Transcript Document

Chapter 5
The
Binomial Probability
Distribution
and
Related Topics
Essential Question:
How are the mean and the
standard deviation determined
from a discrete probability
distribution?
Student Objectives:
The student will distinguish between random and
discrete random variables.
The student will graph discrete probability
distributions.
The student will compute the mean and standard
deviation for a discrete probability distribution.
The student will compute the mean and standard
deviation for a linear function of a random
variable x.
The student will compute the mean and standard
deviation for a linear combination of two
independent random variables.
Terms:
Continuous random variable
Discrete random variable
Linear function of a random variable
Linear function of two independent random
variables
Mean
Probability Distribution
Random variable
Standard deviation
Statistical Experiment
any process by which an
observation
(or measurement)
is obtained
Examples of Statistical
Experiments
• Counting the number of books in the
College Library
• Counting the number of mistakes on a
page of text
• Measuring the amount of rainfall in
your state during the month of June
Random Variable
a quantitative variable that
assumes a value determined by
chance
Discrete Random Variable
A discrete random variable is a
quantitative random variable that can take
on only a finite number of values or a
countable number of values.
Example: the number of books in
the College Library
Continuous Random
Variable
A continuous random variable is a
quantitative random variable that can
take on any of the countless number of
values in a line interval or a
measurable amount.
Example: the amount of rainfall in your
state during the month of June
Continuous vs Discrete
Directions: In problems #1 - 7, identify each of the following as either a discrete
or continuous random variable.
1. The number of people who are in a car.
Discrete
2. The number of miles you drive in one week.
3. The weight of a box of cereal.
Continuous
Continuous
4. The number of boxes of cereal you buy in one year.
5. The length of time you spend eating your lunch.
Discrete
Continuous
6. The number of patients on a psychiatric ward in one day.
7. The volume of blood that is transfused during an operation.
Discrete
Continuous
Probability Distribution
an assignment of probabilities to
the specific values of the random
variable or to a range of values of
the random variable
Probability Distribution of
a Discrete Random
Variable
• A probability is assigned to each value of the
random variable.
• The sum of these probabilities must be 1.
Probability distribution for
the rolling of an ordinary
fair die
x
1
2
3
4
5
6
P( x )
1
6
1
6
1
6
1
6
1
6
1
6
Features of a Probability Distribution
x
1
2
3
4
5
6
P( x)
1
6
1
6
1
6
1
6
1
6
1
6
6
= 1
6
Probabilities must
be between zero
and one (inclusive)
Σ P(x) =1
Mean and standard
deviation of a discrete
probability distribution
Mean =  = expectation or expected
value, the long-run average
Formula:
å( x × P( x))
Standard Deviation
s=
x
m
(
)
å
2
× P( x)
Finding the mean:
x
0
1
2
3
4
Totals
P( x)
0.1
0.3
0.3
0.2
0.1
1.0
x × P( x)
0.0
0.3
m = å( xP( x)) =1.9
0.6
0.6
0.4
1.9
Finding the standard deviation
x
0
1
2
3
4
Totals
P( x) x × P( x) x - m
( x - m)
-1.9
-0.9
0.1
1.1
2.1
3.61
0.81
0.01
1.21
4.41
0.1
0.3
0.3
0.2
0.1
1.0
0
0.3
0.6
0.6
0.4
1.9
2
( x - m)
2
× P ( x)
0.361
0.243
0.003
0.242
0.441
1.29
Standard Deviation
s=
å ((x - m)
s = 1.29
s = 1.1358
2
× P(x))
Probability Distributions
8. In a personality inventory test for passive-aggressive traits, the possible scores are:
1 = extremely passive
2 = moderately passive
3 = neither
4 = moderately aggressive
5 = extremely aggressive
The test was administered to a group of 110 people and the results were as follows:
x (score)
1
2
3
4
5
f (frequency)
19
23
32
26
10
Construct a probability distribution table, calculate the expected value (the mean) and
the standard deviation. Use a histogram to graph the probability distribution.
x
f
1
19
2
23
3
32
4
26
5
10
Sum:
P(x)
Probability Distributions
8. The histogram:
Probability Distributions
8. The chart:
x
f
P(x)
xP(x)
x-µ
(x - µ)2
(x - µ)2P(x)
1
19
0.1727
0.1727
-1.8636
3.4731
0.5999
2
23
0.2091
0.4182
-0.8636
0.7459
0.1560
3
32
0.2909
0.8727
0.1364
0.0186
0.0054
4
26
0.2364
0.9455
1.1364
1.2913
0.3052
5
10
0.0909
0.4545
2.1364
4.5640
0.4149
Sum:
110
1.0000
2.8636
m = 2.8636
1 = extremely passive
2 = moderately passive
3 = neither
4 = moderately aggressive
5 = extremely aggressive
1.4814
s = 1.4814
s = 1.2171
Linear Functions
Let x be a random variable with the mean m and a standard deviation s .
