Download Examples for Chapter 6

Chapter 6 Continuous Probability Distribution
I. Basic Definitions
II. Normal Distribution
• Probability density function and distribution table
• Characteristics - identify a normal distribution
• Compute probability
– The standard normal distribution Z
– Normal distribution
– Applications
III. Normal Distribution Approximates Binomial
Distribution
I. Basic Definitions
Continuous random variable: it takes all values over an
interval.
For continuous probability distribution
• For an individual value of X: P(X=x) = 0
• For an interval of X: 0  P(x1  X  x2)  1
• Probability density function f(x) measures probability
for a neighborhood of x.
- It’s not P(X=x).
- We use probability density function to compute
cumulative probability.
• Cumulative probability P(x1  X  x2) Vs. probability
function f(x): Area and height. In general,
x2
P( x1  X  x2 )   f ( x)dx
x1
II. Normal Distribution
• Probability density function and distribution table
– Probability density function
f ( x) 
1
e
2 

( x )2
2 2
x2
P( x1  X  x2 )   f ( x)dx
x1
– Probability distribution table: for the
standard normal distribution Z - Table 1 (A-4)
• Identify a normal distribution - characteristics
– Symmetric and bell-shaped (p.239 Figure 6.3)
– Follow the empirical rule (p.241 Figure 6.4)
– The total area under the probability curve is 1:
P(- X +) = 1 (p.240)
– Two parameters ( and ) determine the
distribution: X  N(, )
(pp.239-240)
• Compute Probability (Outlines)
1. The standard normal distribution Z  N(0, 1)
What are in Table 1?
Value of Z and P(0  Z  z)
What can we find by using Z-Table?
- Given an interval of Z, find probability.
- Given probability for an interval of Z, find the interval.
2. Normal distribution X  N(, )
Can we use Z-Table?
Z 
X 
x
x
What kind of problems?
- Given an interval of X, find probability.
- Given probability for an interval of X, find the interval.
1. The standard normal distribution Z  N(0, 1)
(1) Given an interval of Z, find probability
Example. P.248 #12
a. P(0  Z  .83) = ?
P(Z  .83) - .5 = .7967 - .5 (Z-Table)
= .2967
b. P(-1.57  Z  0) = ?
(symmetrical)
.5 - P(Z  -1.57 ) = .5 - .0582 (Z-Table)
= .4418
c. P(Z > .44) = ?
1 - P(Z ≤ .44) = 1 - .67
(Z-Table)
0
.44
= .33
d. P (Z  -.23) = ?
1 - P (Z ≤ -.23) = 1 - .4090 (Z-Table)
0
-.23
= .591
Example. P.248 #12
e. P(Z < 1.20) = ?
P(Z < 1.20) = .8849 (Z-Table)
f. P(Z  -.71) = ?
P(Z  -.71) =.2389 (Z-Table)
1.20
0
-.71
0
Example. P.241 #13
a. P(-1.98  Z  .49) = ?
P(Z  .49) – P(Z  -1.98) = .6879 + .0239 (Z-Table)
= .6640
-1.98 0 .49
Homework: p.248 #13
1. The standard normal distribution Z  N(0, 1)
(2) Given probability for an interval of Z, find the
interval
Example. P.249 #15
Find value of z.
a. The area to the left of z is .2119.
P(Z  z) = .2119
Key: which side is z?
0z
.2119
Suppose the first picture is
correct, the probability is
greater than .5 (impossible to be .2119).
z0
The second one is correct.
z = -.80 (Z-Table)
Homework: p.249 #15
P=?
2. Normal distribution X  N(, )
To use Table 1:
Z 
X  x
x
(1) Given an interval of X, find probability
Procedure: X  Z  Z-Table: P(Z)  P(X) = P(Z)
Example 1:
According to a survey, subscribers to The Wall Street
Journal Interactive Edition spend an average of 27
hours per week using the computer at work. Assume
the normal distribution applies and that the standard
deviation is 8 hours.
a. What is the probability a randomly selected
subscriber spends less than 11 hours using the
computer at work?
