: If [c, d] is any interval within [a, b] then
For a = 1, b = 4, c = 2, and d = 3,
otherwise0
b a if1
xabxf
ab
cd
ab
x
dxab
dxxfdXcP
dc
d
c
d
c
1
3
1
14
2332
XP
2011 Pearson Education, Inc.
Chapter 16: Continuous Random Variables
16.1 Continuous Random Variables
Example 3 – Exponential Density FunctionThe exponential density function is defined by
where k is a positive constant, called a parameter, whose value depends on the experiment under consideration. If X is a random variable with this density function, then X is said to have an exponential distribution. Let k = 1. Then f(x) = e−x for x ≥ 0, and f(x) = 0 for x < 0.
0 if0
0 if
x
xkexf
kx
2011 Pearson Education, Inc.
Chapter 16: Continuous Random Variables
16.1 Continuous Random Variables
Example 3 – Exponential Density Function
a. Find P(2 < X < 3).
Solution:
b. Find P(X > 4).
Solution:
086.0)(
32
3223
32
3
2
eeee
edxeXP xx
018.001
lim
lim 4
44
44
eee
dxedxeXP
rr
rx
r
x
2011 Pearson Education, Inc.
Chapter 16: Continuous Random Variables
16.1 Continuous Random Variables
Example 5 – Finding the Mean and Standard DeviationIf X is a random variable with density function given by
find its mean and standard deviation.Solution:
otherwise 0
20 if 21 xx
xf
3
4
6
2
1 Mean,
2
0
32
0
xdxxxdxxxfμ
9
2
9
16
83
4
2
1
2
0
422
0
2222
xdxxxμdxxfxσ
3
2
9
2 Deviation, Standard
2011 Pearson Education, Inc.
Chapter 16: Continuous Random Variables
16.2 The Normal Distribution16.2 The Normal Distribution
• Continuous random variable X has a normal distribution if its density function is given by
called the normal density function.
• Continuous random variable Z has a standard normal distribution if its density function is given by
called the standard normal density function.
xeπσ
xf σμx 2
1 2/2/1
2/2
2
1 zeπσ
zf
2011 Pearson Education, Inc.
Chapter 16: Continuous Random Variables
16.2 The Normal Distribution
Example 1 – Analysis of Test Scores
Let X be a random variable whose values are the scores obtained on a nationwide test given to high school seniors. X is normally distributed with mean 600 and standard deviation 90. Find the probability that X lies (a) within 600 and (b) between 330 and 870.
Solution:
0.95 is 600 of points 180
9022600 within
σP
0.997 is 600 of points 270
9033870 and 330 between
σP
2011 Pearson Education, Inc.
Chapter 16: Continuous Random Variables
16.2 The Normal Distribution
Example 3 – Probabilities for Standard Normal Variable Za. Find P(−2 < Z < −0.5).Solution:
b. Find z0 such that P(−z0 < Z < z0) = 0.9642.Solution: The total area is 0.9642. By symmetry, the area between z = 0 and z = z0 is
Appendix C shows that 0.4821 corresponds to a Z-value of 2.1.
2857.05.02
25.05.02
AA
ZPZP
4821.09642.02
1
2011 Pearson Education, Inc.
Chapter 16: Continuous Random Variables
16.3 The Normal Approximation to the 16.3 The Normal Approximation to the
Binomial DistributionBinomial Distribution
Example 1 – Normal Approximation to a Binomial Distribution
• The probability of x successes is given by
xnxxn qpCxXP
Suppose X is a binomial random variable with n = 100 and p = 0.3. Estimate P(X = 40) by using the normal approximation.Solution: We have We use a normal distribution with
604040100 7.03.040 CXP
58.421)7.0(3.0100
303.0100
npqσ
npμ
2011 Pearson Education, Inc.
Chapter 16: Continuous Random Variables
16.3 The Normal Approximation to the Binomial Distribution
Example 1 – Normal Approximation to a Binomial Distribution
Solution (cont’d):
Converting 39.5 and 40.5 to Z-values gives
Therefore,
0082.0
4808.04890.0
07.229.2
29.207.240
AA
ZPXP
29.221
305.40 and 07.2
21
305.3921
zz