The linear function L = a + bx has a mean, variance, and standard deviation
as follows:
Mean of a linear function:
m L = a + bm
Variance of a linear function:
s =b s
Standard deviation of
sL = b s = bs
a linear function:
2
L
2
2
2
2
Linear Combinations
Let x1 and x2 be independent random variables with respective means
of m1 and m2 and variances of s 1 and s 2 . For the linear combination
W = ax1 + bx2 , the mean variance, and standard deviation are as
follows:
Mean of a linear combination:
mW = am1 + bm2
Variance of a linear combination:
s = a s +b s
Standard deviation of
sW = a s + b s
a linear combination:
2
W
2
2
1
2
2
2
1
2
2
2
2
2
Linear Functions and Combinations
9.
A local automotive repair shop has two work centers. The first center
examines the car with a computer diagnostic machine and the second
center repairs the car. Let x1 and x2 be random variables representing
the time in minutes to diagnose and repair the car. Assume x1 and x2
and are independent random variables. A recent study of the repair
center produced the following data:
Computer diagnose
Repair
m1 = 18.5 minutes and s 1 = 5.6 minutes
x2 : m2 = 137.75 minutes and s 2 = 17.2 minutes
x1 :
a. Suppose it costs $1.25 per minute to diagnose the automobile and
$0.95 per minute to repair the automobile (without parts). Compute
the linear combination equation, mean, variance, and the standard
deviation of W.
Linear Functions and Combinations
9.
A local automotive repair shop has two work centers. The first center
examines the car with a computer diagnostic machine and the second
center repairs the car. Let x1 and x2 be random variables representing
the time in minutes to diagnose and repair the car. Assume x1 and x2
and are independent random variables. A recent study of the repair
center produced the following data:
Computer diagnose
Repair
m1 = 18.5 minutes and s 1 = 5.6 minutes
x2 : m2 = 137.75 minutes and s 2 = 17.2 minutes
x1 :
b. The repair shop charges a flat rate of $0.75 per minute to diagnose the
car and if no repairs are needed, there is an additional $25.00 service
charge. Compute the linear equation, mean, variance, and standard
deviation of L.
Linear Functions and Combinations
9. a. Suppose it costs $1.25 per minute to diagnose the automobile and
$0.95 per minute to repair the automobile (without parts). Compute
the linear combination equation, mean, variance, and the standard
deviation of W.
Computer diagnose x1 : m1 = 18.5 minutes and s 1 = 5.6 minutes
Repair
x2 : m2 = 137.75 minutes and s 2 = 17.2 minutes
W = 1.25x1 + 0.95x2
mW
mW
mW
mW
= 1.25 (18.5 ) + 0.95 (137.75 )
= 23.125 + 130.8625
= 153.9875
= $153.99
Linear Functions and Combinations
9. a. Suppose it costs $1.25 per minute to diagnose the automobile and
$0.95 per minute to repair the automobile (without parts). Compute
the linear combination equation, mean, variance, and the standard
deviation of W.
Computer diagnose
Repair
m1 = 18.5 minutes and s 1 = 5.6 minutes
x2 : m2 = 137.75 minutes and s 2 = 17.2 minutes
x1 :
s = (1.25 ) s + ( 0.95 ) s
2
2
W
2
2
1
2
2
s 2 = 1.5625s 12 + 0.9025s 22
W
s = 1.5625 ( 5.6 ) + 0.9025 (17.2 )
2
2
2
W
s 2 = 1.5625 ( 31.36 ) + 0.9025 ( 295.84 )
W
s 2 = 49 + 266.9956
W
s = 315.9956
2
W
s 2 = $316.00
W
s W = 315.9956
s W = 17.7763
s W = $17.78
Linear Functions and Combinations
9. b. The repair shop charges a flat rate of $0.75 per minute to diagnose the
car and if no repairs are needed, there is an additional $25.00 service
charge. Compute the linear equation, mean, variance, and standard
deviation of L.
Computer diagnose
Repair
m1 = 18.5 minutes and s 1 = 5.6 minutes
x2 : m2 = 137.75 minutes and s 2 = 17.2 minutes
x1 :
L = 0.75x1 + 25
mL
mL
mL
mL
= 25 + 0.75 (18.5 )
= 25 + 13.875
= 38.875
= $38.88
s = ( 0.75 ) s
2
2
L
s = 0.5625s
2
L
2
1
2
1
s = 0.5625 ( 5.6 )
2
2
L
s = 0.5625 ( 31.36 )
2
L
s = 17.64
2
L
s = $17.64
2
L
s L = 17.64
s L = 4.2
s L = $4.20
Homework Assignment
Chapter 5 Section 1
Pages 190 - 195
Exercises: #1 - 19, odd
Exercises: # 2 - 18, even