Example 1: (continued)
b. What percentage of the subscribers spends more
than 40 hours per week using the computer at work?
c. A person is classified as a heavy user if he or she is
in the upper 20% in terms of hours of usage. How
many hours must a subscriber use the computer in
order to be classified as a heavy user?
Solution:
X: the number of hours per week using computer at
work.
X  N(, ):  = 27,  = 8
“a” and “b”: x  P; “c”: P  x.
P=?
a. P(X < 11)
• Z-score for X = 11:
z = (x- )/ = (11 - 27)/8
11
P=?
= -2
• P(Z < -2) = 0.0228 (Z-Table)
-2
• P(X < 11) = P(Z < -2) = .0228
b. P(X > 40)
• Z-score for X = 40:
z = (x- )/ = (40 - 27)/8
= 1.625
• P(Z > 1.625) = 1 – P(Z < 1.625)
1 - .9484 = 0.0516 = 5.16%
27
0.2734
0
P=?
27
0.4484
0
40
P=?
1.625
Example. p.250 #23
The time needed to complete a final examination in a
particular college course is normally distributed with
a mean of 80 minutes and a standard deviation of 10
minutes.
a.What is the probability of completing the exam in
one hour or less?
b.What is the probability that a student will complete
the exam in more than 60 minutes but less than 75
minutes?
c.Assume that the class has 60 students and that the
examination period is 90 minutes in length. How many
students do you expect will be unable to complete the
exam in the allotted time?
Solution:
X: the number of minutes to complete the exam.
X  N(, ):  = 80,  = 10
“a”, “b” and “c”: x  P.
“c”: x  P (% of students)  the number of students:
(60)(% of students)
P=?
a. P(X < 60)
• Z-score for X = 60:
80
z = (x- )/ = (60 - 80)/10
60
0.2734
P=?
= -2
• P(Z < -2) = 0.0228
0
-2
• P(X < 60) = P(Z < -2) = .0228
b. P(60 < X < 75)
P=?
• Z-score for X = 60:
z1 = (x- )/ = (60 - 80)/10
60 75 80
= -2
P=?
• Z-score for X = 75:
z2 = (x- )/ = (75 - 80)/10
-2 -0.5 80
= -0.5
• P(-2 < Z < -.5) = P(Z < -.5) – P(Z < -2)
= .3085 - .0228 = .2857
• P(60 < X < 75) = P(-2 < Z < -.5) = .2857
P=?
c. P(X > 90)
• Z-score for X = 90:
80
90
z = (x- )/ = (90 - 80)/10
0.3413
=1
P=?
• P(Z > 1) = 1 – P(Z < 1)
= 1 - .8413 = .1587
0
1
• P(X > 90) = P(Z > 1) = .1587
The number of students = (60)(.1587) = 9.522.
2. Normal distribution X  N(, )
To use Table 1:
Z 
X  x
x
(2) Given probability for an interval of X, find the
interval
Procedure: X  Z  Table 1: P(Z)  Z  x=x+zx
Example 1: (Continued)
c. A person is classified as a heavy user if he or she is
in the upper 20% in terms of hours of usage. How
many hours must a subscriber use the computer in
order to be classified as a heavy user?
Solution:
X: the number of hours per week using computer at
work.
X  N(, ):  = 27,  = 8 . “c”: P  x.
Solution:
c. P(X > x) = 20%, x=?
• X  Z:
Key: Which side is x?
The z and x are on the same side.
0.20
27
x=?
0.30
0.20
0
• P(Z < z) = 1- P(Z > z) = 1 - .20 = 0.8
z ≈ 0.84 (P(Z < z) = .7995 closest to .8)
• z  x:
x=x+zx
x = 27 + (.84)(8) = 33.72
Homework: p.250 #23
z=?